CommeUnJeu · L1 PCSI

Numerical series

⌚ ~91 min ▢ 11 blocks ✓ 33 exercises ➣ Prerequisites : Real sequences

A numerical series is a formal sum $\sum u_n$ associated with a real or complex sequence $(u_n)$. Studying « the series $\sum u_n$ » means studying the limit of its partial sums $S_n = u_0 + u_1 + \cdots + u_n$ as $n \to +\infty$. When that limit exists, we call it the sum of the series and write $\sum_{n=0}^{+\infty} u_n$. The whole chapter is about deciding when $\sum u_n$ converges, and --- when it does --- about extracting structural information from the partial sums (sign, equivalents, asymptotic behaviour of the remainder).
The chapter has three sections. The Convergence and divergence section sets up the definitions (partial sum, convergence, sum, remainder), proves linearity and the necessary condition $u_n \to 0$, states the sequence-series link (a sequence converges if and only if its associated telescoping series converges --- the conceptual key of the chapter), and closes with the two universal reference series : geometric series and the exponential identity $e^z = \sum z^n / n!$ (stated without proof in Geometric series and the complex exponential, full proof in the Absolute convergence section). The Series of positive terms section treats series with positive terms : the partial sums are monotone, so convergence reduces to bounding $(S_n)$ from above. The whole positive-series toolbox (comparison $0 \le u_n \le v_n$, equivalents, comparison to an integral, Riemann reference series $\sum 1/n^\alpha$) lives in this monotone framework. The Absolute convergence section introduces absolute convergence and the ``abs cv $\Rightarrow$ cv'' Theorem, which converts every sign-variable problem into a positive-term problem, reducing it to the positive-series toolbox.
Three reflexes the reader should leave with :

the sequence-series link : every recurrent-sequence question can be re-phrased as a series-convergence question by passing to the telescoping series $\sum (u_{n+1} - u_n)$ ;
the Riemann reflex : when faced with a positive series $\sum u_n$ of unclear nature, look for an equivalent of the form $C / n^\alpha$ and conclude by Riemann ($\alpha > 1$ converges, $\alpha \le 1$ diverges) ;
absolute convergence first : a sign-variable series should be tested for absolute convergence, which reduces it to the positive-series toolbox.

I Convergence and divergence of a series

I.1 Series$\virgule$ partial sums$\virgule$ sum$\virgule$ remainder

A series is the formal sum-object associated with a sequence : given $(u_n)_{n \ge n_0}$ of real or complex numbers, we form the partial sums $S_n = u_{n_0} + u_{n_0 + 1} + \cdots + u_n$ and ask whether $(S_n)$ has a limit. The series is a piece of bookkeeping --- it bundles, in one notation, the question of « does the sum extend to infinity? » and the answer (when affirmative). Notations vary : $\sum u_n$, $\sum_n u_n$, $\sum_{n \ge n_0} u_n$ all denote the same object ; the starting index $n_0$ is often $0$ or $1$ and matters only for the value of the sum, not for its nature.

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

Let $(u_n)_{n \ge n_0}$ be a sequence of real or complex numbers (with $n_0 \in \mathbb N$). The series of general term $u_n$, denoted $\sum u_n$, is the formal object associated with the sequence of partial sums $$ S_n \ = \ \sum_{k = n_0}^{n} u_k \qquad (n \ge n_0). $$ The series $\sum u_n$ is said to converge if the sequence $(S_n)$ of partial sums has a finite limit ; in that case, the sum of the series is $$ \sum_{n = n_0}^{+\infty} u_n \ = \ \lim_{n \to +\infty} S_n. $$ Otherwise, $\sum u_n$ is said to diverge.

Definition — Nature of a series$\virgule$ remainder

The nature of $\sum u_n$ is the qualifier « convergent » or « divergent ». When $\sum u_n$ converges with sum $S$, the remainder of order $n$ is the difference between $S$ and the $n$-th partial sum : $$ R_n \ = \ S - S_n \ = \ \sum_{k = n + 1}^{+\infty} u_k. $$ By construction $R_n \to 0$ as $n \to +\infty$.

Example — Two basic series

The series $\sum_{n \ge 0} 1$ (constant term equal to $1$) has partial sums $S_n = n + 1 \to +\infty$, hence diverges. The series $\sum_{n=0}^{+\infty} 1/2^n$ has $S_n = 2 - 1/2^n \to 2$ (sum of a geometric progression), hence converges with sum $2$.

Skills to practice

Computing partial sums and remainders

I.2 Linearity of the sum of convergent series

Once convergence is defined as the limit of partial sums, linearity transfers verbatim from the limit-of-a-sequence framework : the sum of two convergent series is the convergent series of the term-by-term sum, and multiplying every term by a scalar multiplies the sum by that scalar. This is the algebraic skeleton of the chapter ; every Proposition that follows builds on it.

