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Power series

⌚ ~75 min ▢ 9 blocks ✓ 17 exercises ➣ Prerequisites : Sequences and series of functions, Numerical and vector series, Complex numbers, Summable families, Numerical series, Differentiability

A power series is the infinite-degree extension of a polynomial: an expression $\sum_{n \ge 0} a_n z^n$ where the coefficient sequence $(a_n)$ takes values in $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$ and the variable $z$ ranges over $\mathbb{K}$. The construction is motivated by the search for solutions of linear differential equations: postulate $f(x) = \sum a_n x^n$, inject into the equation, identify the coefficients --- a method known since Newton and which gives, for instance, $\mathrm{e}^x = \sum_{n=0}^{+\infty} x^n/n!$ as the solution of $y' = y$ with $y(0) = 1$.
The convergence domain of $\sum a_n z^n$ turns out to have a remarkable shape: a disk in $\mathbb{C}$ (an interval in $\mathbb{R}$), characterised by a single number --- the radius of convergence $R \in [0, +\infty]$. Inside the open disk the sum is continuous and, in the real case, of class $\mathcal{C}^\infty$ with differentiation and integration done term by term. The chapter culminates with the catalogue of classical developments (exponential, sine, cosine, hyperbolic, logarithm, arctangent, generalized binomial) that every student must know by heart, and the five practical routes to obtain a development.
Notation. $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$; $(a_n)_{n \in \mathbb{N}} \in \mathbb{K}^\mathbb{N}$ a coefficient sequence; $\sum a_n z^n$ the associated power series; $R \in \overline{\mathbb{R}}_+ = [0, +\infty]$ its radius of convergence; $D(0, R) = \{z \in \mathbb{C} : |z| < R\}$ the open disk and $D_f(0, r) = \{z \in \mathbb{C} : |z| \le r\}$ a closed disk in $\mathbb{C}$; $(-R, R)$ the open interval of convergence in $\mathbb{R}$; $f(z) = \sum_{n=0}^{+\infty} a_n z^n$ the sum function on the convergence domain.

I Power series and radius of convergence

I.1 Power series and convergence domain

A power series is a series of functions $\sum u_n$ with the very specific shape $u_n(z) = a_n z^n$, parametrised by the coefficient sequence $(a_n)$ and the variable $z \in \mathbb{K}$. The function class is so structured that the convergence domain has a remarkable disk shape, controlled by a single radius $R$ --- the focus of \S 1.2.

Definition — Power series

A power series on $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$ is a series of functions of the form $\sum_{n \ge 0} a_n z^n$, where $(a_n)_{n \in \mathbb{N}} \in \mathbb{K}^\mathbb{N}$ is a sequence of coefficients and $z$ is a variable in $\mathbb{K}$. When the numerical series $\sum a_n z^n$ converges for a given $z$, its sum is denoted $f(z) = \sum_{n=0}^{+\infty} a_n z^n$; the function $f$, defined on the set of $z$ where convergence holds, is called the sum function of the power series.

The Definition above (power series) and the one below (convergence domain) are two faces of the same concept: the formal object (the series) and the set on which the sum function is defined. The three Examples following the second Definition jointly illustrate both.

Definition — Convergence domain

The convergence domain of a power series $\sum a_n z^n$ is the set $$ D = \{\, z \in \mathbb{K} \,:\, \text{the numerical series } \sum a_n z^n \text{ converges} \,\}. $$ The sum function $f \colon D \to \mathbb{K}$ is defined on this set.

Example — The geometric series

Take $a_n = 1$ for every $n$. The numerical series $\sum z^n$ converges (geometrically) if and only if $|z| < 1$, with sum $1/(1 - z)$ on this domain (MPSI Numerical series). The convergence domain is the open unit disk $D(0, 1) \subset \mathbb{C}$ (open unit interval $(-1, 1) \subset \mathbb{R}$ in the real case), and $$ \frac{1}{1 - z} = \sum_{n=0}^{+\infty} z^n \qquad \text{for every } z \in \mathbb{C} \text{ with } |z| < 1. $$ The archetypal power series.

Example — The exponential series

Take $a_n = 1/n!$. For every $z \in \mathbb{C}$, the numerical series $\sum z^n/n!$ converges absolutely (d'Alembert: $|u_{n+1}/u_n| = |z|/(n+1) \to 0 < 1$, recalled from MPSI Numerical series). The convergence domain is the whole of $\mathbb{C}$; the sum function will be identified with the complex exponential $\exp(z)$ in \S 4.2.

Example — A series converging only at zero

Take $a_n = n!$. For any $z \ne 0$, $|a_{n+1} z^{n+1}/(a_n z^n)| = (n+1)|z| \to +\infty$, so $a_n z^n$ does not tend to $0$ and $\sum n! \, z^n$ diverges grossly. The convergence domain reduces to the single point $\{0\}$ --- a third regime alongside the geometric series's open disk and the exponential series's full plane. These three Examples illustrate that the convergence domain of a power series is essentially controlled by a single number --- the modulus of $z$ at the boundary --- to be defined and studied in \S 1.2.

Skills to practice

Identifying a power series

I.2 Abel's lemma and the radius of convergence

Abel's lemma is a horizontal comparison principle: if the sequence $(a_n z_0^n)$ is merely bounded at a single point $z_0$, then $\sum a_n z^n$ is absolutely convergent at every point $z$ strictly inside the disk $|z| < |z_0|$. This forces the convergence domain to have a single critical radius $R$ below which everything converges absolutely and above which the series diverges grossly.

Proposition — Abel's lemma

Let $z_0 \in \mathbb{K}$ with $z_0 \ne 0$ and assume the sequence $(a_n z_0^n)_{n \in \mathbb{N}}$ is bounded. Then for every $z \in \mathbb{K}$ with $|z| < |z_0|$, the series $\sum a_n z^n$ is absolutely convergent.

Proof

Let $M \ge 0$ such that $|a_n z_0^n| \le M$ for every $n$. For $|z| < |z_0|$, $$ |a_n z^n| = |a_n z_0^n| \cdot \left| \frac{z}{z_0} \right|^n \le M \left| \frac{z}{z_0} \right|^n. $$ Setting $q = |z/z_0|$, one has $q < 1$ so the geometric series $\sum q^n$ converges. By comparison of non-negative series (recalled from Numerical and vector series, \S 1.4), the series $\sum |a_n z^n|$ converges, hence $\sum a_n z^n$ converges absolutely.

Definition — Radius of convergence

The radius of convergence of a power series $\sum a_n z^n$ is the element of $\overline{\mathbb{R}}_+ = [0, +\infty]$ defined by $$ R = \sup\bigl\{\, r \ge 0 \,:\, (a_n r^n)_{n \in \mathbb{N}} \text{ is bounded} \,\bigr\}. $$

Theorem — Trichotomy of convergence

Let $\sum a_n z^n$ be a power series of radius $R$. For every $z \in \mathbb{K}$:

if $|z| < R$, the series $\sum a_n z^n$ converges absolutely;
if $|z| > R$, the series diverges grossly (the general term $a_n z^n$ does not tend to $0$);
if $|z| = R$, no general conclusion can be drawn.

Proof

Case $|z| < R$. By definition of $R$ as a supremum, there exists $r$ with $|z| < r \le R$ and $(a_n r^n)$ bounded. Since $r \ne 0$ (because $|z| < r$ and $|z| \ge 0$, so $r > 0$), Abel's lemma applies and gives $\sum a_n z^n$ absolutely convergent.
Case $|z| > R$. If $\sum a_n z^n$ converged, then $a_n z^n \to 0$, so $(a_n z^n)$ would be bounded; but $|z| > R$ contradicts the definition of $R$ as a supremum of $r$ such that $(a_n r^n)$ is bounded. So $(a_n z^n)$ is not bounded, in particular does not tend to $0$: gross divergence.

The boundary case $|z| = R$ is undecidable in general; the examples below exhibit each of the three possible behaviours (absolutely convergent, semi-convergent, divergent) at $|z| = R$.

Definition — Open interval and disk of convergence

For a power series of radius $R$:

if $\mathbb{K} = \mathbb{R}$, the set $(-R, R)$ is called the open interval of convergence;
if $\mathbb{K} = \mathbb{C}$, the set $D(0, R) = \{z \in \mathbb{C} : |z| < R\}$ is called the open disk of convergence; and, when $R < +\infty$, the circle $C(0, R) = \{z \in \mathbb{C} : |z| = R\}$ is the circle of convergence (or circle of uncertainty, since boundary behaviour is undecided). When $R = +\infty$ the circle is not defined --- the open disk is the entire plane and there is no boundary.

The trichotomy is illustrated below: first on the real line, then on the complex plane. On the real line, absolute convergence holds on the open interval $(-R, R)$, gross divergence outside, and the two boundary points $\pm R$ are undecided.

