CommeUnJeu · L2 MP

Numerical and vector series

⌚ ~67 min ▢ 8 blocks ✓ 13 exercises ➣ Prerequisites : Numerical series, Normed vector spaces, Compactness, connectedness, finite dimension

From MPSI we know what a numerical series is: the sequence of partial sums of a sequence of real or complex numbers, convergent when that sequence of partial sums has a limit. This chapter does two things. It extends the notion of convergent series from $\mathbb{R}$ and $\mathbb{C}$ to a finite-dimensional normed vector space $E$, so that terms may be vectors, matrices or polynomials. And it sharpens the numerical toolbox of MPSI with three asymptotic instruments: d'Alembert's rule, the practical use of series-integral comparison to estimate partial sums of divergent series and remainders of convergent ones, and the summation of comparison relations together with its Cesàro corollary.
Series are an instrument, not the subject. Their typical use is to extract equivalents and asymptotic developments from partial sums or remainders --- and most of this chapter is calibrated to that use.

Conventions

Throughout this chapter $\mathbb{K}$ denotes $\mathbb{R}$ or $\mathbb{C}$ and $E$ a finite-dimensional $\mathbb{K}$-normed vector space with norm $\|\cdot\|$. The dimension of $E$ is $p \ge 1$; a basis of $E$, when fixed, will be written $\mathcal{B} = (e_1, \dots, e_p)$. A sequence $(u_n)_{n \in \mathbb{N}}$ of $E$ gives rise to its series, written $\sum u_n$; when the series converges its sum is denoted $\sum_{n=0}^{+\infty} u_n \in E$. Numerical series are the case $E = \mathbb{K}$, $\|\cdot\| = |\cdot|$, and the entire MPSI theory applies --- it is recalled when it is used, never re-derived. Properties of finite-dimensional normed spaces that come up --- equivalence of all norms, automatic continuity of linear maps, componentwise convergence for a product norm --- are recalled from Normed vector spaces and Compactness, connectedness, finite dimension where they are first needed.

I Series in a finite-dimensional normed space

I.1 Partial sums$\virgule$ convergence$\virgule$ sum$\virgule$ remainder

A vector series is the same object as a scalar series, with the scalar terms replaced by vectors of $E$. The definition that follows is the MPSI definition copied verbatim, with ``number'' read as ``vector of $E$'' --- only the convergence step now asks for a limit of vectors, in the sense of the norm $\|\cdot\|$.

Definition — Series$\virgule$ partial sums$\virgule$ convergence$\virgule$ sum$\virgule$ remainder

Let $(u_n)_{n \in \mathbb{N}}$ be a sequence of $E$.

The partial sum of order $n$ is $S_n = \sum_{k=0}^{n} u_k \in E$. The series of general term $u_n$ is the sequence $(S_n)_{n \in \mathbb{N}}$, written $\sum u_n$.
The series $\sum u_n$ converges when the sequence $(S_n)$ has a limit in $E$; otherwise it diverges. When it converges, the sum of the series is $$ \sum_{n=0}^{+\infty} u_n \;=\; \lim_{n \to +\infty} S_n \;\in\; E. $$
For a convergent series, the remainder of order $n$ is $R_n = \sum_{k=n+1}^{+\infty} u_k \in E$; one has $S = S_n + R_n$ where $S$ denotes the sum.

For $E = \mathbb{K}$ these are the MPSI definitions of a numerical series, with $|\cdot|$ in place of $\|\cdot\|$.

Example — A geometric-type vector series in $\mathbb{R}^2$

Take $u_n = \bigl((-1/2)^n,\; (1/3)^n\bigr) \in \mathbb{R}^2$. The two coordinate series are geometric of common ratio $-1/2$ and $1/3$, hence convergent (MPSI) with sums $1/(1+1/2) = 2/3$ and $1/(1-1/3) = 3/2$. The partial sum $S_n$ is the pair of partial sums of the coordinates, and convergence in $\mathbb{R}^2$ is coordinate-by-coordinate (MPSI; formalized for series in \S 1.3 below), so $S_n \to (2/3,\, 3/2)$ and $$ \sum_{n=0}^{+\infty} u_n \;=\; \bigl(2/3,\; 3/2\bigr). $$

Example — A telescoping vector series

Let $v_n = \bigl(1/(n+1),\; 1/2^n\bigr) \in \mathbb{R}^2$ and set $u_n = v_{n+1} - v_n$. The partial sum telescopes: $$ S_n \;=\; \sum_{k=0}^{n} (v_{k+1} - v_k) \;=\; v_{n+1} - v_0 \;\longrightarrow\; (0,0) - (1,1) \;=\; (-1, -1). $$ So $\sum u_n$ converges with sum $(-1, -1)$.

Example — A matrix series

Take $u_n = \frac{1}{2^n}\,J$ in $\mathcal{M}_2(\mathbb{R})$ where $J = \begin{psmallmatrix} 1 & 1 \\ 0 & 1 \end{psmallmatrix}$. The partial sum factors as $S_n = \bigl(\sum_{k=0}^{n} 1/2^k\bigr) J$. The scalar factor tends to $2$ (geometric series), and the limit of a sequence of matrices is taken entry by entry, so $S_n \to 2J = \begin{psmallmatrix} 2 & 2 \\ 0 & 2 \end{psmallmatrix}$.

Proposition — The remainder tends to zero

If $\sum u_n$ converges, with sum $S$ and remainders $(R_n)$, then $R_n \to 0$.

Proof

By definition $R_n = S - S_n$, and $S_n \to S$, so $R_n \to 0$.

Proposition — Necessary condition for convergence

If $\sum u_n$ converges, then $u_n \to 0$.

Proof

Set $S = \sum_{n=0}^{+\infty} u_n$. For $n \ge 1$, $u_n = S_n - S_{n-1}$. Both partial sums tend to $S$, hence $u_n \to S - S = 0$.