Proposition — Linearity

Let $\sum u_n$ and $\sum v_n$ be two convergent series with sums $U$ and $V$, and let $\lambda, \mu \in \mathbb K$ (where $\mathbb K = \mathbb R$ or $\mathbb C$). Then the series $\sum (\lambda u_n + \mu v_n)$ converges, with sum $$ \sum_{n = n_0}^{+\infty} (\lambda u_n + \mu v_n) \ = \ \lambda U + \mu V. $$

Proof

By linearity of finite sums, the partial sums of $\sum (\lambda u_n + \mu v_n)$ are $$ \sum_{k = n_0}^{n} (\lambda u_k + \mu v_k) \ = \ \lambda \sum_{k = n_0}^{n} u_k + \mu \sum_{k = n_0}^{n} v_k \ = \ \lambda S_n^u + \mu S_n^v. $$ By hypothesis $S_n^u \to U$ and $S_n^v \to V$. By linearity of the limit of sequences, $\lambda S_n^u + \mu S_n^v \to \lambda U + \mu V$. Hence $\sum (\lambda u_n + \mu v_n)$ converges with sum $\lambda U + \mu V$.

Example — Linearity in action

Both $\sum_{n=0}^{+\infty} 1/2^n = 2$ and $\sum_{n=0}^{+\infty} 1/3^n = 3/2$ converge (geometric series, see the Geometric series and the complex exponential subsection). By linearity, $$ \sum_{n=0}^{+\infty} \left(2 \cdot \frac{1}{2^n} - \frac{1}{3^n}\right) \ = \ 2 \cdot 2 - \frac{3}{2} \ = \ \frac{5}{2}. $$

Skills to practice

Applying linearity

I.3 Necessary condition$\virgule$ gross divergence

A series that converges has a general term tending to $0$ : this is the elementary necessary condition, the very first test one applies to a candidate series. The converse is famously false : the harmonic series $\sum 1/n$ has $u_n \to 0$ yet diverges (proven in the Series-integral comparison and Riemann series subsection via the Riemann reference). When the necessary condition fails, we say the series diverges grossly : a one-line refutation by inspection of the general term.

Proposition — Necessary condition for convergence

If the series $\sum u_n$ converges, then $u_n \to 0$ as $n \to +\infty$.

Proof

Let $S$ denote the sum of $\sum u_n$. For $n \ge n_0 + 1$, $$ u_n \ = \ S_n - S_{n-1}. $$ Both $S_n$ and $S_{n-1}$ tend to $S$ as $n \to +\infty$ (the second is a shift of the first), hence $u_n = S_n - S_{n-1} \to S - S = 0$.

Definition — Gross divergence

A series $\sum u_n$ is said to diverge grossly (in French : diverger grossièrement) when $u_n \not\to 0$. The contrapositive of the previous Proposition says : if $\sum u_n$ diverges grossly, then $\sum u_n$ diverges.

Method — The gross-divergence reflex

Before any other test, ask : does $u_n \to 0$ ?

If $u_n \not\to 0$ (the limit is nonzero, infinite, or does not exist), conclude immediately : the series diverges grossly, hence diverges.
If $u_n \to 0$, the necessary condition is satisfied but the convergence is not guaranteed --- proceed to a finer test : the positive-series toolbox (next section) when the terms are positive, or absolute convergence (last section) when the terms have variable sign.

Example — Gross divergence in three flavors

(i) $\sum_{n \ge 0} (-1)^n$ : the general term oscillates between $\pm 1$, no limit, so $u_n \not\to 0$ --- diverges grossly.
(ii) $\sum_{n \ge 1} n / (n + 1)$ : $u_n \to 1 \ne 0$ --- diverges grossly.
(iii) $\sum_{n \ge 1} \cos(1/n)$ : $\cos(1/n) \to \cos 0 = 1 \ne 0$ --- diverges grossly.
Counter-illustration (necessary $\ne$ sufficient). The harmonic series $\sum_{n \ge 1} 1/n$ has $u_n = 1/n \to 0$, so it passes the necessary condition ; yet it diverges (proven in the Series-integral comparison and Riemann series subsection by comparison to an integral).

Skills to practice

Recognising gross divergence

I.4 Sequence-series link

The single most useful structural fact about series is the sequence-series link : a sequence $(u_n)$ converges if and only if its associated telescoping series $\sum (u_{n+1} - u_n)$ converges. This is the bidirectional bridge between the « limit of a sequence » framework of chapter SuitesReelles and the « convergence of a series » framework of the present chapter. Almost every recurrent-sequence problem can be rephrased as a series-convergence problem (and vice versa) by traversing this bridge.