In the complex plane, the absolute-convergence region is a full open disk of radius $R$, the divergence region is its exterior, and the boundary circle $C(0, R)$ is the «cercle d'incertitude».

Example — Three boundary regimes at $R \equal 1$

The three series $\sum z^n$, $\sum z^n/n^2$, $\sum z^n/n$ all have radius $R = 1$ (by d'Alembert below), yet their boundary behaviours differ:

$\sum z^n$ at $|z| = 1$: $|z^n| = 1$ does not tend to $0$, gross divergence everywhere on the unit circle;
$\sum z^n/n^2$ at $|z| = 1$: $|z^n/n^2| = 1/n^2$ defines a Riemann-convergent numerical series, absolute convergence everywhere on the unit circle;
$\sum z^n/n$ at $z = 1$: harmonic series, divergence; at $z = -1$: alternating-harmonic series, semi-convergence (converges by Leibniz, not absolutely). The behaviour is genuinely point-dependent on the circle.

Proposition — Encadrement of $R$

Let $\sum a_n z^n$ be a power series of radius $R$, and let $z_0 \in \mathbb{K}$. Then:

if $\sum a_n z_0^n$ converges, $|z_0| \le R$;
if $\sum a_n z_0^n$ diverges, $|z_0| \ge R$;
if $\sum a_n z_0^n$ is semi-convergent (converges but not absolutely), $|z_0| = R$ exactly.

Proof

Direct from the trichotomy. If $\sum a_n z_0^n$ converges, then $|z_0| \not> R$ (gross divergence would contradict convergence), so $|z_0| \le R$. If $\sum a_n z_0^n$ diverges, then $|z_0| \not< R$ (absolute convergence would contradict divergence), so $|z_0| \ge R$. Combining: a semi-convergent series at $z_0$ is convergent (gives $|z_0| \le R$) but not absolutely convergent, so $|z_0| \not< R$ (else trichotomy gives absolute convergence) --- hence $|z_0| = R$.

Proposition — Comparison rules for the radius

Let $\sum a_n z^n$ and $\sum b_n z^n$ be two power series of radii $R_a$ and $R_b$.

If $a_n = \mathrm{O}(b_n)$ as $n \to +\infty$ (in particular if $|a_n| \le |b_n|$ from some rank), then $R_a \ge R_b$.
If $a_n = \mathrm{o}(b_n)$, then $R_a \ge R_b$.
If $a_n \sim b_n$, then $R_a = R_b$.

Proof

We use the equivalent characterisation $R(\sum c_n z^n) = \sup\{r \ge 0 : (c_n r^n) \text{ bounded}\}$ (Definition).
If $a_n = \mathrm{O}(b_n)$, there exists $C \ge 0$ such that $|a_n| \le C |b_n|$ from some rank $N$. For every $r$ with $(b_n r^n)$ bounded (so $r \le R_b$), $|a_n r^n| \le C |b_n r^n|$ from rank $N$, hence $(a_n r^n)$ is bounded; this gives $r \le R_a$ for every $r < R_b$, so $R_b \le R_a$. The $\mathrm{o}$ case is contained in the $\mathrm{O}$ case ($\mathrm{o}(b_n) \subset \mathrm{O}(b_n)$). For the equivalence $a_n \sim b_n$: it is symmetric, so the previous step applied both ways gives $R_a = R_b$.

Proposition — Radius of $\sum n a_n z^n$

The series $\sum a_n z^n$ and $\sum n a_n z^n$ have the same radius of convergence. In particular, for every $\alpha \in \mathbb{R}$, the series $\sum_{n \ge 1} n^\alpha a_n z^n$ has the same radius as $\sum a_n z^n$ (the index starts at $n \ge 1$ to avoid the ill-defined $0^\alpha$ for $\alpha < 0$).

Proof

If $\alpha \ge 0$: from $(n^\alpha a_n r^n)$ bounded, $|a_n r^n| = n^{-\alpha} \cdot |n^\alpha a_n r^n| \le n^{-\alpha} \cdot M$ for some $M$; the factor $n^{-\alpha} \le 1$ for $n \ge 1$ (since $\alpha \ge 0$), so $(a_n r^n)$ is bounded by $M$. Hence $r \le R$, and $R_\alpha \le R$.
If $\alpha < 0$: from $(n^\alpha a_n r^n)$ bounded by $M$, pick $r' < r$. Then $$ \begin{aligned} |a_n (r')^n| &= n^{-\alpha} \cdot |n^\alpha a_n r^n| \cdot (r'/r)^n && \text{(rewriting } (r')^n = r^n (r'/r)^n) \\ &\le n^{-\alpha} \cdot M \cdot (r'/r)^n && \text{(using } |n^\alpha a_n r^n| \le M) \\ &= M \cdot n^{-\alpha} (r'/r)^n && \text{(rearrange).} \end{aligned} $$ The factor $n^{-\alpha}$ grows polynomially (since $-\alpha > 0$), but $(r'/r)^n$ decays geometrically since $r'/r < 1$, so the product tends to $0$ and is bounded. Hence $r' \le R$ for every $r' < r$, giving $r \le R$; $r$ arbitrary $< R_\alpha$ yields $R_\alpha \le R$.

The headline statement is the $\alpha = 1$ case; the general $\alpha \in \mathbb{R}$ extension is the in-particular claim, proved by the same chain with $n$ replaced by $n^\alpha$ throughout (and the $\alpha < 0$ refinement above).

Method — Compute the radius of convergence

(a) By encadrement (Proposition). If you know the behaviour of $\sum a_n z_0^n$ at one sample $z_0$ --- convergence gives $|z_0| \le R$, divergence gives $|z_0| \ge R$, semi-convergence gives $|z_0| = R$ exactly. Combining two samples often pins $R$ down.
(b) By d'Alembert. Assume $(a_n)_{n \ge N}$ is eventually non-zero. If $|a_{n+1}/a_n| \to \ell \in [0, +\infty]$, then $R = 1/\ell$ (convention: $1/0 = +\infty$, $1/{+\infty} = 0$). Proof: apply d'Alembert to the numerical series $\sum a_n z^n$, $|u_{n+1}/u_n| = |a_{n+1}/a_n| \cdot |z| \to \ell |z|$; absolute convergence for $\ell |z| < 1$, gross divergence for $\ell |z| > 1$.
(c) By comparison (Proposition). Use $a_n = \mathrm{O}(b_n)$ or $a_n \sim b_n$ to transfer the radius from a sister series of known radius.
(d) Recognise $\sum n^\alpha a_n z^n$ as having the same radius as $\sum a_n z^n$.
Lacunary series (some $a_n$ are $0$): d'Alembert in its raw form does not apply --- the ratio is undefined infinitely often. Instead recognise the series as a sub-series in a power of $z$: e.g.\ $\sum a_{2k} z^{2k}$ is a power series in $w = z^2$; apply d'Alembert in $w$, then back-substitute.

Example — $\sum x^n/n$ has radius $1$

With $a_n = 1/n$ for $n \ge 1$ (and $a_0 = 0$), $|a_{n+1}/a_n| = n/(n+1) \to 1$. By d'Alembert, $R = 1/1 = 1$.

Example — Lacunary $\sum x^{2n}/2^n$ has radius $\sqrt{2}$

The coefficient of $x^k$ is $0$ if $k$ is odd, $1/2^{k/2}$ if $k$ is even, so the sequence is lacunary and raw d'Alembert fails. Recognise the series as a power series in $w = x^2$: $$ \sum_{n \ge 0} \frac{x^{2n}}{2^n} = \sum_{n \ge 0} \biggl(\frac{w}{2}\biggr)^n, \qquad w = x^2. $$ This is the geometric series in $w/2$, converging for $|w/2| < 1$, i.e.\ $|w| < 2$, i.e.\ $x^2 < 2$, i.e.\ $|x| < \sqrt{2}$. Hence $R = \sqrt{2}$.

Skills to practice

Computing the radius of convergence

I.3 Operations on power series

Power series form an algebra: sum (coefficient-wise) and product (Cauchy product) preserve the power-series structure. The radius behaviour follows from the trichotomy for the sum, and from the Cauchy-product theorem for absolutely convergent series (recalled from MPSI Summable families) for the product.

Proposition — Sum of two power series

Let $\sum a_n z^n$ and $\sum b_n z^n$ be two power series of radii $R_a$ and $R_b$. The sum series $\sum (a_n + b_n) z^n$ is a power series of radius $R$ satisfying $$ R = \min(R_a, R_b) \quad \text{if } R_a \ne R_b, \qquad R \ge R_a \quad \text{if } R_a = R_b. $$

Proposition — Scalar multiplication

Let $\sum a_n z^n$ be a power series of radius $R_a$ and let $\lambda \in \mathbb{K}$. Then $\sum \lambda a_n z^n$ is a power series whose radius equals $R_a$ if $\lambda \ne 0$, and $+\infty$ (the zero series) if $\lambda = 0$.