Method — The gross-divergence reflex

Before any other tool, check whether the general term tends to $0$. If $u_n \not\to 0$, the series diverges grossly --- nothing more to do. The converse is false: $\sum 1/n$ has general term tending to $0$ yet diverges (the harmonic series, recalled from MPSI). So $u_n \to 0$ buys nothing on its own; what matters is the speed.

Skills to practice

Computing partial sums and sums

I.2 Linearity$\virgule$ sequence-series link$\virgule$ telescoping

Two structural facts that travel with the definition. The convergent series of $E$ form a $\mathbb{K}$-vector space, and the sum is linear on it: this is the algebraic side. The sequence-series link --- $(v_n)$ converges if and only if its telescoping series $\sum (v_{n+1} - v_n)$ converges --- is the duality between sequences and series that one uses both ways: a hard sequence can be studied via its (easier) telescoping series, and a convergent telescoping series produces a known limit.

Proposition — Linearity

Let $\sum u_n$ and $\sum v_n$ be two convergent series of $E$, and $\lambda, \mu \in \mathbb{K}$. Then $\sum (\lambda u_n + \mu v_n)$ converges and $$ \sum_{n=0}^{+\infty} (\lambda u_n + \mu v_n) \;=\; \lambda \sum_{n=0}^{+\infty} u_n + \mu \sum_{n=0}^{+\infty} v_n. $$ In particular, the convergent series of $E$ form a $\mathbb{K}$-vector subspace of $E^{\mathbb{N}}$, and the sum is a linear map from this subspace to $E$.

Proof

Write $T_n = \sum_{k=0}^{n} (\lambda u_k + \mu v_k) = \lambda S_n + \mu S_n'$ where $S_n, S_n'$ are the partial sums of $\sum u_n, \sum v_n$. By assumption $S_n \to S$ and $S_n' \to S'$. Linearity of limits in a normed space gives $T_n \to \lambda S + \mu S'$.

Definition — Telescoping series

The telescoping series associated with a sequence $(v_n)_{n \in \mathbb{N}}$ of $E$ is the series $\sum (v_{n+1} - v_n)$.

Proposition — Sequence-series link

Let $(v_n)_{n \in \mathbb{N}}$ be a sequence of $E$. Then $(v_n)$ converges if and only if its telescoping series $\sum (v_{n+1} - v_n)$ converges, and in case of convergence $$ \sum_{n=0}^{+\infty} (v_{n+1} - v_n) \;=\; \bigl(\lim_{n \to +\infty} v_n\bigr) - v_0. $$

Proof

The partial sum of order $n$ of the telescoping series is, by telescoping, $$ T_n \;=\; \sum_{k=0}^{n} (v_{k+1} - v_k) \;=\; v_{n+1} - v_0. $$ So $(T_n)$ converges if and only if $(v_{n+1})$ does, equivalently if and only if $(v_n)$ does, and the limit of $T_n$ is then $\lim v_n - v_0$.

Method — Studying a sequence through its telescoping series

To prove a sequence $(v_n)$ converges, it suffices to prove its telescoping series $\sum (v_{n+1} - v_n)$ converges --- which is sometimes easier, since one then has the entire numerical-series toolbox at one's disposal (estimates on the general term $v_{n+1} - v_n$, comparisons with $\sum 1/n^\alpha$, etc.). Conversely, if one has a closed expression for a sum that telescopes, the limit comes for free.

Example — A sequence proved convergent via its telescoping series

Consider $w_n = \sum_{k=1}^{n} 1/k^2$. The differences $w_{n+1} - w_n = 1/(n+1)^2$ form the general term of $\sum 1/k^2$, which converges (Riemann series with $\alpha = 2 > 1$, recalled from MPSI). By the sequence-series link, $(w_n)$ converges --- the value of $\sum_{n=1}^{+\infty} 1/n^2$ (Euler's $\pi^2/6$) is out of reach here, but its existence is established.

Proposition — Image of a convergent series by a continuous linear map

Let $E, F$ be two finite-dimensional $\mathbb{K}$-normed spaces and $L \in \mathcal{L}(E, F)$. Let $\sum u_n$ be a convergent series of $E$. Then $\sum L(u_n)$ is a convergent series of $F$ and $$ \sum_{n=0}^{+\infty} L(u_n) \;=\; L\!\left(\sum_{n=0}^{+\infty} u_n\right). $$

Proof

A linear map between finite-dimensional normed spaces is continuous (recalled from Compactness, connectedness, finite dimension). Set $S = \sum_{n=0}^{+\infty} u_n$ and $S_n = \sum_{k=0}^{n} u_k$. Then $\sum_{k=0}^{n} L(u_k) = L(S_n)$ by linearity, and $L(S_n) \to L(S)$ by continuity of $L$ at $S$. So $\sum L(u_n)$ converges with sum $L(S)$.

Example — Pushing a convergent series through a coordinate form

Take the convergent series of Example 1 in $\mathbb{R}^2$, with sum $S = (2/3, 3/2)$, and the linear form $\varphi : (x, y) \mapsto 2x - y$. Then $\sum \varphi(u_n)$ converges and $$ \sum_{n=0}^{+\infty} \varphi(u_n) \;=\; \varphi(S) \;=\; 2 \cdot 2/3 - 3/2 \;=\; 4/3 - 3/2 \;=\; -1/6. $$

Skills to practice

Exploiting the sequence-series link

I.3 Study through coordinates in a basis

The bridge between a vector series and the MPSI scalar theory: fix a basis of $E$ and read off the coordinate sequences. A vector series converges if and only if each of its $p$ scalar coordinate series converges, and the sum is read coordinatewise. This is the lever that brings every MPSI result on numerical series to bear on a vector series.