Proposition — Sequence-series link

For every real or complex sequence $(u_n)_{n \ge n_0}$, the sequence $(u_n)$ converges if and only if the telescoping series $\sum (u_{n+1} - u_n)$ converges. When this is the case, $$ \sum_{n = n_0}^{+\infty} (u_{n+1} - u_n) \ = \ \lim_{n \to +\infty} u_n - u_{n_0}. $$

Proof

By telescoping, $$ \begin{aligned} \sum_{k = n_0}^{n} (u_{k+1} - u_k) \ &= \ (u_{n_0 + 1} - u_{n_0}) + (u_{n_0 + 2} - u_{n_0 + 1}) + \cdots + (u_{n + 1} - u_n) && \text{(extended form)} \\ &= \ u_{n+1} - u_{n_0} && \text{(internal cancellations).} \end{aligned} $$ The partial sums of the telescoping series are $u_{n+1} - u_{n_0}$. They have a finite limit when $n \to +\infty$ if and only if $u_{n+1}$ has a finite limit, that is, $(u_n)$ converges. In that case, the limit of the partial sums is $\lim u_n - u_{n_0}$, which is the announced value of the sum.

Method — Studying a sequence via its telescoping series

To study the convergence of a sequence $(u_n)$ that is hard to handle directly :

Form the telescoping series $\sum (u_{n+1} - u_n)$ of its consecutive differences.
Estimate $u_{n+1} - u_n$ (by DL, by direct algebra, by an equivalent), often producing a term of the form $O(1/n^\alpha)$ or geometric decay.
Apply the positive-series and absolute-convergence toolboxes to the telescoping series. If it converges, so does $(u_n)$ ; the sum gives the limit.

Example — A pure telescoping convergence

Let $u_n = 1 - 1/2^n$ for $n \ge 0$. The telescoping series is $$ u_{n+1} - u_n \ = \ \left(1 - \frac{1}{2^{n+1}}\right) - \left(1 - \frac{1}{2^n}\right) \ = \ \frac{1}{2^n} - \frac{1}{2^{n+1}} \ = \ \frac{1}{2^{n+1}}. $$ The series $\sum 1/2^{n+1}$ is geometric of ratio $1/2$, hence converges (see the Geometric series and the complex exponential subsection). By the sequence-series link, $(u_n)$ converges. Its limit, given by $\lim u_n - u_0 = \sum_{n=0}^{+\infty} 1/2^{n+1} = 1$, plus $u_0 = 0$, gives $\lim u_n = 1$ --- consistent with the direct calculation $1 - 1/2^n \to 1$.

Example — Computing $\sum 1/(k(k+1))$ by partial fractions

For $k \ge 1$, partial-fraction decomposition gives $$ \frac{1}{k(k+1)} \ = \ \frac{1}{k} - \frac{1}{k+1}. $$ The partial sums of $\sum_{k=1}^n 1/(k(k+1))$ are, by telescoping, $$ \sum_{k = 1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) \ = \ 1 - \frac{1}{n+1} \ \underset{n \to +\infty}{\longrightarrow} \ 1. $$ Hence $\sum_{k=1}^{+\infty} \frac{1}{k(k+1)} = 1$.

Example — Convergence of $H_n - \ln n$ (Euler constant $\gamma$)

Define $H_n = \sum_{k=1}^n 1/k$ and set $u_n = H_n - \ln n$ for $n \ge 1$. Compute the telescoping difference : $$ u_{n+1} - u_n \ = \ \frac{1}{n+1} - \bigl( \ln(n+1) - \ln n \bigr) \ = \ \frac{1}{n+1} - \ln\!\left( 1 + \frac{1}{n} \right). $$ Using the expansion $\ln(1 + x) = x - x^2/2 + O(x^3)$ at $x = 1/n$ and $1/(n+1) = 1/n - 1/n^2 + O(1/n^3)$ : $$ u_{n+1} - u_n \ = \ \left( \frac{1}{n} - \frac{1}{n^2} \right) - \left( \frac{1}{n} - \frac{1}{2 n^2} \right) + O\!\left(\frac{1}{n^3}\right) \ = \ -\frac{1}{2 n^2} + O\!\left(\frac{1}{n^3}\right). $$ Hence $|u_{n+1} - u_n| = O(1/n^2)$. The Riemann series $\sum 1/n^2$ converges, so, by comparison by $O$, $\sum |u_{n+1} - u_n|$ converges, so $\sum (u_{n+1} - u_n)$ converges absolutely, hence converges. By the sequence-series link, $(u_n)$ converges. Its limit is the Euler-Mascheroni constant $\gamma \approx 0{,}5772156\ldots$ already encountered in chapter AnalyseAsymptotique.

Skills to practice

Studying a sequence via its telescoping series

I.5 Geometric series and the complex exponential

We close this section with the two universal reference series : geometric series $\sum z^n$ (the simplest example of a non-trivial convergence question, with a fully explicit nature criterion) and the exponential identity $e^z = \sum z^n / n!$ on $\mathbb C$ (the simplest example of a convergent series of a slightly more elaborate nature, and the gateway to the complex exponential's properties). Both will be used as comparison references throughout the rest of the chapter.