Proof

Sum. For $|z| < \min(R_a, R_b)$, the two series $\sum a_n z^n$ and $\sum b_n z^n$ are absolutely convergent (by the trichotomy applied to each), so the sum series $\sum (a_n + b_n) z^n$ is absolutely convergent (by the triangle inequality). Hence $R \ge \min(R_a, R_b)$.
For the sharper «$=$» when $R_a \ne R_b$ (WLOG $R_a < R_b$): for every $z$ with $R_a < |z| < R_b$, $a_n z^n \not\to 0$ (gross divergence of $\sum a_n z^n$ since $|z| > R_a$), while $b_n z^n \to 0$ (absolute convergence of $\sum b_n z^n$ since $|z| < R_b$); their sum $a_n z^n + b_n z^n \not\to 0$, so $\sum (a_n + b_n) z^n$ diverges grossly at every such $z$. This forces $R \le |z|$ for every $|z| \in (R_a, R_b)$. Taking the infimum of $|z|$ over this open interval yields $R \le R_a$.
Scalar multiplication. If $\lambda = 0$, $\sum 0 \cdot z^n = 0$ converges for every $z$, radius $+\infty$. If $\lambda \ne 0$, $|\lambda a_n r^n| = |\lambda| \cdot |a_n r^n|$ is bounded iff $(a_n r^n)$ is, so the radii coincide.

Proposition — Cauchy product

Let $\sum a_n z^n$ and $\sum b_n z^n$ be two power series of radii $R_a$ and $R_b$, and let $$ c_n = \sum_{k=0}^n a_k b_{n - k}. $$ Then $\sum c_n z^n$ is a power series of radius $R_c \ge \min(R_a, R_b)$, and for every $z$ with $|z| < \min(R_a, R_b)$, $$ \biggl( \sum_{n=0}^{+\infty} a_n z^n \biggr) \biggl( \sum_{n=0}^{+\infty} b_n z^n \biggr) = \sum_{n=0}^{+\infty} c_n z^n. $$

Proof

For $|z| < \min(R_a, R_b)$, the two numerical series $\sum a_n z^n$ and $\sum b_n z^n$ are absolutely convergent (trichotomy), i.e.\ the sequences $(a_n z^n)$ and $(b_n z^n)$ belong to $\ell^1(\mathbb{N})$. By the Cauchy product theorem for two $\ell^1$ sequences (recalled from MPSI Summable families), the Cauchy product of these two sequences, namely $$ \begin{aligned} p_n &= \sum_{k=0}^n (a_k z^k)(b_{n-k} z^{n-k}) && \text{(definition of the Cauchy product)} \\ &= \sum_{k=0}^n a_k b_{n-k} \cdot z^k \cdot z^{n-k} && \text{(commutativity of multiplication)} \\ &= \biggl( \sum_{k=0}^n a_k b_{n - k} \biggr) z^n && \text{(factor out } z^n = z^k z^{n-k}) \\ &= c_n z^n && \text{(definition of } c_n), \end{aligned} $$ belongs to $\ell^1(\mathbb{N})$ and satisfies $$ \begin{aligned} \sum_{n=0}^{+\infty} c_n z^n &= \sum_{n=0}^{+\infty} p_n && \text{(equality } p_n = c_n z^n) \\ &= \biggl( \sum_{n=0}^{+\infty} a_n z^n \biggr) \biggl( \sum_{n=0}^{+\infty} b_n z^n \biggr) && \text{(Cauchy product theorem, MPSI Summable families).} \end{aligned} $$ In particular $\sum |c_n z^n|$ converges for every $|z| < \min(R_a, R_b)$, hence the radius $R_c \ge \min(R_a, R_b)$.

Example — $1/(1-z)^2$ via Cauchy product

Take $a_n = b_n = 1$ (so $R_a = R_b = 1$). The Cauchy product gives $c_n = \sum_{k=0}^n 1 \cdot 1 = n + 1$. By the Cauchy-product Proposition, for every $|z| < 1$, $$ \frac{1}{(1 - z)^2} = \biggl( \sum_{n=0}^{+\infty} z^n \biggr)^2 = \sum_{n=0}^{+\infty} (n + 1) z^n. $$ A satisfying derivation of an identity that can also be obtained by term-by-term differentiation (\S 3.1, alternative derivation).

Method — Combine power series

For the sum $\sum (a_n + b_n) z^n$: the radius is $\ge \min(R_a, R_b)$, with equality when $R_a \ne R_b$. On the common open disk $|z| < \min(R_a, R_b)$, the identity is $\sum (a_n + b_n) z^n = \sum a_n z^n + \sum b_n z^n$ by term-by-term linearity.
For the Cauchy product $\sum c_n z^n$ with $c_n = \sum_k a_k b_{n-k}$: the radius is $\ge \min(R_a, R_b)$ only --- cancellations can make $R_c$ strictly larger (e.g.\ the Cauchy product of $\sum z^n$ (radius $1$) and the polynomial $1 - z$ viewed as the power series $b_0 = 1$, $b_1 = -1$, $b_n = 0$ for $n \ge 2$ (radius $+\infty$) collapses to the constant $1$, of radius $+\infty > 1$). On the common open disk, the identity is $(\sum a_n z^n)(\sum b_n z^n) = \sum c_n z^n$.

Skills to practice

Combining power series

II Continuity of the sum (complex variable)

II.1 Normal convergence on every closed sub-disk

The trichotomy gives absolute convergence on the open disk of convergence, but the sum function $z \mapsto \sum a_n z^n$ is continuous because of a stronger mode of convergence: normal convergence on every closed sub-disk strictly inside the open disk. The «whole open disk» itself does not in general support normal convergence; the «every closed sub-disk» formulation is the natural one for power series.

Theorem — Normal convergence on every closed sub-disk

Let $\sum a_n z^n$ be a power series of radius $R$. For every $r$ with $0 \le r < R$, the series of functions $z \mapsto a_n z^n$ converges normally on the closed disk $D_f(0, r)$ (resp.\ on the closed interval $[-r, r]$ in the real case), hence uniformly on it.

Proof

For $z \in D_f(0, r)$, $|a_n z^n| \le |a_n| r^n$, so $\|u_n\|_\infty \le |a_n| r^n$ where $u_n(z) = a_n z^n$. The numerical series $\sum |a_n| r^n$ is the series $\sum a_n z^n$ evaluated at $z = r < R$, in absolute value: by the trichotomy, $|r| < R$ gives absolute convergence, i.e.\ $\sum |a_n| r^n$ converges. Hence $\sum \|u_n\|_\infty$ converges, which is the definition of normal convergence (recalled from Sequences and series of functions, \S 2.1). Normal $\Rightarrow$ uniform, by the same chapter.

Caveat. Normal convergence on the whole open disk $D(0, R)$ fails in general. Take $\sum z^n$ on the open unit disk: $\|u_n\|_\infty = \sup_{|z| < 1} |z|^n = 1$, and $\sum 1$ diverges. The natural hypothesis is «on every closed sub-disk $D_f(0, r)$ with $r < R$»; the «whole open disk» version is strictly stronger and usually unavailable.

Corollary — Continuity on the open disk of convergence

The sum $f(z) = \sum_{n=0}^{+\infty} a_n z^n$ of a power series of radius $R > 0$ is continuous on the open disk of convergence $D(0, R)$ (resp.\ on the open interval $(-R, R)$ in the real case).

Proof

Each map $z \mapsto a_n z^n$ is continuous on $\mathbb{K}$ (polynomial). To prove continuity of $f$ at an arbitrary point $z_0 \in D(0, R)$, pick $s$ with $|z_0| < s < R$: then $z_0$ lies in the closed sub-disk $D_f(0, s) \subset D(0, R)$, and on this closed sub-disk the series $\sum a_n z^n$ converges uniformly (Theorem above). A uniform limit of continuous functions is continuous (recalled from Sequences and series of functions, \S 1.2 --- the proof is the classical $\varepsilon/3$ argument, which works identically on the closed sub-disk $D_f(0, s)$ as on a real segment). Hence $f$ is continuous on $D_f(0, s)$, in particular at $z_0$. As $z_0$ was arbitrary in $D(0, R)$, $f$ is continuous on the open disk of convergence. Same proof for the real case with closed sub-intervals $[-s, s]$.

Example — $\exp(z)$ continuous on $\mathbb{C}$

The series $\sum z^n/n!$ has radius $R = +\infty$ (Example of \S 1.1, d'Alembert). By the Corollary, its sum (which will be the complex exponential $\exp(z)$ of \S 4.2) is continuous on $D(0, +\infty) = \mathbb{C}$.

