---
title: Series and Convergence Tests
module: Sequences and Series
moduleNumber: 2
lessonNumber: 5
order: 205
summary: >
  A series converges when its sequence of partial sums does, so every fact about
  sequences transfers. Geometric and telescoping series sum in closed form; the
  n-th term test rejects series whose terms miss zero, though the harmonic
  series shows the converse fails; and the comparison test against the geometric
  and p-series benchmarks settles most nonnegative-term series.
topics: [Sequences and Series]
sources:
  - book: Lebl
    ref: "§2.5 Series"
  - book: Rosenlicht
    ref: "Ch. 2 — The real number system"
draft: false
---

A series adds infinitely many numbers. The sum is given meaning by the sequence
of finite partial sums, so the whole theory of sequences applies to a series
through that sequence. The motivation for building analysis rigorously was
largely to understand series, since solutions of differential equations and
expansions of functions arrive as series.

## Series and partial sums

> **Definition (Series).** Given a sequence $\{x_n\}_{n=1}^\infty$, the formal
> object $\sum_{n=1}^\infty x_n$ is a **series**. Its **partial sums** are
> $$
> s_k = \sum_{n=1}^k x_n = x_1 + x_2 + \cdots + x_k.
> $$
> The series **converges** if the sequence $\{s_k\}_{k=1}^\infty$ converges, in
> which case $\sum_{n=1}^\infty x_n = \lim_{k\to\infty} s_k$. Otherwise the
> series **diverges** and the symbol denotes no number.

The notation $\sum_{n=1}^\infty x_n$ does double duty: it names the series and,
when convergence holds, the limit of the partial sums. Context decides which.
The index may start anywhere convenient; $\sum_{n=0}^\infty r^n$ and
$\sum_{n=1}^\infty r^{n-1}$ denote the same series.

> **Worked example.** The series $\sum_{n=1}^\infty 1/2^n$ converges to $1$.
> Induction on $k$ gives the identity
> $$
> \sum_{n=1}^k \frac{1}{2^n} + \frac{1}{2^k} = 1,
> $$
> so the partial sums are $s_k = 1 - 1/2^k$. Then $|1 - s_k| = 1/2^k \to 0$, and
> $s_k \to 1$.[^lebl-254]

$$
% caption: The partial sums of a convergent series form a sequence; the series'
% value is that sequence's limit. Here the sums climb to 1.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (9.4,0) node[right, black] {$k$};
\draw[black] (0,0) -- (0,3.5);
\draw[acc, thick] (0,3.0) -- (9.4,3.0) node[right, black] {series value};
\foreach \k/\y in {1/1.5, 2/2.25, 3/2.625, 4/2.8125, 5/2.906, 6/2.953, 7/2.977, 8/2.988}
  {\fill[acc] (\k*1.1,\y) circle (2.2pt); \draw[black, thin] (\k*1.1,0) -- (\k*1.1,\y);}
\node[acc, anchor=west] at (1.2,1.35) {$s_k$};
\end{tikzpicture}
$$

## The geometric series

The one family that can be summed in closed form is the geometric series, and it
underlies most convergence tests.

> **Proposition (Geometric series).** For $-1 < r < 1$, the series
> $\sum_{n=0}^\infty r^n$ converges and
> $$
> \sum_{n=0}^\infty r^n = \frac{1}{1 - r}.
> $$

> **Proof.** The partial sum $\sum_{n=0}^{k-1} r^n = \frac{1 - r^k}{1 - r}$ is
> the finite geometric sum. Since $|r| < 1$ gives $r^k \to 0$, the limit is
> $\frac{1}{1 - r}$.[^lebl-geo] $\square$

If $|r| \ge 1$ the terms $r^n$ do not go to zero, and the series diverges. The
picture is a unit length repeatedly halved (for $r = 1/2$): the pieces shrink
fast enough that their total is finite.

$$
% caption: The geometric series with ratio one half. Successive pieces halve, and
% the whole collection fills exactly one unit.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) rectangle (8,0.8);
\draw[black] (4,0) -- (4,0.8);
\draw[black] (6,0) -- (6,0.8);
\draw[black] (7,0) -- (7,0.8);
\draw[black] (7.5,0) -- (7.5,0.8);
\draw[black] (7.75,0) -- (7.75,0.8);
\fill[acc!14] (0,0) rectangle (4,0.8);
\node at (2,0.4) {$\frac{1}{2}$};
\node at (5,0.4) {$\frac{1}{4}$};
\node at (6.5,0.4) {$\frac{1}{8}$};
\node[font=\scriptsize] at (7.25,0.4) {$\frac{1}{16}$};
\node[black, anchor=south, font=\scriptsize] at (7.9,0.75) {. . .};
\draw[black, <->] (0,-0.35) -- (8,-0.35) node[midway, below] {total $= 1$};
\end{tikzpicture}
$$

A second family collapses because consecutive terms cancel: whenever
$x_n = c_n - c_{n+1}$ for a convergent sequence $\{c_n\}$, the series
**telescopes** and sums to $c_1 - \lim_n c_n$.

> **Worked example.** $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$. The partial
> fraction decomposition $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ makes
> the partial sum telescope:
> $$
> s_k = \sum_{n=1}^k \left( \frac{1}{n} - \frac{1}{n+1} \right) = 1 - \frac{1}{k+1} \longrightarrow 1.
> $$

The correspondence runs both ways: the terms recover from the partial sums as
first differences, $x_n = s_n - s_{n-1}$.