Theorem — Geometric series

Let $z \in \mathbb C$. The series $\sum_{n \ge 0} z^n$ converges if and only if $|z| < 1$. In that case, $$ \sum_{n = 0}^{+\infty} z^n \ = \ \frac{1}{1 - z}. $$

Proof

Case $z = 1$. The partial sums are $S_n = n + 1 \to +\infty$, hence the series diverges.
Case $z \ne 1$. The geometric-sum formula (from chapter CalculAlgebrique) gives, for every $n \ge 0$, $$ S_n \ = \ \sum_{k = 0}^{n} z^k \ = \ \frac{1 - z^{n+1}}{1 - z}. $$ Hence $S_n$ has a finite limit if and only if $z^{n+1}$ has a finite limit. Now $|z^{n+1}| = |z|^{n+1}$ :
- if $|z| < 1$, $|z|^{n+1} \to 0$, so $z^{n+1} \to 0$ ; the limit of $S_n$ is $1/(1-z)$ --- convergence ;
- if $|z| > 1$, $|z|^{n+1} \to +\infty$, so $|S_n| \to +\infty$ --- divergence ;
- if $|z| = 1$, then $|z^n| = 1$ for all $n$, so $z^n \not\to 0$ ; by gross divergence (Theorem of the Necessary condition, gross divergence subsection), the series diverges.

Theorem — Complex exponential as a series

For every $z \in \mathbb C$, the positive series $\sum_{n \ge 0} |z|^n / n!$ converges. Consequently, the complex series $\sum_{n \ge 0} z^n / n!$ converges, and its sum is the complex exponential : $$ e^z \ = \ \sum_{n = 0}^{+\infty} \frac{z^n}{n!}. $$ (In the terminology of the Absolute convergence section, $\sum z^n / n!$ is absolutely convergent.)

Proof

Admitted at this stage. Full proof in the Series of positive terms section and the Definition and main theorem subsection of Absolute convergence. The convergence of the positive series $\sum |z|^n / n!$ uses the term-by-term comparison Theorem (subsection Term-by-term comparison) applied to a geometric majoration (sketch: for $n \ge 2 |z|$, the ratio $|z|/(n+1) \le 1/2$, so $a_n = |z|^n / n!$ is bounded above, from rank $N = \lceil 2 |z| \rceil$, by $a_N / 2^{n - N}$ --- a convergent geometric reference). The convergence of the complex series $\sum z^n / n!$ then follows from the « abs cv $\Rightarrow$ cv » Theorem of the same Absolute convergence subsection once the positive-series toolbox is in place. The identification of the sum with the complex exponential $e^z$ defined in chapter FonctionsUsuelles (via $e^z = e^x (\cos y + i \sin y)$ for $z = x + i y$) follows from the differential-equation characterisation of $e^z$ or from the Taylor series of $e^x$ for real $x$.

Method — Recognising a geometric series

To recognise that a series is geometric :

Try to write the general term in the form $u_n = a \, r^n$ (or $a \, r^{n - n_0}$ if the series starts at index $n_0$).
If $|r| \ge 1$, the series diverges (grossly if $|r| > 1$, or by $z^n \not\to 0$ if $|r| = 1$, $r \ne 1$).
If $|r| < 1$, the series converges with sum $a \, r^{n_0} / (1 - r)$.

Example — Three geometric sums

(i) $\sum_{n=0}^{+\infty} (1/3)^n = 1/(1 - 1/3) = 3/2$.
(ii) $\sum_{n=2}^{+\infty} (1/2)^n = (1/2)^2 / (1 - 1/2) = 1/2$ (start index $n_0 = 2$).
(iii) $\sum_{n=0}^{+\infty} (-1/2)^n = 1 / (1 - (-1/2)) = 1 / (3/2) = 2/3$.

Example — The exponential at $z \equal i\pi$

The exponential identity gives, at $z = i\pi$, $$ e^{i \pi} \ = \ \sum_{n = 0}^{+\infty} \frac{(i \pi)^n}{n!} \ = \ -1 $$ (Euler's celebrated identity, equivalent to $\cos \pi = -1$ and $\sin \pi = 0$). The partial sums of $\sum (i\pi)^n / n!$ converge absolutely to $-1$ --- a strong test of the convergence framework.

Skills to practice

Recognising geometric series

II Series of positive terms

Finite initial terms do not change the nature

Adding, deleting, or changing finitely many terms of a series does not change its nature. When the series converges, it changes the value of the sum, but never the qualifier « convergent » vs.\ « divergent ». This justifies all the « for $n$ large enough » statements that follow : we always have the option of truncating the series to its tail from a suitable rank $N$.