Method — Prove continuity of a power-series sum

Invoke the Corollary above; only the radius $R > 0$ needs to be checked. Continuity on the open disk (resp.\ open interval) is automatic --- no need to verify uniform convergence by hand: the chapter has done it once and for all.

Skills to practice

Proving continuity on the open disk

II.2 Abel's radial convergence theorem

The Corollary of \S 2.1 gives continuity on the open disk, but the boundary $|z| = R$ is delicate. In the real case, Abel's radial convergence theorem gives the boundary value as a one-sided limit, when the boundary series converges. The theorem statement is in the program; its proof is hors programme (« démonstration non exigible »). We include the proof for the curious reader, after the in-place marker.

Theorem — Abel's radial convergence

Let $\sum a_n x^n$ be a real power series of radius $R \in (0, +\infty)$. If the numerical series $\sum a_n R^n$ converges, then $$ \sum_{n=0}^{+\infty} a_n x^n \xrightarrow[x \to R^-]{} \sum_{n=0}^{+\infty} a_n R^n. $$

Proof

Reduce to $R = 1$ by the substitution $x = R y$: the power series $\sum (a_n R^n) y^n$ has radius $1$ and the same convergence at $y = 1$ as $\sum a_n R^n$. So assume $R = 1$ and $\sum a_n$ converges; let $S = \sum_{n=0}^{+\infty} a_n$ and $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ on $[0, 1)$.
Set $S_n = \sum_{k=0}^n a_k$ (partial sums) and $R_n = S - S_n = \sum_{k > n} a_k$ (remainders), so $a_n = R_{n-1} - R_n$ for $n \ge 1$. The KEY IDENTITY: for every $x \in [0, 1)$, $$ f(x) - S = (x - 1) \sum_{n=0}^{+\infty} R_n x^n. $$ The RHS series converges absolutely on $[0, 1)$: $(R_n)$ is bounded (it tends to $0$ as $n \to +\infty$, since $\sum a_n$ converges), and $\sum x^n$ converges geometrically.
Derivation of the identity. For finite $p \ge 1$ and $x \in [0, 1)$, Abel summation by parts gives $$ \begin{aligned} \sum_{n=1}^p a_n x^n - \sum_{n=1}^p a_n &= \sum_{n=1}^p a_n (x^n - 1) \\ &= \sum_{n=1}^p (R_{n-1} - R_n)(x^n - 1) \\ &= \sum_{n=1}^p R_{n-1}(x^n - 1) - \sum_{n=1}^p R_n (x^n - 1). \end{aligned} $$ Re-index the first sum with $m = n - 1$ ($m$ from $0$ to $p - 1$): $$ \sum_{n=1}^p R_{n-1}(x^n - 1) = \sum_{m=0}^{p-1} R_m (x^{m+1} - 1). $$ Combine the two sums term by term. The shared index range $1 \le m \le p - 1$ contributes $R_m \bigl[ (x^{m+1} - 1) - (x^m - 1) \bigr] = R_m \, x^m (x - 1)$. The boundary terms are $R_0 (x - 1)$ (from $m = 0$ in the first sum only) and $-R_p (x^p - 1) = R_p (1 - x^p)$ (from $m = p$ in the second sum only). Hence $$ \sum_{n=1}^p a_n x^n - \sum_{n=1}^p a_n = (x - 1) \sum_{n=0}^{p-1} R_n x^n + (1 - x^p) R_p. $$ Adding $a_0 (x^0 - 1) = 0$ on the LHS doesn't change anything (since $x^0 = 1$). Letting $p \to +\infty$: the LHS tends to $f(x) - S$ (with $a_0$ added back), the first term on the RHS is the partial sum of an absolutely convergent series and tends to $(x-1) \sum_{n=0}^{+\infty} R_n x^n$, and $(1 - x^p) R_p \to 0$ (since $R_p \to 0$ and $1 - x^p$ is bounded). The identity follows.
Conclusion. Given $\varepsilon > 0$, pick $N$ so that $|R_n| < \varepsilon$ for $n \ge N$. Then $$ \begin{aligned} |f(x) - S| &= |1 - x| \cdot \biggl| \sum_{n=0}^{+\infty} R_n x^n \biggr| \\ &\le (1 - x) \sum_{n=0}^{N-1} |R_n| + (1 - x) \varepsilon \sum_{n \ge N} x^n \\ &\le (1 - x) \cdot N \cdot \max_{n < N} |R_n| + (1 - x) \cdot \varepsilon \cdot \frac{x^N}{1 - x} \\ &\le N \cdot \max_{n < N} |R_n| \cdot (1 - x) + \varepsilon. \end{aligned} $$ The first term tends to $0$ as $x \to 1^-$, so for $x$ close enough to $1^-$, $|f(x) - S| < 2\varepsilon$. Hence $f(x) \to S$ as $x \to 1^-$. (Adapted from Prost, Séries entières, p. 8.)

La démonstration de ce théorème est hors programme. The proof below is included for the curious reader; the student is responsible only for the statement.

Corollary — Continuity on the closed convergence interval

Let $\sum a_n x^n$ be a real power series of radius $R \in (0, +\infty)$. The sum $f$ is continuous on the open interval $(-R, R)$ (by \S 2.1) AND at every boundary endpoint $\pm R$ where the boundary numerical series $\sum a_n (\pm R)^n$ converges.

Proof

On $(-R, R)$, continuity is the Corollary of \S 2.1. At the right endpoint $R$, when $\sum a_n R^n$ converges, Abel's radial gives $f(x) \to \sum a_n R^n = f(R)$ as $x \to R^-$, i.e.\ left-continuity at $R$.
At the left endpoint $-R$, when $\sum a_n (-R)^n$ converges: set $b_n = (-1)^n a_n$. The series $\sum b_n y^n$ has the same radius as $\sum a_n y^n$ (since $|b_n| = |a_n|$), namely $R$, and $\sum b_n R^n = \sum a_n (-R)^n$ converges by hypothesis. By Abel's radial applied to $\sum b_n y^n$, $\sum b_n y^n \to \sum b_n R^n$ as $y \to R^-$. Setting $y = -x$: $\sum a_n (-y)^n = \sum b_n y^n$, so $f(-y) \to \sum a_n (-R)^n$ as $y \to R^-$, i.e.\ $f(x) \to f(-R)$ as $x = -y \to -R^+$. Right-continuity at $-R$.

Example — Alternating harmonic series equals $\ln 2$

Take $f(x) = \sum_{n \ge 1} x^n/n$, which equals $-\ln(1 - x)$ on $(-1, 1)$ (proved in \S 3.2 by term-by-term integration). At $x = -1$, the series $\sum (-1)^n/n$ converges (alternating, Leibniz). By the Corollary above (Abel's radial applied at the left endpoint $-R = -1$), $f$ extends continuously to $x = -1$: $$ \sum_{n=1}^{+\infty} \frac{(-1)^n}{n} = f(-1) = \lim_{x \to -1^+} -\ln(1 - x) = -\ln 2. $$ Hence $\sum_{n=1}^{+\infty} (-1)^{n+1}/n = \ln 2$.

Skills to practice

Applying Abel's radial convergence

III Regularity of the sum on the open interval (real variable)

III.1 Term-by-term differentiation$\virgule$ class $\mathcal{C}^\infty$

On the open interval $(-R, R)$ in the real case, the sum $f(x) = \sum a_n x^n$ is differentiable, and its derivative is obtained by differentiating term by term: $f'(x) = \sum n a_n x^{n-1}$. Iterating, $f$ is of class $\mathcal{C}^\infty$, with the $k$-th derivative obtained by repeated term-by-term differentiation. The key inputs are the «$\sum n a_n z^n$ same radius» Proposition of \S 1.2 and the term-by-term differentiation theorem of Sequences and series of functions \S 2.2.

Theorem — Term-by-term differentiation

Let $\sum a_n x^n$ be a real power series of radius $R$, with sum $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ on $(-R, R)$. Then $f$ is of class $\mathcal{C}^1$ on $(-R, R)$ and $$ f'(x) = \sum_{n=1}^{+\infty} n a_n x^{n-1} \qquad \text{for every } x \in (-R, R). $$ The derived series $\sum n a_n x^{n-1}$ has the same radius $R$ as the original.