Proposition — Coordinate criterion

Fix a basis $\mathcal{B} = (e_1, \dots, e_p)$ of $E$ and write each vector $u_n$ in coordinates: $u_n = \sum_{j=1}^{p} u_n^{(j)} e_j$ with $u_n^{(j)} \in \mathbb{K}$. Then the vector series $\sum u_n$ converges if and only if each scalar coordinate series $\sum u_n^{(j)}$ converges (for $j = 1, \dots, p$), and in case of convergence $$ \sum_{n=0}^{+\infty} u_n \;=\; \sum_{j=1}^{p} \left(\sum_{n=0}^{+\infty} u_n^{(j)}\right) e_j. $$

Proof

Apply the criterion to the partial-sum sequence $(S_n)$ rather than to the series. Both directions rest only on continuity of linear maps in finite dimension; no Cauchy criterion is used.

($\Rightarrow$). The coordinate forms $\varphi_j : x \mapsto x^{(j)}$ are linear, hence continuous since $E$ is finite-dimensional (recalled from Compactness, connectedness, finite dimension). If $S_n \to S$ in $E$, then $\varphi_j(S_n) \to \varphi_j(S)$ in $\mathbb{K}$ for each $j$. But $\varphi_j(S_n) = \sum_{k=0}^{n} u_k^{(j)}$ is the partial sum of the $j$-th coordinate series, so each $\sum u_n^{(j)}$ converges with sum $\varphi_j(S) = S^{(j)}$.
($\Leftarrow$). Conversely, suppose each $\sum u_n^{(j)}$ converges, with sum $\sigma_j \in \mathbb{K}$. Then in $\mathbb{K}^p$ the vector $(S_n^{(1)}, \dots, S_n^{(p)})$ of coordinate partial sums converges to $(\sigma_1, \dots, \sigma_p)$ (componentwise convergence for a product norm, recalled from Normed vector spaces). The reconstruction map $\psi : (a_1, \dots, a_p) \mapsto \sum_{j=1}^{p} a_j e_j$ is linear from $\mathbb{K}^p$ to $E$, both finite-dimensional, hence continuous. Therefore $S_n = \psi(S_n^{(1)}, \dots, S_n^{(p)}) \to \psi(\sigma_1, \dots, \sigma_p) = \sum_j \sigma_j e_j$.

The two directions together yield the equivalence and the announced formula.

Example — Re/Im criterion for a complex series

View $\mathbb{C}$ as a real vector space with basis $(1, i)$. A complex sequence $u_n \in \mathbb{C}$ has coordinates $u_n^{(1)} = \operatorname{Re}(u_n)$ and $u_n^{(2)} = \operatorname{Im}(u_n)$ in this real basis. The coordinate criterion specialises to: $\sum u_n$ (with $u_n \in \mathbb{C}$) converges as a real-vector series if and only if both $\sum \operatorname{Re}(u_n)$ and $\sum \operatorname{Im}(u_n)$ converge, with $\sum u_n = \sum \operatorname{Re}(u_n) + i \sum \operatorname{Im}(u_n)$ --- exactly the MPSI scalar criterion. Note: $\operatorname{Re}$ and $\operatorname{Im}$ are not $\mathbb{C}$-linear, so the criterion is genuinely an $\mathbb{R}$-basis statement.

Example — A matrix series read coefficient by coefficient

The canonical basis of $\mathcal{M}_p(\mathbb{K})$ being the family $(E_{i,j})_{1 \le i, j \le p}$ of elementary matrices, a matrix series $\sum A_n$ in $\mathcal{M}_p(\mathbb{K})$ converges if and only if its $p^2$ entry series $\sum (A_n)_{i,j}$ converge, and the entry of order $(i,j)$ of the sum is the sum of the corresponding entry series: $$ \left(\sum_{n=0}^{+\infty} A_n\right)_{\!\!i,j} \;=\; \sum_{n=0}^{+\infty} (A_n)_{i,j}. $$ Convergence of a matrix series is just the simultaneous convergence of $p^2$ numerical series.

Method — Componentwise study of a vector / matrix / complex series

Whenever a basis of $E$ is at hand --- almost always: $\mathbb{R}^p$, $\mathbb{C}^p$, $\mathcal{M}_p(\mathbb{K})$, $\mathbb{K}_n[X]$ all carry a canonical basis --- the coordinate criterion reduces the convergence problem to $p$ numerical questions. One then uses the MPSI absolute-convergence toolbox (Riemann series, geometric, equivalents, $O$/$o$ comparison) on each coordinate independently.

Skills to practice

Studying a vector series through its coordinates

I.4 Absolute convergence

Absolute convergence is the most powerful sufficient condition for a vector series to converge: it reduces a vector problem to a single positive numerical series, the series of norms. In a finite-dimensional space, absolute convergence implies convergence --- and the proof goes through coordinates, never through a Cauchy criterion. The Cauchy criterion is out of the program scope, and the study of semi-convergent series is not an objective of the program --- two phrases from the official program reproduced verbatim at the point where they apply.

Definition — Absolute convergence

The series $\sum u_n$ of $E$ is absolutely convergent when the positive numerical series $\sum \|u_n\|$ converges. Since all norms on the finite-dimensional space $E$ are equivalent (recalled from Compactness, connectedness, finite dimension), absolute convergence does not depend on the choice of norm.

Theorem — Absolute convergence implies convergence

If $\sum u_n$ is an absolutely convergent series of the finite-dimensional normed space $E$, then $\sum u_n$ converges.