II.1 Convergence criterion for positive-term series

When $u_n \ge 0$ for $n$ large, the partial sums $(S_n)$ form a non-decreasing sequence. A non-decreasing sequence converges if and only if it is bounded above. This single observation yields a one-step necessary-and-sufficient condition for the convergence of a positive-term series and is the foundation on which every other positive-series test rests.

Theorem — Convergence criterion for positive-term series

Let $\sum_{n \ge n_0} u_n$ be a series with $u_n \ge 0$ for every $n \ge n_0$. Then $\sum u_n$ converges if and only if the sequence of partial sums $(S_n)$ is bounded above. In that case, $$ \sum_{n = n_0}^{+\infty} u_n \ = \ \sup_{n \ge n_0} S_n. $$

Proof

Since $u_n \ge 0$ for every $n \ge n_0$, the partial sums satisfy $S_{n+1} - S_n = u_{n+1} \ge 0$, so $(S_n)_{n \ge n_0}$ is non-decreasing. By a standard result of chapter SuitesReelles, a non-decreasing sequence converges if and only if it is bounded above ; in that case its limit equals its supremum. Hence $(S_n)$ has a finite limit iff $(S_n)$ is bounded above, and the limit equals $\sup_{n \ge n_0} S_n$.

Method — Direct majoration of partial sums

For a positive-term series $\sum u_n$ :

Either exhibit an explicit upper bound $S_n \le M$ for all $n$ --- then the series converges and its sum is at most $M$ ;
Or show that $S_n \to +\infty$ --- then the series diverges (the only possible divergence for a positive-term series).

Skills to practice

Showing convergence by direct majoration of partial sums

II.2 Term-by-term comparison

The first concrete application of the monotone-convergence framework (subsection Convergence criterion for positive-term series) is term-by-term comparison : if $0 \le u_n \le v_n$, then $S_n^u \le S_n^v$, so any upper bound on $S_n^v$ also bounds $S_n^u$. This « hierarchy of comparisons » is the workhorse for proving convergence of an unknown positive series by majorising it by a known convergent reference (geometric, Riemann), and conversely, for proving divergence by minorising it by a known divergent reference.

Theorem — Term-by-term comparison

Let $\sum u_n$ and $\sum v_n$ be two series with $0 \le u_n \le v_n$ for $n$ large enough.

If $\sum v_n$ converges, then $\sum u_n$ converges.
If $\sum u_n$ diverges, then $\sum v_n$ diverges.

Proof

Without loss of generality, $0 \le u_n \le v_n$ for $n \ge n_1$. Summing for $k = n_1, \ldots, n$ : $$ 0 \ \le \ \sum_{k = n_1}^{n} u_k \ \le \ \sum_{k = n_1}^{n} v_k. $$ (1) If $\sum v_n$ converges, the right-hand sequence is bounded above by $\sum_{k = n_1}^{+\infty} v_k$. Hence the partial sums of $\sum u_n$ (from rank $n_1$) are bounded above. By the positive-series convergence criterion, $\sum u_n$ converges.
(2) Contrapositive of (1) : if $\sum u_n$ diverges and $\sum v_n$ converged, then $\sum u_n$ would converge --- contradiction. So $\sum v_n$ diverges.

Method — Standard comparison framings

To determine the nature of a positive series $\sum u_n$ by comparison :

For convergence : find a known convergent positive series $\sum v_n$ (Riemann $1/n^\alpha$ with $\alpha > 1$, geometric with $|r| < 1$, $\sum 1/n!$, etc.) such that $u_n \le v_n$ for $n$ large.
For divergence : find a known divergent positive series $\sum w_n$ (harmonic $1/n$, Riemann $1/n^\alpha$ with $\alpha \le 1$) such that $u_n \ge w_n$ for $n$ large.

Skills to practice

Determining the nature of a positive-term series by direct majoration or comparison

II.3 Comparison by equivalent and by $O$

The term-by-term comparison is « pointwise » : it requires an explicit inequality $u_n \le v_n$. In practice, that inequality is rarely literal --- instead, one knows $u_n \sim v_n$ at $+\infty$ (the two general terms have the same dominant behaviour) or $u_n = O(v_n)$ (one is at most of the size of the other). The equivalent-and-$O$ tests reduce the asymptotic-comparison problem to the term-by-term comparison framework. They are the most-used positive-series tests at this level.

Theorem — Equivalents for positive-term series

Let $\sum u_n$ and $\sum v_n$ be two series with $u_n \ge 0$ and $v_n > 0$ for $n$ large enough, and $u_n \sim v_n$ at $+\infty$. Then $\sum u_n$ and $\sum v_n$ have the same nature (both converge or both diverge).