Proof

Set $u_n(x) = a_n x^n$ for $n \ge 0$; then $u_n'(x) = n a_n x^{n-1}$ for $n \ge 1$ (and $u_0' = 0$). We verify the three hypotheses of the term-by-term differentiation theorem of Sequences and series of functions \S 2.2:

Each $u_n$ is of class $\mathcal{C}^1$ on $(-R, R)$ (polynomial).
The series $\sum u_n$ converges at one point, e.g.\ $x_0 = 0$ (trivially, all terms are $0$ except the constant $a_0$).
The series of derivatives $\sum_{n \ge 1} n a_n x^{n-1}$ is a power series in $x$: re-indexing with $m = n - 1$, it becomes $\sum_{m \ge 0} (m + 1) a_{m+1} x^m$. This is a power series whose coefficients $b_m = (m + 1) a_{m+1}$ differ from $a_{m+1}$ by the polynomial factor $(m + 1)$; by the «same radius» Proposition of \S 1.2, $R(\sum b_m x^m) = R(\sum a_{m+1} x^m)$, and the index shift $m \to m+1$ in the second series doesn't change the radius (the series $\sum_{m \ge 0} a_{m+1} x^m$ has the same radius as $\sum_{n \ge 0} a_n x^n$, by direct application of the trichotomy: convergence at a given $x_0 \ne 0$ transfers between the two via the factor $x_0$). Hence the derived series $\sum_{n \ge 1} n a_n x^{n-1}$ has radius $R$. On every $[-r, r] \subset (-R, R)$, it converges normally hence uniformly (\S 2.1).

The three hypotheses are met. By the term-by-term differentiation theorem, $\sum u_n$ converges uniformly on every $[-r, r] \subset (-R, R)$, its sum $f$ is of class $\mathcal{C}^1$ on $(-R, R)$, and $$ f'(x) = \sum_{n=0}^{+\infty} u_n'(x) = \sum_{n=1}^{+\infty} n a_n x^{n-1}. $$

Corollary — The sum is $\mathcal{C}^\infty$ on the open interval

Let $\sum a_n x^n$ be a real power series of radius $R$, with sum $f$ on $(-R, R)$. Then $f$ is of class $\mathcal{C}^\infty$ on $(-R, R)$, and for every $k \ge 0$ and every $x \in (-R, R)$, $$ f^{(k)}(x) = \sum_{n=k}^{+\infty} n (n-1) \cdots (n - k + 1) \, a_n \, x^{n - k}. $$ The $k$-th derived series has the same radius $R$.

Proof

Induction on $k$ applying the Theorem. At $k = 0$, $f$ is $\mathcal{C}^0$ (continuous) by \S 2.1, with the formula $f(x) = \sum a_n x^n$. Heredity: assume $f$ is $\mathcal{C}^k$ on $(-R, R)$ with the displayed formula. Applying the Theorem to the power series $\sum_{n \ge k} n (n-1) \cdots (n - k + 1) a_n x^{n-k}$ (of same radius $R$, by repeated application of the same-R Proposition of \S 1.2), its sum $f^{(k)}$ is of class $\mathcal{C}^1$, and $$ (f^{(k)})'(x) = \sum_{n = k+1}^{+\infty} (n - k) \cdot n(n-1) \cdots (n - k + 1) \, a_n \, x^{n - k - 1} = \sum_{n = k+1}^{+\infty} n (n-1) \cdots (n - k) \, a_n \, x^{n - k - 1}. $$ Re-indexing or recognising this as the formula at step $k + 1$: $f^{(k+1)}(x) = \sum_{n \ge k+1} n(n-1)\cdots(n - k) a_n x^{n - k - 1}$. Hence $f$ is $\mathcal{C}^{k+1}$, closing the induction.

Example — $\sum n x^n \equal x/(1 - x)^2$

Start from the geometric series $\sum_{n=0}^{+\infty} x^n = 1/(1 - x)$ on $(-1, 1)$ (\S 1.1). By term-by-term differentiation, $$ \sum_{n=1}^{+\infty} n x^{n-1} = \frac{1}{(1 - x)^2} \qquad \text{on } (-1, 1). $$ Multiplying by $x$ (or re-indexing $m = n - 1$), $$ \sum_{n=1}^{+\infty} n x^n = \frac{x}{(1 - x)^2} \qquad \text{on } (-1, 1). $$ A classical identity, obtained here by one differentiation step.

Method — Differentiate a power series term by term

On the open interval of convergence $(-R, R)$, take $(\sum a_n x^n)' = \sum n a_n x^{n-1}$, with the index of the leading term shifting to $n = 1$ (the constant term $a_0$ vanishes). For the $k$-th derivative, shift the leading index to $n = k$ and multiply by $n(n-1) \cdots (n - k + 1)$. The radius is preserved at every step.

Skills to practice

Differentiating a power series term by term

III.2 Term-by-term integration

Integration is the inverse operation: the primitive of $f(x) = \sum a_n x^n$ vanishing at $0$ is obtained by integrating term by term, giving the same radius of convergence. This is the natural route to the developments of $\ln(1+x)$ and $\arctan x$ from the geometric series.

Theorem — Term-by-term integration

Let $\sum a_n x^n$ be a real power series of radius $R$, with sum $f$ on $(-R, R)$. The series $\sum_{n \ge 0} (a_n / (n+1)) \, x^{n+1}$ has the same radius $R$; its sum $F$ is the primitive of $f$ vanishing at $0$: $$ F(x) = \int_0^x f(t) \, \mathrm{d}t = \sum_{n=0}^{+\infty} \frac{a_n}{n + 1} \, x^{n+1} \qquad \text{for every } x \in (-R, R). $$

Proof

(i) Radius equality. The power series $\sum c_n z^n$ with $c_n = a_n/(n+1)$ has the same radius as $\sum a_n z^n$, by a two-step chain. First, the «same radius» Proposition of \S 1.2 applied to $(c_n)$ gives $R(\sum c_n z^n) = R(\sum n c_n z^n)$. Second, the comparison rule of \S 1.2 applied to the equivalent $n c_n \sim (n+1) c_n$ as $n \to +\infty$ (since $n/(n+1) \to 1$) gives $R(\sum n c_n z^n) = R(\sum (n+1) c_n z^n)$. But $(n+1) c_n = a_n$, hence the right-hand side is $R(\sum a_n z^n) = R$. Chaining: $R(\sum c_n z^n) = R$. The shifted-by-one series $\sum (a_n/(n+1)) x^{n+1} = x \cdot \sum (a_n/(n+1)) x^n$ has the same radius (multiplying by $x$ does not change convergence).
(ii) Identification with $\int_0^x f$. On every $[-r, r] \subset (-R, R)$, the series $\sum a_n x^n$ converges normally hence uniformly (\S 2.1). By the term-by-term integration theorem of Sequences and series of functions \S 2.2, for every $x \in (-R, R)$, $$ \int_0^x f(t) \, \mathrm{d}t = \int_0^x \biggl( \sum_{n=0}^{+\infty} a_n t^n \biggr) \mathrm{d}t = \sum_{n=0}^{+\infty} \int_0^x a_n t^n \, \mathrm{d}t = \sum_{n=0}^{+\infty} \frac{a_n}{n + 1} x^{n+1}, $$ which is the announced identity. $F(0) = 0$ trivially.

Example — $\sum x^n/n \equal -\ln(1 - x)$

Start from $\sum_{n=0}^{+\infty} x^n = 1/(1 - x)$ on $(-1, 1)$. By term-by-term integration, $$ \sum_{n=0}^{+\infty} \frac{x^{n+1}}{n + 1} = \int_0^x \frac{\mathrm{d}t}{1 - t} = -\ln(1 - x). $$ Re-indexing $m = n + 1$: $$ \sum_{n=1}^{+\infty} \frac{x^n}{n} = -\ln(1 - x) \qquad \text{on } (-1, 1). $$

Example — $\arctan x$ series

Start from $\sum_{n=0}^{+\infty} (-1)^n x^{2n} = 1/(1 + x^2)$ on $(-1, 1)$ (geometric series in $-x^2$, radius $1$). By term-by-term integration, $$ \arctan x = \int_0^x \frac{\mathrm{d}t}{1 + t^2} = \sum_{n=0}^{+\infty} \frac{(-1)^n}{2n + 1} \, x^{2n + 1} \qquad \text{on } (-1, 1). $$ At $x = 1$, the series $\sum (-1)^n/(2n+1)$ converges (alternating, Leibniz); by Abel's radial (\S 2.2), the identity extends to $x = 1$: $$ \frac{\pi}{4} = \arctan 1 = \sum_{n=0}^{+\infty} \frac{(-1)^n}{2n + 1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots, $$ Leibniz's formula for $\pi/4$.

Method — Integrate a power series term by term

The primitive of $f(x) = \sum a_n x^n$ vanishing at $0$ is obtained by dividing the coefficient $a_n$ by $n + 1$ and shifting the exponent to $n + 1$; the radius is preserved; a non-zero integration constant (a different primitive) must be supplied externally. Useful for deriving the $\ln$ and $\arctan$ series from the geometric series, and more generally for primitive computations of rational functions.