Proof

The proof goes through coordinates --- the Cauchy criterion is out of the program scope, and not needed.
Fix a basis $(e_1, \dots, e_p)$ of $E$ and write $u_n = \sum_j u_n^{(j)} e_j$. The coordinate forms $\varphi_j : x \mapsto x^{(j)}$ are linear, hence continuous (finite dimension). The continuity criterion for a linear map --- recalled from Limits and continuity in a normed space --- gives, for each $j$, a constant $C_j \ge 0$ such that $\forall x \in E,\; |\varphi_j(x)| \le C_j\,\|x\|$. Set $C = \max_{1 \le j \le p} C_j$, so $$ \forall x \in E, \;\forall j \in \{1, \dots, p\}, \quad |\varphi_j(x)| \;\le\; C \, \|x\|. $$ Applied to $x = u_n$, this gives $|u_n^{(j)}| \le C \, \|u_n\|$ for every $j$ and every $n$. By comparison with the positive convergent numerical series $C \sum \|u_n\|$, the positive numerical series $\sum |u_n^{(j)}|$ converges, so the scalar series $\sum u_n^{(j)}$ is absolutely convergent in $\mathbb{K}$ --- and MPSI's $AC \Rightarrow C$ for numerical series gives its convergence in $\mathbb{K}$. The coordinate criterion of \S 1.3 then assembles these $p$ scalar convergences into the convergence of $\sum u_n$ in $E$.

Proposition — Triangle inequality for an absolutely convergent series

For an absolutely convergent series $\sum u_n$ of $E$, $$ \left\| \sum_{n=0}^{+\infty} u_n \right\| \;\le\; \sum_{n=0}^{+\infty} \|u_n\|. $$

Proof

For every $n \in \mathbb{N}$, the finite triangle inequality gives $\|S_n\| = \|\sum_{k=0}^{n} u_k\| \le \sum_{k=0}^{n} \|u_k\|$. Pass to the limit $n \to +\infty$: the left-hand side tends to $\|S\|$ by continuity of the norm, and the right-hand side tends to $\sum_{n=0}^{+\infty} \|u_n\|$ (assumed finite). The inequality between limits is preserved.

Method — Proving convergence by absolute convergence

To establish convergence of a vector series $\sum u_n$: bound $\|u_n\|$ (with $\le$, $O$, $o$, $\sim$) by the general term of a known positive convergent series --- a Riemann series $\sum 1/n^\alpha$ with $\alpha > 1$, a geometric series with ratio of modulus $< 1$, an exponential series $\sum 1/n!$. By comparison, $\sum \|u_n\|$ converges, the series is absolutely convergent, hence convergent by the theorem above. This is the workhorse: most vector / matrix series met in practice converge absolutely, and one rarely needs anything more delicate.

Example — Matrix Neumann-type series $\sum A^n$ for $\|A\| < 1$

Equip $\mathcal{M}_p(\mathbb{K})$ with a subordinate operator norm $\|\cdot\|$ --- recalled from Limits and continuity in a normed space (program \S 4f); such a norm is submultiplicative: $\|AB\| \le \|A\| \, \|B\|$ for every $A, B$. For $A \in \mathcal{M}_p(\mathbb{K})$ with $\|A\| < 1$, induction gives $\|A^n\| \le \|A\|^n$. The series $\sum \|A\|^n$ is a convergent geometric series (ratio in $[0, 1[$), so $\sum A^n$ converges absolutely, hence converges. Multiply the partial sum on the left by $I_p - A$: $$ (I_p - A)\!\sum_{k=0}^{n} A^k \;=\; \sum_{k=0}^{n} A^k - \sum_{k=1}^{n+1} A^k \;=\; I_p - A^{n+1} \;\longrightarrow\; I_p, $$ since $\|A^{n+1}\| \le \|A\|^{n+1} \to 0$. Passing to the limit (continuity of left multiplication by $I_p - A$, which is linear hence continuous in finite dimension): $$ (I_p - A)\!\sum_{n=0}^{+\infty} A^n \;=\; I_p, \qquad \sum_{n=0}^{+\infty} A^n \;=\; (I_p - A)^{-1}. $$

Example — The matrix exponential series $\sum A^n / n!$

For every $A \in \mathcal{M}_p(\mathbb{K})$, the series $\sum A^n / n!$ converges absolutely. Indeed, in a subordinate operator norm $\|A^n / n!\| \le \|A\|^n / n!$, and the scalar series $\sum \|A\|^n / n!$ converges with sum $\mathrm{e}^{\|A\|}$ (the MPSI exponential series, not the d'Alembert rule of \S 2.1, which has not yet been established). The sum, traditionally written $\exp(A)$ or $\mathrm{e}^A$, is the matrix exponential --- it is the central object of a later chapter, Matrix exponential.

Skills to practice

Proving convergence by absolute convergence

II Complements on numerical series

II.1 d'Alembert's rule

d'Alembert's rule is a ratio test for positive-term numerical series: when the ratio of two consecutive terms has a limit, the series's behaviour is decided by whether that limit sits strictly below $1$ (convergence) or strictly above (divergence). The rule was stated as non-examinable in the MPSI numerical-series chapter; the MP program \S 5b makes it exigible, so the proof appears here in full. It is the natural tool for series whose general term is built from factorials and powers.

Theorem — d'Alembert's rule

Let $(u_n)$ be a sequence of strictly positive reals from some rank on, such that the ratio $u_{n+1}/u_n$ has a limit $\ell \in [0, +\infty]$. Then:

if $\ell < 1$, the series $\sum u_n$ converges;
if $\ell > 1$, the sequence $(u_n)$ tends to $+\infty$ and $\sum u_n$ diverges grossly;
if $\ell = 1$, the rule is inconclusive --- no conclusion can be drawn from the ratio alone.

Proof

The two non-trivial cases are $\ell < 1$ and $\ell > 1$. The $\ell = 1$ inconclusive case is completed by the two counterexamples of the « inconclusive case » Example block below ($\sum 1/n$ and $\sum 1/n^2$, both with ratio limit $1$ but opposite natures), which together prove that no general conclusion follows from $\ell = 1$.