Proof

By definition of $u_n \sim v_n$ (with $v_n > 0$ for $n$ large), $u_n / v_n \to 1$ at $+\infty$. Fix $\varepsilon = 1/2$ : there exists $N$ such that for $n \ge N$, $$ \frac{1}{2} \ \le \ \frac{u_n}{v_n} \ \le \ \frac{3}{2}, \qquad \text{hence} \qquad \frac{1}{2} v_n \ \le \ u_n \ \le \ \frac{3}{2} v_n. $$ By linearity (subsection Linearity of the sum of convergent series), the nature of $\sum v_n$, $\sum \frac{1}{2} v_n$, and $\sum \frac{3}{2} v_n$ are identical (multiplying a convergent series by a non-zero scalar produces a convergent series, and the reverse follows by multiplying by the inverse scalar). By the term-by-term comparison Theorem applied to $u_n$ majored above by $\frac{3}{2} v_n$ and minored below by $\frac{1}{2} v_n$ from rank $N$ :

if $\sum v_n$ converges, then $\sum u_n$ converges (majoration by $\frac{3}{2} v_n$) ;
if $\sum v_n$ diverges, then $\sum u_n$ diverges (minoration by $\frac{1}{2} v_n$).

Proposition — Comparison by $O$

Let $\sum u_n$ and $\sum v_n$ be two series with $v_n \ge 0$ for $n$ large enough and $u_n = O(v_n)$ at $+\infty$. If $\sum v_n$ converges, then $\sum |u_n|$ converges. (In the Absolute convergence section we will rephrase this as : $\sum u_n$ converges absolutely.)

Proof

By definition of $u_n = O(v_n)$, there exist $K > 0$ and $N$ such that $|u_n| \le K v_n$ for $n \ge N$. Both $|u_n|$ and $K v_n$ are non-negative. The series $\sum K v_n$ converges by linearity (subsection Linearity of the sum of convergent series), since $\sum v_n$ does. By the term-by-term comparison Theorem applied to $0 \le |u_n| \le K v_n$ for $n \ge N$, $\sum |u_n|$ converges.

Method — Equivalent or $O$ reflex

To determine the nature of a positive-term series $\sum u_n$ with messy general term :

Factor the dominant term. Write $u_n$ as a product, isolate the dominant factor at $+\infty$.
Compare via equivalents or $O$. If $u_n \sim C / n^\alpha$ with $C > 0$, conclude by the Riemann Theorem (subsection Series-integral comparison and Riemann series) : converges iff $\alpha > 1$. If $u_n = O(v_n)$ with $\sum v_n$ a known convergent positive series, conclude convergence.
Growth-comparison reflex (croissances comparées). Use $\ln^\beta n = o(n^a)$ and $n^a = o(b^n)$ to simplify.

Example — Nature by equivalent

Consider $\sum_{n \ge 1} \frac{n + \cos n}{n^3}$. The general term is positive (numerator $\ge n - 1 > 0$ for $n \ge 2$). The equivalent at $+\infty$ is $$ \frac{n + \cos n}{n^3} \ \sim \ \frac{n}{n^3} \ = \ \frac{1}{n^2}. $$ Both general terms are positive for $n$ large. The Riemann series $\sum 1/n^2$ converges ($\alpha = 2 > 1$), hence $\sum (n + \cos n)/n^3$ converges (by the equivalents-for-positive-series Theorem).

Example — Nature by growth comparison

Consider $\sum_{n \ge 2} \frac{\ln n}{n^{3/2}}$. The general term is positive for $n \ge 2$. By comparative growth (croissance comparée), $\ln n = o(n^{1/4})$, so $\ln n / n^{3/2} = o(1 / n^{5/4})$ at $+\infty$. The Riemann series $\sum 1/n^{5/4}$ converges ($\alpha = 5/4 > 1$). By comparison by $O$ (the ``$o$'' is a stronger ``$O$'') applied to a positive series, $\sum \ln n / n^{3/2}$ converges.

Skills to practice

Determining the nature via equivalent or $O$

II.4 Series-integral comparison and Riemann series

The last positive-series tool --- and the one that grounds the Riemann reference family --- is the series-integral comparison : for a positive, monotone-decreasing continuous function $f$, the series $\sum f(k)$ and the integral $\int^{+\infty} f$ have the same nature, framed by the rectangles below and rectangles above construction. This is the standard technique that proves the Riemann series $\sum 1/n^\alpha$ converges iff $\alpha > 1$ --- by comparison to $\int 1/x^\alpha \, dx$.