Skills to practice

Integrating a power series term by term

IV Functions developable in power series

IV.1 Definition$\virgule$ necessary condition$\virgule$ uniqueness

The inverse problem: given a function $f$, can it be written as a power series $\sum a_n x^n$ near $0$? Such a function is called developable in power series at $0$. The necessary condition in the real case is that $f$ be of class $\mathcal{C}^\infty$ on a neighborhood of $0$, with the development forced to be the Taylor series of $f$ at $0$. Sufficiency requires more --- the $\mathcal{C}^\infty$-but-not-developable counterexample below shows that $\mathcal{C}^\infty$ regularity is necessary but not sufficient.

Definition — Function developable in power series

A function $f$ is developable in power series at $0$ on $(-r, r)$ (resp.\ $D(0, r)$ in $\mathbb{C}$), with $r > 0$, if there exists a power series $\sum a_n x^n$ of radius $R \ge r$ such that $$ f(x) = \sum_{n=0}^{+\infty} a_n x^n \qquad \text{for every } x \in (-r, r) \text{ (resp.\ } D(0, r)\text{).} $$

Theorem — Necessary condition$\virgule$ real case

If $f$ is developable in power series on $(-r, r)$, with $r > 0$, then $f$ is of class $\mathcal{C}^\infty$ on $(-r, r)$, the development is unique, and is the Taylor series of $f$ at $0$: $$ f(x) = \sum_{n=0}^{+\infty} \frac{f^{(n)}(0)}{n!} \, x^n \qquad \text{for every } x \in (-r, r). $$

Proof

Suppose $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ on $(-r, r)$. By the Corollary of \S 3.1, $f$ is of class $\mathcal{C}^\infty$ on $(-r, r)$ with $$ f^{(k)}(x) = \sum_{n=k}^{+\infty} n(n-1) \cdots (n - k + 1) a_n x^{n - k}. $$ Evaluating at $x = 0$: only the term $n = k$ survives (all others have a factor $x^{n-k}$ with $n > k$, hence $0$), giving $f^{(k)}(0) = k! \, a_k$. So $a_k = f^{(k)}(0)/k!$, the Taylor coefficient of $f$ at $0$.

Corollary — Uniqueness of the development

If $\sum a_n x^n = \sum b_n x^n$ for every $x$ in a non-empty open interval around $0$, then $a_n = b_n$ for every $n \ge 0$.

Proof

Both series represent the same function $f$ on a positive-radius neighborhood of $0$. By the Theorem, $a_n = f^{(n)}(0)/n! = b_n$ for every $n$.

Example — $\mathrm{e}^{-1/x^2}$ is $\mathcal{C}^\infty$ but not developable

Let $f \colon \mathbb{R} \to \mathbb{R}$ be $f(x) = \mathrm{e}^{-1/x^2}$ for $x \ne 0$ and $f(0) = 0$. Then $f$ is of class $\mathcal{C}^\infty$ on $\mathbb{R}$ with $f^{(n)}(0) = 0$ for every $n \ge 0$. Hence the Taylor series of $f$ at $0$ is identically $0$, distinct from $f$ on every neighborhood of $0$ (since $f(x) > 0$ for $x \ne 0$). So $f$ is $\mathcal{C}^\infty$ but not developable in power series at $0$.
Proof by induction:

Hypothesis at step $n$. $f^{(n)}$ exists on $\mathbb{R}$ with $f^{(n)}(0) = 0$, and for $x \ne 0$, $f^{(n)}(x) = P_n(1/x) \cdot \mathrm{e}^{-1/x^2}$ for some polynomial $P_n$.
Initialization at $n = 0$. $f(0) = 0$ by definition, and for $x \ne 0$, $f(x) = 1 \cdot \mathrm{e}^{-1/x^2}$ with $P_0 = 1$.
Induction step. Assume the hypothesis at step $n$. For $x \ne 0$, differentiate: $$ f^{(n+1)}(x) = \frac{\mathrm{d}}{\mathrm{d}x} \bigl( P_n(1/x) \mathrm{e}^{-1/x^2} \bigr) = \bigl[ -P_n'(1/x) \cdot 1/x^2 + P_n(1/x) \cdot 2/x^3 \bigr] \mathrm{e}^{-1/x^2}, $$ which is again $P_{n+1}(1/x) \cdot \mathrm{e}^{-1/x^2}$ for a polynomial $P_{n+1}(t) = 2 t^3 P_n(t) - t^2 P_n'(t)$ in $t = 1/x$. For the value at $0$: the difference quotient of $f^{(n)}$ at $0$ is $$ \frac{f^{(n)}(x) - f^{(n)}(0)}{x} = \frac{P_n(1/x) \mathrm{e}^{-1/x^2}}{x} = Q(1/x) \mathrm{e}^{-1/x^2} \qquad (x \ne 0), $$ with $Q(t) = t \cdot P_n(t)$ still a polynomial. As $x \to 0$, $1/x^2 \to +\infty$ so $\mathrm{e}^{-1/x^2}$ decays faster than any polynomial in $1/x$; the difference quotient tends to $0$. Hence $f^{(n)}$ is differentiable at $0$ with $(f^{(n)})'(0) = 0$, i.e.\ $f^{(n+1)}(0) = 0$. The hypothesis at step $n + 1$ is established.
Conclusion. By induction, $f \in \mathcal{C}^\infty(\mathbb{R})$ with $f^{(n)}(0) = 0$ for every $n$. Continuity of each $f^{(n)}$ is automatic: polynomial-times-exponential is continuous on $\mathbb{R}^*$, and $f^{(n)}(x) \to 0 = f^{(n)}(0)$ as $x \to 0$ by the same fast-decay argument.

The Taylor series $\sum f^{(n)}(0) x^n/n! = 0$ identically does not equal $f(x) = \mathrm{e}^{-1/x^2} > 0$ for $x \ne 0$. So $f$ is $\mathcal{C}^\infty$ but NOT developable in power series at $0$.

Method — Prove that $f$ is developable in power series

(a) Show that $f$ is of class $\mathcal{C}^\infty$ on a neighborhood of $0$ (necessary condition).
(b) Compute the Taylor series $\sum f^{(n)}(0) x^n/n!$ of $f$ at $0$, and its radius.
(c) Verify that the Taylor series actually converges to $f$ by one of the five routes of \S 4.2: Taylor-Lagrange remainder bound, term-by-term differentiation / integration of a known series, combination of classical developments via the operations of \S 1.3, partial-fraction decomposition (for rational $f$), or analyse-synthèse via an ODE.

Skills to practice

Identifying the Taylor series and proving developability

IV.2 Classical developments and the 5 practical routes

The catalogue of developments every student must know by heart, organised by the technique that establishes each: Taylor-Lagrange for $\exp$, $\cos$, $\sin$; linear combinations of the exp series for $\mathrm{ch}$, $\mathrm{sh}$; term-by-term integration for $\ln(1+x)$ and $\arctan x$; analyse-synthèse via an ODE for $(1+x)^\alpha$. Each carries its radius and convergence domain.

Theorem — Classical developments: $\exp$$\virgule$ $\cos$$\virgule$ $\sin$$\virgule$ $\mathrm{ch}$$\virgule$ $\mathrm{sh}$

For every $z \in \mathbb{C}$: $$ \exp(z) = \sum_{n=0}^{+\infty} \frac{z^n}{n!} \qquad (R = +\infty). $$ For every $x \in \mathbb{R}$: $$ \begin{aligned} \cos x &= \sum_{n=0}^{+\infty} \frac{(-1)^n x^{2n}}{(2n)!} & (R = +\infty), \\ \sin x &= \sum_{n=0}^{+\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} & (R = +\infty), \\ \mathrm{ch}\, x &= \sum_{n=0}^{+\infty} \frac{x^{2n}}{(2n)!} & (R = +\infty), \\ \mathrm{sh}\, x &= \sum_{n=0}^{+\infty} \frac{x^{2n+1}}{(2n+1)!} & (R = +\infty). \end{aligned} $$