Case $\ell < 1$. Pick $\rho$ with $\ell < \rho < 1$ --- for instance $\rho = (\ell + 1)/2$, which sits strictly between $\ell$ and $1$. Since $u_{n+1}/u_n \to \ell < \rho$, there exists a rank $N$ from which $u_{n+1}/u_n \le \rho$. For $n \ge N$, telescoping the ratios gives $$ u_n \;=\; u_N \cdot \frac{u_{N+1}}{u_N} \cdot \frac{u_{N+2}}{u_{N+1}} \cdots \frac{u_n}{u_{n-1}} \;\le\; u_N \, \rho^{n-N}. $$ By comparison with the convergent geometric series $\sum u_N \, \rho^{n-N}$ (ratio $\rho \in [0, 1[$), the positive-term series $\sum u_n$ converges.
Case $\ell > 1$. Pick $\rho$ with $1 < \rho < \ell$ --- for instance $\rho = (\ell + 1)/2$ if $\ell$ is finite, or any $\rho > 1$ if $\ell = +\infty$. From some rank $N$, $u_{n+1}/u_n \ge \rho > 1$, and the same telescoping gives $u_n \ge u_N \rho^{n-N} \to +\infty$. So $u_n \not\to 0$, and $\sum u_n$ diverges grossly.

Method — Using d'Alembert

Reach for d'Alembert when the general term combines factorials (e.g. $n!$, $(2n)!$) and powers ($x^n$, $a^n$): the ratio $u_{n+1}/u_n$ often telescopes most of the formula, leaving a tractable limit. For a sign-variable or vector series whose terms are nonzero from some rank, apply the rule to $\|u_{n+1}\|/\|u_n\|$: if this ratio tends to $\ell < 1$, the series of norms converges (absolute convergence), hence the original series converges. If $\ell = 1$, d'Alembert says nothing; one falls back on equivalents, $O$, $o$, or series-integral comparison.

Example — $\sum n! / n^n$

Set $u_n = n!/n^n$ for $n \ge 1$. The ratio is $$ \frac{u_{n+1}}{u_n} \;=\; \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} \;=\; \frac{n+1}{n+1} \cdot \frac{n^n}{(n+1)^n} \;=\; \left(\frac{n}{n+1}\right)^n \;=\; \left(1 + \frac{1}{n}\right)^{\!-n} \;\longrightarrow\; \mathrm{e}^{-1}. $$ The limit $\mathrm{e}^{-1} < 1$, so $\sum n!/n^n$ converges by d'Alembert.

Example — Absolute convergence of $\sum z^n / n!$ for every $z \in \mathbb{C}$

For $z \neq 0$, apply d'Alembert to the moduli $|z|^n / n!$: $$ \frac{|z|^{n+1}/(n+1)!}{|z|^n / n!} \;=\; \frac{|z|}{n+1} \;\longrightarrow\; 0 \;<\; 1. $$ So $\sum |z|^n / n!$ converges; the complex series $\sum z^n / n!$ is therefore absolutely convergent. (For $z = 0$ the series reduces to the single term $1$, trivially convergent.) The sum, of course, is $\mathrm{e}^z$ --- recalled from MPSI.

Example — The inconclusive case $\ell = 1$

For $u_n = 1/n$, $u_{n+1}/u_n = n/(n+1) \to 1$, and $\sum 1/n$ diverges (harmonic series). For $v_n = 1/n^2$, $v_{n+1}/v_n = n^2/(n+1)^2 \to 1$, and $\sum 1/n^2$ converges (Riemann series, $\alpha = 2 > 1$). Two series of opposite natures with ratio limit $1$: d'Alembert cannot separate them, and one must use a finer tool (equivalents, $O$, series-integral comparison).

Skills to practice

Applying d'Alembert's rule

II.2 Series-integral comparison

Series-integral comparison was already a convergence criterion in MPSI: for $f$ continuous, positive and decreasing on $[a, +\infty[$, the series $\sum f(n)$ and the integral $\int_a^{+\infty} f$ share the same nature, with the framing $\int_a^{n+1} f \le \sum_{k=a}^{n} f(k) \le f(a) + \int_a^{n} f$ for the partial sums. The MP sharpening is the use of the same idea on the tail: when the integral converges, one gets the symmetric framing for the remainder $\sum_{k=n+1}^{+\infty} f(k)$. The two framings together let one estimate partial sums of divergent series and remainders of convergent series --- the named capacity of the program \S 5b.

Theorem — Series-integral comparison

Let $a \in \mathbb{N}$ and $f : [a, +\infty[ \,\to \mathbb{R}$ continuous, positive, and decreasing. Then:

the per-term framing $\forall k \in \mathbb{N},\, k \ge a, \quad f(k+1) \le \int_k^{k+1} f \le f(k)$ holds;
the series $\sum_{n \ge a} f(n)$ and the improper integral $\int_a^{+\infty} f$ are of the same nature;
(partial sums, divergent case --- recalled from MPSI.) For all $n \ge a$, $$ \int_a^{n+1} f \;\le\; \sum_{k=a}^{n} f(k) \;\le\; f(a) + \int_a^{n} f; $$
(remainders, convergent case --- the new MP sharpening.) When the integral converges, for all $n \ge a$, $$ \int_{n+1}^{+\infty} f \;\le\; \sum_{k=n+1}^{+\infty} f(k) \;\le\; \int_{n}^{+\infty} f. $$

Proof

The proof uses a single index convention throughout: the MPSI form $f(k+1) \le \int_k^{k+1} f \le f(k)$ for $k \ge a$.