Theorem — Series-integral comparison

Let $n_0 \in \mathbb N$ and $f : [n_0, +\infty[ \to \mathbb R_+$ a continuous, non-increasing function. Then the series $\sum_{k \ge n_0} f(k)$ and the integral $\int_{n_0}^{+\infty} f(t) \, dt$ have the same nature. Moreover, for every $n \ge n_0$, $$ \int_{n_0}^{n + 1} f(t) \, dt \ \le \ \sum_{k = n_0}^{n} f(k) \ \le \ f(n_0) + \int_{n_0}^{n} f(t) \, dt. $$

Proof

Rectangles framing. For $k \ge n_0$ and $t \in [k, k+1]$, $f$ being non-increasing, $f(k+1) \le f(t) \le f(k)$. Integrating on $[k, k+1]$ (an interval of length $1$) : $$ f(k + 1) \ \le \ \int_k^{k+1} f(t) \, dt \ \le \ f(k). $$ Summing. For $k = n_0, \ldots, n$ in the right-hand inequality, then $k = n_0, \ldots, n - 1$ in the left-hand inequality : $$ \begin{aligned} \int_{n_0}^{n+1} f(t) \, dt \ &= \ \sum_{k = n_0}^{n} \int_k^{k+1} f(t) \, dt \ \le \ \sum_{k = n_0}^{n} f(k), \\ \sum_{k = n_0 + 1}^{n} f(k) \ &\le \ \sum_{k = n_0}^{n - 1} \int_k^{k+1} f(t) \, dt \ = \ \int_{n_0}^{n} f(t) \, dt. \end{aligned} $$ Adding $f(n_0)$ to the second inequality gives $\sum_{k = n_0}^n f(k) \le f(n_0) + \int_{n_0}^n f(t) \, dt$, the desired upper bound.
Same nature. We argue by double implication on bounded-above sequences.

If $\int_{n_0}^{+\infty} f(t) \, dt$ converges, the upper bound $\sum_{k=n_0}^n f(k) \le f(n_0) + \int_{n_0}^n f(t) \, dt$ shows that the partial sums of $\sum f(k)$ are bounded above. Since the series has positive terms (positive-series convergence criterion), it converges.
Conversely, if $\sum_{k \ge n_0} f(k)$ converges, its partial sums are bounded above. The lower bound $\int_{n_0}^{n+1} f(t) \, dt \le \sum_{k = n_0}^n f(k)$ shows that the truncated integrals are bounded above. Since they form a non-decreasing function of the upper bound (because $f \ge 0$), the improper integral $\int_{n_0}^{+\infty} f$ converges.

Theorem — Riemann series

Let $\alpha \in \mathbb R$. The series $\sum_{n \ge 1} 1/n^\alpha$ converges if and only if $\alpha > 1$.

Proof

Case $\alpha \le 0$. If $\alpha < 0$, $u_n = n^{-\alpha} \to +\infty$ ; if $\alpha = 0$, $u_n = 1$ for all $n$, hence $u_n \to 1 \ne 0$. In either subcase $u_n \not\to 0$ --- gross divergence.
Case $\alpha > 0$. The function $f(x) = 1/x^\alpha$ is continuous and decreasing on $[1, +\infty[$, with $f(x) \ge 0$. Apply the series-integral comparison Theorem. The integral $\int_1^{+\infty} dx / x^\alpha$ is, from chapter Primitives, $$ \int_1^M \frac{dx}{x^\alpha} \ = \ \begin{cases} \ln M & \text{if } \alpha = 1, \\ \dfrac{M^{1 - \alpha} - 1}{1 - \alpha} & \text{if } \alpha \ne 1.\end{cases} $$ As $M \to +\infty$ : if $0 < \alpha \le 1$, the integral tends to $+\infty$, so the series diverges ; if $\alpha > 1$, the integral tends to $1/(\alpha - 1)$ (finite), so the series converges.

Method — The Riemann reflex

For an unfamiliar positive series $\sum u_n$, the first reflex is the Riemann comparison :

Look for an equivalent $u_n \sim C / n^\alpha$ at $+\infty$ with $C > 0$.
Conclude by Riemann : $\sum u_n$ converges iff $\alpha > 1$.

When no clean equivalent exists, fall back to direct majoration or to comparison by $O$.

Example — Riemann at a glance

$\sum_{n \ge 1} 1/n$ (harmonic series) diverges ($\alpha = 1$, the borderline case). $\sum_{n \ge 1} 1/n^2$ converges ($\alpha = 2 > 1$). $\sum_{n \ge 1} 1/\sqrt n$ diverges ($\alpha = 1/2 \le 1$). $\sum_{n \ge 1} 1/n^{3/2}$ converges ($\alpha = 3/2 > 1$).

Skills to practice

Determining the nature by comparison to an integral --- Riemann series

III Absolute convergence

III.1 Definition and main theorem

The positive-series toolbox handles positive-term series. To handle sign-variable series ($u_n \in \mathbb R$ with no sign constraint, or $u_n \in \mathbb C$), the standard technique is to look first at the series of absolute values $\sum |u_n|$ : if it converges, then so does $\sum u_n$ itself. This is the conceptual core of the present section : « absolute convergence implies convergence ». It turns every sign-variable problem into a positive-term problem, accessible by the positive-series toolbox.