Proof

For $\exp$ on $\mathbb{R}$. Taylor-Lagrange at order $n$ on $[0, x]$ (recalled from MPSI Differentiability) gives, for every $x \in \mathbb{R}$, $$ \left| \mathrm{e}^x - \sum_{k=0}^n \frac{x^k}{k!} \right| \le M \cdot \frac{|x|^{n+1}}{(n+1)!}, \qquad M = \sup_{t \in [0, |x|]} |\mathrm{e}^t| = \mathrm{e}^{|x|}, $$ with $M$ fixed in $n$. Since $|x|^{n+1}/(n+1)! \to 0$ as $n \to +\infty$, the remainder tends to $0$ and the Taylor series converges to $\mathrm{e}^x$.
For $\exp$ on $\mathbb{C}$. Define $\exp(z) := \sum_{n=0}^{+\infty} z^n/n!$ on $\mathbb{C}$ as the complex exponential. The series has radius $R = +\infty$ (d'Alembert: $|a_{n+1}/a_n| = 1/(n+1) \to 0$), so $\exp$ is well defined on $\mathbb{C}$ and continuous (\S 2.1). The restriction $z = x \in \mathbb{R}$ recovers the usual real $\exp$ (same power series). The functional equation $\exp(z) \exp(z') = \exp(z + z')$ follows from Cauchy product (\S 1.3): for every $z, z' \in \mathbb{C}$, $$ \exp(z) \exp(z') = \sum_{n=0}^{+\infty} c_n, \qquad c_n = \sum_{k=0}^n \frac{z^k}{k!} \cdot \frac{(z')^{n-k}}{(n-k)!} = \frac{(z + z')^n}{n!} $$ by the binomial identity. So $\exp(z) \exp(z') = \sum_{n=0}^{+\infty} (z+z')^n/n! = \exp(z + z')$.
For $\cos$ and $\sin$. All derivatives of $\cos$ and $\sin$ are among $\pm \cos, \pm \sin$, so $|\cos^{(n)}(t)| \le 1$ and $|\sin^{(n)}(t)| \le 1$ for every $n$ and every $t \in \mathbb{R}$. Taylor-Lagrange at order $n$ on $[0, x]$ gives, for every $x \in \mathbb{R}$, $$ \left| \cos x - \sum_{k=0}^n \cos^{(k)}(0) \frac{x^k}{k!} \right| \le 1 \cdot \frac{|x|^{n+1}}{(n+1)!} \to 0 \quad (n \to +\infty). $$ With $\cos^{(2k)}(0) = (-1)^k$ and $\cos^{(2k+1)}(0) = 0$ (MPSI), the Taylor series of $\cos$ at $0$ is $\sum (-1)^k x^{2k}/(2k)!$, and the limit identifies it with $\cos x$. Similarly for $\sin$, with $\sin^{(2k)}(0) = 0$ and $\sin^{(2k+1)}(0) = (-1)^k$.
For $\mathrm{ch}$ and $\mathrm{sh}$. The MPSI definitions are $\mathrm{ch}\, x = (\mathrm{e}^x + \mathrm{e}^{-x})/2$ and $\mathrm{sh}\, x = (\mathrm{e}^x - \mathrm{e}^{-x})/2$. Using the exp series on $\mathbb{R}$ just established and the linear-combination Proposition of \S 1.3 (which applies because the two exp series have the same radius $+\infty$): $$ \begin{aligned} \mathrm{ch}\, x &= \frac{1}{2} \biggl( \sum_{n=0}^{+\infty} \frac{x^n}{n!} + \sum_{n=0}^{+\infty} \frac{(-x)^n}{n!} \biggr) = \sum_{n=0}^{+\infty} \frac{x^{2n}}{(2n)!} \quad \text{(odd terms cancel)}, \\ \mathrm{sh}\, x &= \frac{1}{2} \biggl( \sum_{n=0}^{+\infty} \frac{x^n}{n!} - \sum_{n=0}^{+\infty} \frac{(-x)^n}{n!} \biggr) = \sum_{n=0}^{+\infty} \frac{x^{2n+1}}{(2n+1)!} \quad \text{(even terms cancel)}. \end{aligned} $$ Radii are $+\infty$.
Remark. For $\cos$ and $\sin$, the Euler identities $\cos x = (\mathrm{e}^{\mathrm{i}x} + \mathrm{e}^{-\mathrm{i}x})/2$, $\sin x = (\mathrm{e}^{\mathrm{i}x} - \mathrm{e}^{-\mathrm{i}x})/(2\mathrm{i})$ are consequences of the established series (not premises) --- at MPSI cos and sin are defined geometrically, so using these identities as premises would be circular. For $\mathrm{ch}$ and $\mathrm{sh}$, the analogous identities are the MPSI definitions, so we used them directly.

Example — Euler's formula $\mathrm{e}^{x + \mathrm{i}y} \equal \mathrm{e}^x(\cos y + \mathrm{i} \sin y)$

For every $x, y \in \mathbb{R}$, apply the functional equation $\exp(z + z') = \exp(z) \exp(z')$ (just proved) with $z = x$ and $z' = \mathrm{i}y$: $$ \mathrm{e}^{x + \mathrm{i}y} = \mathrm{e}^x \cdot \exp(\mathrm{i}y). $$ Now $\exp(\mathrm{i}y) = \sum_{n=0}^{+\infty} (\mathrm{i}y)^n/n!$. Separating even and odd terms: $$ \exp(\mathrm{i}y) = \sum_{n=0}^{+\infty} \frac{\mathrm{i}^{2n} y^{2n}}{(2n)!} + \sum_{n=0}^{+\infty} \frac{\mathrm{i}^{2n+1} y^{2n+1}}{(2n+1)!} = \sum_{n=0}^{+\infty} \frac{(-1)^n y^{2n}}{(2n)!} + \mathrm{i} \sum_{n=0}^{+\infty} \frac{(-1)^n y^{2n+1}}{(2n+1)!}, $$ using $\mathrm{i}^{2n} = (-1)^n$ and $\mathrm{i}^{2n+1} = \mathrm{i}(-1)^n$. By the cos and sin series above, this is $\cos y + \mathrm{i} \sin y$. Hence $$ \mathrm{e}^{x + \mathrm{i}y} = \mathrm{e}^x (\cos y + \mathrm{i} \sin y) \qquad \text{for every } x, y \in \mathbb{R}. $$ This recovers the polar form of a non-zero complex number and connects the Taylor-series picture (this chapter) with the trigonometric picture (MPSI Complex numbers).

Proposition — $\ln(1+x)$ and $\arctan x$ developments

For every $x \in (-1, 1)$: $$ \ln(1 + x) = \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{n} \, x^n \qquad (R = 1), $$ $$ \arctan x = \sum_{n=0}^{+\infty} \frac{(-1)^n}{2n + 1} \, x^{2n+1} \qquad (R = 1). $$

Proof

$\ln(1+x)$. Start from $1/(1 + x) = \sum_{n=0}^{+\infty} (-x)^n = \sum (-1)^n x^n$ on $(-1, 1)$ (geometric series in $-x$, radius $1$). By term-by-term integration (\S 3.2), $$ \ln(1 + x) = \int_0^x \frac{\mathrm{d}t}{1 + t} = \sum_{n=0}^{+\infty} \frac{(-1)^n}{n + 1} x^{n+1} = \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{n} x^n. $$ At $x = 1$, the series $\sum (-1)^{n+1}/n$ converges (alternating); by Abel's radial (\S 2.2), the identity extends to $x = 1$: $\ln 2 = \sum (-1)^{n+1}/n$.
$\arctan x$. Start from $1/(1 + x^2) = \sum (-1)^n x^{2n}$ on $(-1, 1)$ (geometric in $-x^2$). By term-by-term integration, $$ \arctan x = \int_0^x \frac{\mathrm{d}t}{1 + t^2} = \sum_{n=0}^{+\infty} \frac{(-1)^n}{2n + 1} x^{2n + 1}. $$ At $x = 1$, the series $\sum (-1)^n/(2n+1)$ converges (alternating); by Abel's radial, $\pi/4 = \sum (-1)^n/(2n+1)$ (Leibniz's formula). At $x = -1$, the development extends by oddness of $\arctan$ (the series is odd in $x$, both sides flip sign), or equivalently by Abel's radial applied at the left endpoint via the $-R$ sub-case of \S 2.2.

Boundary identities

The boundary extensions yield two famous identities: $$ \ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{n}, $$ $$ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \sum_{n=0}^{+\infty} \frac{(-1)^n}{2n + 1} \quad \text{(Leibniz)}. $$ Both follow from Abel's radial applied at the relevant endpoint.

Proposition — $(1+x)^\alpha$ development

For every $\alpha \in \mathbb{R}$ and every $x \in (-1, 1)$: $$ (1 + x)^\alpha = 1 + \sum_{n=1}^{+\infty} \frac{\alpha (\alpha - 1) \cdots (\alpha - n + 1)}{n!} \, x^n, $$ with $R = 1$ if $\alpha \notin \mathbb{N}$, and $R = +\infty$ if $\alpha \in \mathbb{N}$ (the series is finite --- the binomial formula).