Per-term framing. Decreasingness of $f$ on $[k, k+1]$ gives $f(k+1) \le f(t) \le f(k)$ for $t \in [k, k+1]$. Integrating on $[k, k+1]$ (length $1$) yields the per-term framing $f(k+1) \le \int_k^{k+1} f \le f(k)$ for every $k \ge a$.
Same nature. Sum the per-term lower bound $f(k+1) \le \int_k^{k+1} f$ for $k = a, \dots, n-1$: $\sum_{k=a+1}^{n} f(k) \le \int_a^{n} f$. So if $\int_a^{+\infty} f$ converges, the positive-term series's partial sums are bounded, hence the series converges. Conversely, summing the upper bound $\int_k^{k+1} f \le f(k)$ for $k = a, \dots, n-1$ gives $\int_a^{n} f \le \sum_{k=a}^{n-1} f(k)$, so convergence of the series forces boundedness of $(\int_a^{x} f)_{x \ge a}$ (increasing in $x$), hence convergence of $\int_a^{+\infty} f$.
Partial-sum framing. Lower bound. Sum the upper per-term inequality $\int_k^{k+1} f \le f(k)$ for $k = a, \dots, n$: $$ \int_a^{n+1} f \;=\; \sum_{k=a}^{n} \int_k^{k+1} f \;\le\; \sum_{k=a}^{n} f(k). $$ Upper bound. Sum the lower per-term inequality $f(k+1) \le \int_k^{k+1} f$ for $k = a, \dots, n-1$: $$ \sum_{k=a}^{n-1} f(k+1) \;\le\; \int_a^{n} f, \quad\text{i.e.}\quad \sum_{j=a+1}^{n} f(j) \;\le\; \int_a^{n} f, $$ hence $\sum_{j=a}^{n} f(j) - f(a) \le \int_a^{n} f$, so $\sum_{k=a}^{n} f(k) \le f(a) + \int_a^{n} f$.
Remainder framing. Assume $\int_a^{+\infty} f$ converges. Sum the per-term framing for $k = n+1, \dots, N$ ($n \ge a$): $$ \sum_{k=n+1}^{N} f(k+1) \;\le\; \sum_{k=n+1}^{N} \int_k^{k+1} f \;=\; \int_{n+1}^{N+1} f \;\le\; \sum_{k=n+1}^{N} f(k). $$ The leftmost sum re-indexes as $\sum_{j=n+2}^{N+1} f(j)$; let $N \to +\infty$, both sides tend to finite limits (since the series and the integral converge), and one obtains $R_{n+1} \le \int_{n+1}^{+\infty} f \le R_n$. Replacing $n$ by $n-1$ in the left inequality (or simply re-running the same summation starting at $k = n$) gives the announced framing $\int_{n+1}^{+\infty} f \le R_n \le \int_n^{+\infty} f$.

Method — Estimating partial sums and remainders by series-integral comparison

Faced with a positive decreasing $f(n)$:

If $\sum f(n)$ diverges, frame the partial sum $\sum_{k=a}^{n} f(k)$ between $\int_a^{n+1} f$ and $f(a) + \int_a^{n} f$. The two integrals usually have a closed primitive; the framing gives an equivalent of the partial sum (most often, $\sim \int_a^{n} f$).
If $\sum f(n)$ converges, frame the remainder $R_n = \sum_{k > n} f(k)$ between $\int_{n+1}^{+\infty} f$ and $\int_n^{+\infty} f$. The two integrals have a closed primitive; the framing gives an equivalent of the remainder.

In both directions the rough recipe is the same: an integral is easier than a sum, so trade the discrete object for its continuous twin and read off the asymptotic.

Example — Partial sum: harmonic series and Euler's constant

Apply the partial-sum framing to $f(x) = 1/x$ on $[1, +\infty[$: $$ \int_1^{n+1} \frac{\mathrm{d}t}{t} \;\le\; \sum_{k=1}^{n} \frac{1}{k} \;\le\; 1 + \int_1^{n} \frac{\mathrm{d}t}{t}, \quad\text{i.e.}\quad \ln(n+1) \;\le\; H_n \;\le\; 1 + \ln n, $$ where $H_n = \sum_{k=1}^n 1/k$. Dividing by $\ln n$ and taking $n \to +\infty$ gives $H_n / \ln n \to 1$, so $$ H_n \;\underset{n \to +\infty}{\sim}\; \ln n. $$ The series-integral framing also gives the existence of Euler's constant $\gamma$: setting $u_n = H_n - \ln n$, one checks that $u_{n+1} - u_n = 1/(n+1) - \ln(1 + 1/n) = O(1/n^2)$, so $\sum (u_{n+1} - u_n)$ converges absolutely; by the sequence-series link of \S 1.2, $(u_n)$ converges to some limit $\gamma$, and $H_n = \ln n + \gamma + o(1)$.

Example — Remainder: convergent Riemann series with $\alpha = 2$

Apply the remainder framing to $f(x) = 1/x^2$ on $[1, +\infty[$: $$ \int_{n+1}^{+\infty} \frac{\mathrm{d}t}{t^2} \;\le\; \sum_{k=n+1}^{+\infty} \frac{1}{k^2} \;\le\; \int_{n}^{+\infty} \frac{\mathrm{d}t}{t^2}, \quad\text{i.e.}\quad \frac{1}{n+1} \;\le\; R_n \;\le\; \frac{1}{n}. $$ Dividing through by $1/n$ gives $R_n \cdot n \to 1$, so $$ \sum_{k=n+1}^{+\infty} \frac{1}{k^2} \;\underset{n \to +\infty}{\sim}\; \frac{1}{n}. $$

Skills to practice

Estimating sums and remainders by series-integral comparison

II.3 Summation of comparison relations

The three comparison relations of MPSI --- domination ($O$), negligibility ($o$) and equivalence ($\sim$) --- transfer from the general terms of two series to the remainders (when the reference series converges) or to the partial sums (when it diverges). The constraint is mild but real: the reference series must be of constant sign (positive in this chapter); the compared term may be sign-variable or even $\mathbb{K}$-valued for the $O$ and $o$ relations, since the proof rests only on the triangle inequality $|\sum u_k| \le \sum |u_k|$, valid on $\mathbb{C}$ as on $\mathbb{R}$.