Definition — Absolute convergence

A series $\sum u_n$ (with $u_n \in \mathbb R$ or $\mathbb C$) is absolutely convergent if the series of absolute values $\sum |u_n|$ converges.

Theorem — Absolute convergence implies convergence

Every absolutely convergent series is convergent. Moreover, $$ \left| \sum_{n = n_0}^{+\infty} u_n \right| \ \le \ \sum_{n = n_0}^{+\infty} |u_n|. $$

Proof

Case $u_n \in \mathbb R$ (real series). Decompose $u_n = u_n^+ - u_n^-$ where $u_n^+ = \max(u_n, 0)$ and $u_n^- = \max(-u_n, 0)$ are the positive and negative parts. Both are $\ge 0$ and bounded by $|u_n|$ : $$ 0 \ \le \ u_n^\pm \ \le \ |u_n|. $$ Since $\sum |u_n|$ converges by hypothesis, the term-by-term comparison Theorem gives that $\sum u_n^+$ and $\sum u_n^-$ both converge. By linearity, $\sum u_n = \sum (u_n^+ - u_n^-)$ converges. The triangle inequality applied to partial sums gives $|S_n| \le \sum_{k = n_0}^n |u_k|$, and passing to the limit yields the announced majoration.
Case $u_n \in \mathbb C$ (complex series). Write $u_n = a_n + i \, b_n$ with $a_n, b_n \in \mathbb R$. Then $|a_n|, |b_n| \le |u_n|$, so by term-by-term comparison the real series $\sum |a_n|$ and $\sum |b_n|$ converge. By the real case above, $\sum a_n$ and $\sum b_n$ converge. By linearity, $\sum u_n = \sum a_n + i \sum b_n$ converges. The triangle inequality, again applied to partial sums and passed to the limit, gives $|\sum u_n| \le \sum |u_n|$.

Example — Absolutely convergent Riemann-alternated

Consider $\sum_{n \ge 1} (-1)^n / n^2$. The series of absolute values is $\sum 1/n^2$, a Riemann series with $\alpha = 2 > 1$, hence convergent. Therefore $\sum (-1)^n / n^2$ converges absolutely (and a fortiori converges).

Skills to practice

Showing absolute convergence

III.2 Comparison by $O$ for sign-variable series

The ``abs cv $\Rightarrow$ cv'' Theorem, combined with the Comparison-by-$O$ Proposition (subsection Comparison by equivalent and by $O$), gives the standard reflex for sign-variable series : majorise $|u_n|$ by a positive-term reference series. This subsection closes with the d'Alembert ratio test --- non-examinable, but useful downstream (power series in the second year).

Method — Reduction to the positive case

To determine the nature of a sign-variable series $\sum u_n$ :

Study $\sum |u_n|$ (a positive-term series) using the positive-series toolbox.
If $\sum |u_n|$ converges, then $\sum u_n$ converges absolutely, hence converges, and $\left|\sum u_n\right| \le \sum |u_n|$.
If $\sum |u_n|$ diverges, absolute convergence fails. This reduction is the main systematic tool for a sign-variable series at this level (outside the special case of a telescoping series, treated via the sequence-series link) : when it does not apply, the nature of $\sum u_n$ is left undecided here.

Example — Sign-variable made positive

Consider $\sum_{n \ge 1} \frac{\sin n}{n^2}$. The general term is sign-variable (since $\sin n$ oscillates). Take absolute values : $|\sin n / n^2| \le 1/n^2$. The Riemann series $\sum 1/n^2$ converges, so by the term-by-term comparison Theorem, $\sum |\sin n / n^2|$ converges, that is $\sum \sin n / n^2$ converges absolutely. Therefore it converges.

d'Alembert ratio test --- non-examinable complement

The following statement is not in the 2021 program and is therefore non-examinable. It is stated for cultural reference, because it is the standard tool for the convergence of power series (séries entières) in second-year curricula and frequently shortens positive-series convergence proofs in practice. (Statement, admitted without proof.) Let $\sum u_n$ be a series with $u_n > 0$ for $n$ large enough, and assume the ratio $u_{n+1}/u_n$ has a limit $\ell \in [0, +\infty]$ as $n \to +\infty$.

If $\ell < 1$, then $\sum u_n$ converges.
If $\ell > 1$ (including $\ell = +\infty$), then $\sum u_n$ diverges (in fact, $u_n \not\to 0$, gross divergence).
If $\ell = 1$, the test is inconclusive.

In this course, when the d'Alembert ratio test is the natural tool (e.g.\ for $\sum z^n / n!$, $\sum n^k / r^n$), we re-derive the conclusion by a direct geometric majoration (cf.\ the proof of $e^z$ convergence in the Geometric series and the complex exponential subsection).

Skills to practice

Reducing a sign-variable series to a positive reference