Proof

Analyse-synthèse via the Cauchy problem $(1 + x) y' = \alpha y$ with $y(0) = 1$ on $(-1, +\infty)$, uniquely solved by $y(x) = (1 + x)^\alpha$ (recalled from MPSI Linear ODE of order 1).
Analysis. Postulate $y(x) = \sum_{n \ge 0} a_n x^n$ with $a_0 = y(0) = 1$. Then $y'(x) = \sum_{n \ge 1} n a_n x^{n-1}$ and $(1 + x) y'(x) = \sum_{n \ge 1} n a_n x^{n-1} + \sum_{n \ge 1} n a_n x^n$. Re-indexing the first sum (let $m = n - 1$): $$ (1 + x) y'(x) = \sum_{m \ge 0} (m+1) a_{m+1} x^m + \sum_{n \ge 1} n a_n x^n = a_1 + \sum_{n \ge 1} [(n+1) a_{n+1} + n a_n] x^n. $$ The equation $(1+x) y' = \alpha y$ gives, comparing coefficients of $x^n$: $$ a_1 = \alpha a_0 = \alpha \quad ; \quad (n+1) a_{n+1} + n a_n = \alpha a_n \quad \text{for } n \ge 1, $$ i.e.\ $(n+1) a_{n+1} = (\alpha - n) a_n$. By induction, $a_n = \dfrac{\alpha(\alpha-1) \cdots (\alpha - n + 1)}{n!}$ for $n \ge 0$ (with $a_0 = 1$).
Synthesis. Verify the candidate series has positive radius. If $\alpha \in \mathbb{N}$, then $a_n = 0$ for $n > \alpha$ (in the recurrence $(n+1) a_{n+1} = (\alpha - n) a_n$, the factor $\alpha - n$ vanishes at $n = \alpha$, giving $a_{\alpha + 1} = 0$ and then $a_n = 0$ for every $n > \alpha$), so the series is finite ($R = +\infty$). If $\alpha \notin \mathbb{N}$, then $a_n \ne 0$ for every $n$, and by d'Alembert: $$ \left| \frac{a_{n+1}}{a_n} \right| = \frac{|\alpha - n|}{n + 1} \xrightarrow[n \to +\infty]{} 1, $$ so $R = 1$.
Either way, the candidate $y$ defined by the series solves $(1 + x) y' = \alpha y$ on the open interval of convergence and satisfies $y(0) = 1$. By Cauchy uniqueness on $(-1, 1)$ (recalled from MPSI Linear ODE of order 1), this $y$ equals $(1 + x)^\alpha$.

Example — $\sqrt{1 + x}$ first terms

Take $\alpha = 1/2$ in the Proposition. The coefficients: $$ a_1 = \frac{1}{2}, \quad a_2 = \frac{(1/2)(-1/2)}{2!} = -\frac{1}{8}, \quad a_3 = \frac{(1/2)(-1/2)(-3/2)}{3!} = \frac{1}{16}, \quad a_4 = -\frac{5}{128}. $$ Hence on $(-1, 1)$: $$ \sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \frac{5 x^4}{128} + \cdots \qquad (R = 1). $$ A classical handy expansion --- useful for numerical estimates and asymptotic expansions.

Synthesis table

The classical-developments cheat sheet to know by heart: \renewcommand{\arraystretch}{1.4}

Series	Domain	$R$
$\sum_{n \ge 0} z^n = \dfrac{1}{1 - z}$	$D(0, 1) \subset \mathbb{C}$	$1$
$\sum_{n \ge 0} \dfrac{z^n}{n!} = \mathrm{e}^z$	$\mathbb{C}$	$+\infty$
$\sum_{n \ge 0} \dfrac{(-1)^n x^{2n}}{(2n)!} = \cos x$	$\mathbb{R}$	$+\infty$
$\sum_{n \ge 0} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} = \sin x$	$\mathbb{R}$	$+\infty$
$\sum_{n \ge 0} \dfrac{x^{2n}}{(2n)!} = \mathrm{ch}\, x$	$\mathbb{R}$	$+\infty$
$\sum_{n \ge 0} \dfrac{x^{2n+1}}{(2n+1)!} = \mathrm{sh}\, x$	$\mathbb{R}$	$+\infty$
$\sum_{n \ge 1} \dfrac{(-1)^{n+1} x^n}{n} = \ln(1+x)$	$(-1, 1]$	$1$
$\sum_{n \ge 0} \dfrac{(-1)^n x^{2n+1}}{2n+1} = \arctan x$	$[-1, 1]$	$1$
$1 + \sum_{n \ge 1} \dfrac{\alpha(\alpha-1) \cdots (\alpha - n + 1)}{n!} x^n = (1+x)^\alpha$	$(-1, 1)$ for $\alpha \notin \mathbb{N}$ (interior)	$1$

Note on $(1+x)^\alpha$. The Proposition's statement above is for the open interval $(-1, 1)$. Boundary behaviour at $\pm 1$ depends on $\alpha$ — convergence at $1$ holds when $\alpha > -1$, at $-1$ when $\alpha > 0$ (by Abel's radial combined with the Raabe-Duhamel criterion for the boundary series, out of program scope); the table records only the interior domain to stay in-program.

Method — Obtain a power-series development --- the 5 routes

(1) Classical developments + operations. Combine the table above via the linear-combination and Cauchy-product Propositions of \S 1.3. Example: $x \ln(1 - x) + 2 \mathrm{e}^x$ assembles linear combinations of $\ln(1 - x)$ and $\mathrm{e}^x$.
(2) Term-by-term differentiation or integration. Use \S 3.1 or \S 3.2 to derive the development of $f' \to f$ or $\int f$ from a known one.
(3) Taylor-Lagrange estimate. For $f$ of class $\mathcal{C}^\infty$ near $0$ with $|f^{(n+1)}(t)| \le M_{n+1}$ on a neighborhood, the remainder of the order-$n$ Taylor expansion is $\le M_{n+1} \, |x|^{n+1}/(n+1)!$; if this tends to $0$ as $n \to +\infty$, $f$ equals its Taylor series. Used for $\exp$, $\cos$, $\sin$ above.
(4) Partial-fraction decomposition for a rational fraction $P/Q$, after cancelling any common factor of $P$ and $Q$: the reduced denominator $\widetilde{Q}$ must satisfy $\widetilde{Q}(0) \ne 0$ (no pole at $0$). Decompose $P/\widetilde{Q}$ into simple elements $c/(x - a)^k$ with $a \ne 0$, then expand each $1/(x - a) = (-1/a) \sum (x/a)^n$ on $|x| < |a|$ and differentiate $k - 1$ times. The radius of the result equals the distance from $0$ to the nearest pole. (If $\widetilde{Q}(0) = 0$, $P/Q$ has a genuine pole at $0$ and is not developable in power series there.)
(5) ODE analyse-synthèse. Postulate $y = \sum a_n x^n$, inject into a linear ODE satisfied by $f$, derive the coefficient recurrence by identification, solve the recurrence, check the candidate series has positive radius (often by d'Alembert), then identify with $f$ by Cauchy uniqueness. Used for $(1+x)^\alpha$ above.

Going further

Functions developable in power series at every point of an open subset of $\mathbb{R}$ (resp.\ $\mathbb{C}$) are called analytic; analytic functions are the bridge from this chapter to complex analysis (out of program). The $\exp$ function as defined here on $\mathbb{C}$ is the same series-defined object that Exponential of a matrix (chapter 23) generalises to $\exp(tA)$ for a square matrix $A$.

Skills to practice

Obtaining a classical or rational-fraction development

Series	Domain	\(R\)
\(\sum_{n \ge 0} z^n = \dfrac{1}{1 - z}\)	\(D(0, 1) \subset \mathbb{C}\)	\(1\)
\(\sum_{n \ge 0} \dfrac{z^n}{n!} = \mathrm{e}^z\)	\(\mathbb{C}\)	\(+\infty\)
\(\sum_{n \ge 0} \dfrac{(-1)^n x^{2n}}{(2n)!} = \cos x\)	\(\mathbb{R}\)	\(+\infty\)
\(\sum_{n \ge 0} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} = \sin x\)	\(\mathbb{R}\)	\(+\infty\)
\(\sum_{n \ge 0} \dfrac{x^{2n}}{(2n)!} = \mathrm{ch}\, x\)	\(\mathbb{R}\)	\(+\infty\)
\(\sum_{n \ge 0} \dfrac{x^{2n+1}}{(2n+1)!} = \mathrm{sh}\, x\)	\(\mathbb{R}\)	\(+\infty\)
\(\sum_{n \ge 1} \dfrac{(-1)^{n+1} x^n}{n} = \ln(1+x)\)	\((-1, 1]\)	\(1\)
\(\sum_{n \ge 0} \dfrac{(-1)^n x^{2n+1}}{2n+1} = \arctan x\)	\([-1, 1]\)	\(1\)
\(1 + \sum_{n \ge 1} \dfrac{\alpha(\alpha-1) \cdots (\alpha - n + 1)}{n!} x^n = (1+x)^\alpha\)	\((-1, 1)\) for \(\alpha \notin \mathbb{N}\) (interior)	\(1\)