Theorem — Summation of comparison relations -- convergent case

Let $\sum v_n$ be a positive-term convergent numerical series, with remainder $R_n' = \sum_{k > n} v_k$. Let $(u_n)$ be a $\mathbb{K}$-valued sequence ($\mathbb{R}$ or $\mathbb{C}$). Then, in the program's order:

if $u_n = O(v_n)$, then $\sum u_n$ converges and its remainder $R_n$ satisfies $R_n = O(R_n')$;
if $u_n = o(v_n)$, then $\sum u_n$ converges and $R_n = o(R_n')$;
if $u_n \sim v_n$ --- in this $\sim$ case $u_n$ is assumed real, and the relation forces $u_n > 0$ from some rank on, since $v_n > 0$ --- then $\sum u_n$ converges and $R_n \sim R_n'$.

Convergence of $\sum u_n$ is a conclusion, not a hypothesis.

Proof

Case $O$. There exists $M \ge 0$ and a rank $n_0$ with $|u_n| \le M v_n$ for $n \ge n_0$. By comparison with the convergent positive-term series $\sum v_n$, the series $\sum |u_n|$ converges; hence $\sum u_n$ converges absolutely (MPSI), and in particular it converges. For the remainder, the triangle inequality and the bound $|u_n| \le M v_n$ give, for $p \ge n_0$, $$ |R_p| \;=\; \left|\sum_{n > p} u_n\right| \;\le\; \sum_{n > p} |u_n| \;\le\; M \sum_{n > p} v_n \;=\; M R_p', $$ so $R_n = O(R_n')$.
Case $o$. Fix $\varepsilon > 0$; from some rank $n_0$, $|u_n| \le \varepsilon v_n$. The same triangle-inequality argument bounds $|R_p| \le \varepsilon R_p'$ for $p \ge n_0$, i.e. $R_n = o(R_n')$. Convergence of $\sum u_n$ follows as in the $O$ case (with $M = 1$, say).
Case $\sim$. Write $u_n = v_n + (u_n - v_n)$ with $u_n - v_n = o(v_n)$ by hypothesis. Both $\sum v_n$ and $\sum (u_n - v_n)$ converge (the latter by the $o$ case just proved), so $\sum u_n$ converges by linearity. Their remainders satisfy $R_n = R_n' + (R_n - R_n')$ with $R_n - R_n' = o(R_n')$ (from the $o$ case applied to $u_n - v_n$). Dividing by $R_n'$ gives $R_n / R_n' \to 1$, i.e. $R_n \sim R_n'$.

Theorem — Summation of comparison relations -- divergent case

Let $\sum v_n$ be a positive-term divergent numerical series, with partial sum $S_n' = \sum_{k=0}^{n} v_k$. Let $(u_n)$ be a $\mathbb{K}$-valued sequence. Then, in the program's order:

if $u_n = O(v_n)$, the partial sum $S_n = \sum_{k=0}^{n} u_k$ satisfies $S_n = O(S_n')$;
if $u_n = o(v_n)$, then $S_n = o(S_n')$;
if $u_n \sim v_n$ --- in this $\sim$ case $u_n$ is assumed real, and the relation forces $u_n > 0$ from some rank on, since $v_n > 0$ --- then $\sum u_n$ also diverges and $S_n \sim S_n'$.

Proof

Case $o$. Fix $\varepsilon > 0$; from rank $n_0$, $|u_n| \le \varepsilon v_n$. For $n \ge n_0$ split the partial sum at $n_0$: $$ |S_n| \;\le\; \underbrace{\sum_{k=0}^{n_0} |u_k|}_{\text{constant } C} \;+\; \sum_{k=n_0+1}^{n} |u_k| \;\le\; C + \varepsilon \sum_{k=n_0+1}^{n} v_k \;\le\; C + \varepsilon S_n'. $$ Since $S_n' \to +\infty$, there exists $n_1$ with $C \le \varepsilon S_n'$ for $n \ge n_1$, so $|S_n| \le 2\varepsilon S_n'$ for $n \ge \max(n_0, n_1)$. As $\varepsilon$ was arbitrary, $S_n = o(S_n')$.
Case $O$. Same telescoping with $\varepsilon$ replaced by a fixed $M$: $|S_n| \le C + M S_n'$, and $C / S_n' \to 0$, so $S_n = O(S_n')$.
Case $\sim$. Write $u_n = v_n + (u_n - v_n)$ with $u_n - v_n = o(v_n)$. The case $o$ yields $\sum_{k \le n}(u_k - v_k) = o(S_n')$, so $S_n - S_n' = o(S_n')$, equivalently $S_n \sim S_n'$. The hypothesis $u_n \sim v_n$ with $v_n > 0$ also gives that $u_n$ is positive from some rank, and $S_n \to +\infty$ (divergence of $\sum u_n$).

Method — Summing a comparison relation

The recipe in both cases is the same: pick a positive-sign reference series $\sum v_n$ whose nature you know --- converging Riemann $\sum 1/n^\alpha$ ($\alpha > 1$), diverging Riemann ($\alpha \le 1$), geometric, etc. --- compare the general term $u_n$ to $v_n$ via $\sim$, $o$ or $O$, then transfer the comparison to remainders if $\sum v_n$ converges, to partial sums if it diverges. The framework gives equivalents of remainders and partial sums almost for free, once the right reference is identified.

Example — Remainder equivalent by $\sim$

Take $u_n = 1/n^2$ and $v_n = 1/(n(n+1))$. One has $u_n \sim v_n$ (both $\sim 1/n^2$), both positive, both giving convergent series. The remainder of $\sum v_n$ is computable directly by telescoping: $v_n = 1/n - 1/(n+1)$, so $$ \sum_{k=n+1}^{+\infty} \frac{1}{k(k+1)} \;=\; \frac{1}{n+1}. $$ By summation of $\sim$ on convergent remainders, $$ R_n \;=\; \sum_{k=n+1}^{+\infty} \frac{1}{k^2} \;\sim\; \frac{1}{n+1} \;\sim\; \frac{1}{n}, $$ recovering the estimate of \S 2.2 without explicit series-integral comparison.

Example — Partial sum equivalent: $\sum 1/\sqrt{k}$

The Riemann series $\sum 1/k^{1/2}$ diverges ($\alpha = 1/2 \le 1$). The reference $v_n = 1/\sqrt{n}$ has partial sum $S_n' = \sum_{k=1}^{n} 1/\sqrt{k}$ (starting at $k = 1$ to avoid $1/\sqrt{0}$; the starting index of a series does not affect its asymptotic behaviour, as the missing initial terms form a finite constant), which one can frame by series-integral comparison: $S_n' \sim 2\sqrt{n}$. Now take $u_n = 1/\sqrt{n} \cdot (1 + 1/n)$; then $u_n \sim 1/\sqrt{n} = v_n$, positive. By summation of $\sim$ in the divergent case, $$ \sum_{k=1}^{n} \frac{1}{\sqrt{k}}\left(1 + \frac{1}{k}\right) \;\underset{n \to +\infty}{\sim}\; 2\sqrt{n}. $$

Skills to practice

Summing comparison relations

II.4 Cesàro's theorem

Cesàro's theorem is one specific consequence of the divergent-case summation, important enough to deserve its own name: if a sequence converges (in $\mathbb{K}$), then its arithmetic means converge to the same limit. For real sequences the result persists with $\ell = \pm \infty$, by a direct argument that does not go through the summation theorem.

Theorem — Cesàro

Let $(u_n)$ be a $\mathbb{K}$-valued sequence ($\mathbb{R}$ or $\mathbb{C}$) with $u_n \to \ell \in \mathbb{K}$. Then $$ \frac{1}{n+1} \sum_{k=0}^{n} u_k \;\xrightarrow[n \to +\infty]{}\; \ell. $$ For a real sequence, the result persists for $\ell = \pm\infty$ (with the convention that the running average then tends to $\pm \infty$).

Proof

Finite $\ell \in \mathbb{K}$. Set $w_n = u_n - \ell$, so $w_n \to 0$, i.e. $w_n = o(1)$ against the positive divergent reference $v_n = 1$. The divergent-case $o$-summation --- valid for a $\mathbb{K}$-valued $w_n$, as the proof of \S 2.3 uses only the triangle inequality --- gives $\sum_{k=0}^{n} w_k = o(\sum_{k=0}^{n} 1) = o(n+1)$. Therefore $$ \frac{1}{n+1} \sum_{k=0}^{n} u_k \;=\; \frac{1}{n+1} \sum_{k=0}^{n} (\ell + w_k) \;=\; \ell + \frac{1}{n+1} \sum_{k=0}^{n} w_k \;=\; \ell + o(1) \;\longrightarrow\; \ell. $$
Real $\ell = +\infty$. Given $A \in \mathbb{R}$, there exists $n_0$ with $u_n \ge A$ for $n \ge n_0$. For $n \ge n_0$, $$ \frac{1}{n+1} \sum_{k=0}^{n} u_k \;=\; \frac{1}{n+1}\sum_{k=0}^{n_0 - 1} u_k \;+\; \frac{1}{n+1}\sum_{k=n_0}^{n} u_k \;\ge\; \frac{C}{n+1} \;+\; \frac{n - n_0 + 1}{n+1} A, $$ where $C = \sum_{k=0}^{n_0-1} u_k$ is a constant. As $n \to +\infty$, $C/(n+1) \to 0$ and $(n - n_0 + 1)/(n+1) \to 1$, so the right-hand side tends to $A$; in particular $\liminf_{n \to +\infty} \frac{1}{n+1} \sum_{k=0}^{n} u_k \ge A$. Since $A \in \mathbb{R}$ was arbitrary, this $\liminf$ is $+\infty$, hence the running average tends to $+\infty$. The case $\ell = -\infty$ is identical with reversed inequalities.

Method — Recognising a Cesàro situation

Whenever an average $\frac{1}{n+1}\sum_{k=0}^{n} u_k$ appears with a sequence $(u_n)$ known to have a limit, the average has the same limit --- no further work needed. Be aware that the converse of Cesàro is false: a sequence whose Cesàro mean converges need not converge itself (counterexample $u_n = (-1)^n$, average $\to 0$ while $(u_n)$ has no limit).

Example — Application of Cesàro to a sequence of squares

Let $(u_n)$ be real with $u_n \to \ell \in \mathbb{R}$. Then $u_n^2 \to \ell^2$ by continuity of squaring, so by Cesàro $$ \frac{1}{n+1} \sum_{k=0}^{n} u_k^2 \;\longrightarrow\; \ell^2. $$ Concrete case: define $u_0 = 1$ and $u_n = 1 + 1/n$ for $n \ge 1$ (any value of $u_0$ works; a finite initial block does not affect the Cesàro limit). Then $u_n \to 1 = \ell$, $u_n^2 = 1 + 2/n + 1/n^2 \to 1$, and the running average $\frac{1}{n+1}\sum_{k=0}^{n} u_k^2$ tends to $1$.

Skills to practice

Applying Cesàro's theorem