> **Worked example.** Suppose a series has partial sums $s_k = \frac{k}{k+1}$.
> Then $x_1 = s_1 = \frac{1}{2}$ and, for $n \ge 2$,
> $$
> x_n = s_n - s_{n-1} = \frac{n}{n+1} - \frac{n-1}{n} = \frac{n^2 - (n^2-1)}{n(n+1)} = \frac{1}{n(n+1)},
> $$
> so the series is $\sum \frac{1}{n(n+1)}$ again. It converges because
> $\{s_k\}$ converges: $\lim_{k\to\infty} \frac{k}{k+1} = 1$.[^lebl-ex251]

## Tails, the Cauchy criterion, and linearity

Three properties transfer directly from sequences to series.

- **Tails.** For any $M$, the series $\sum_{n=1}^\infty x_n$ and
  $\sum_{n=M}^\infty x_n$ converge together, differing only by the fixed finite
  sum $\sum_{n=1}^{M-1} x_n$. Convergence is a property of the tail.[^lebl-tail]
- **Cauchy criterion.** Since a real sequence converges iff it is Cauchy, a
  series converges iff its partial sums are Cauchy: for every $\epsilon > 0$
  there is $M$ with
  $$
  \left| \sum_{i=n+1}^{k} x_i \right| < \epsilon \qquad \text{for all } k > n \ge M.
  $$
  This says every sufficiently late block of consecutive terms has small
  sum.[^lebl-cauchy]
- **Linearity.** If $\sum x_n$ and $\sum y_n$ converge and $\alpha \in
  \mathbb{R}$, then $\sum \alpha x_n = \alpha \sum x_n$ and $\sum (x_n + y_n) =
  \sum x_n + \sum y_n$. Each follows by taking limits of the corresponding
  identity for partial sums.[^lebl-lin]

> **Worked example.** Linearity sums any tail of the geometric series in closed
> form: for $|r| < 1$ and $i \in \mathbb{N}$,
> $$
> \sum_{n=i}^\infty r^n = r^i \sum_{n=0}^\infty r^n = \frac{r^i}{1-r},
> $$
> multiplying the geometric series by the constant $r^i$ and reindexing
> ($r^i \cdot r^n = r^{n+i}$).[^lebl-lin]

## The n-th term test

The Cauchy criterion with $k = n + 1$ isolates a single term, yielding a fast
necessary condition for convergence.

> **Proposition (n-th term test).** If $\sum_{n=1}^\infty x_n$ converges, then
> $\lim_{n\to\infty} x_n = 0$.

> **Proof.** Convergence makes the partial sums Cauchy: given $\epsilon > 0$
> there is $M$ with $|x_{n+1}| = \left| \sum_{i=n+1}^{n+1} x_i \right| <
> \epsilon$ for $n \ge M$. Hence $x_n \to 0$.[^lebl-nth] $\square$

The test is a divergence test: if the terms do not go to zero, the series
diverges. It gives at once that $\sum_{n=0}^\infty r^n$ diverges for $|r| \ge 1$.
The converse is false, as the harmonic series shows.

## The harmonic series diverges

> **Proposition.** The harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges,
> even though $\frac{1}{n} \to 0$.

> **Proof.** Group the terms in blocks of length $2^{k-1}$ ending at $2^k$. Each
> block is bounded below by replacing its terms with the smallest:
> $$
> \sum_{m = 2^{k-1}+1}^{2^k} \frac{1}{m} \ge 2^{k-1} \cdot \frac{1}{2^k} = \frac{1}{2}.
> $$
> Summing $k$ such blocks, $s_{2^k} \ge 1 + \frac{k}{2}$. Since $\{k/2\}$ is
> unbounded (the Archimedean property), the partial sums are unbounded and the
> series diverges.[^lebl-harm] $\square$

$$
% caption: Oresme's grouping. Blocks of 2, 4, 8 consecutive terms of the harmonic
% series each total at least one half, so the partial sums exceed 1 + k/2.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (10.6,0);
% bars for 1/1 .. 1/8, height = 3/m, width .9
\foreach \m/\x in {1/0, 2/1.05, 3/2.1, 4/3.15, 5/4.2, 6/5.25, 7/6.3, 8/7.35} {
  \draw[acc] (\x,0) rectangle (\x+0.9,{3.0/\m});
  \fill[acc!12] (\x,0) rectangle (\x+0.9,{3.0/\m});
}
\node at (0.45,-0.35) {$1$};
\node at (1.5,-0.35) {$\frac{1}{2}$};
\node at (2.55,-0.35) {$\frac{1}{3}$};
\node at (3.6,-0.35) {$\frac{1}{4}$};
\node at (4.65,-0.35) {$\frac{1}{5}$};
\node at (5.7,-0.35) {$\frac{1}{6}$};
\node at (6.75,-0.35) {$\frac{1}{7}$};
\node at (7.8,-0.35) {$\frac{1}{8}$};
% hand-drawn group brackets above
\draw[black] (2.1,1.25) -- (2.1,1.45) -- (4.05,1.45) -- (4.05,1.25);
\node[black, anchor=south] at (3.08,1.5) {at least $\frac{1}{2}$};
\draw[black] (4.2,0.95) -- (4.2,1.15) -- (8.25,1.15) -- (8.25,0.95);
\node[black, anchor=south] at (6.22,1.2) {at least $\frac{1}{2}$};
\node[black, anchor=west] at (8.6,0.5) {. . . unbounded};
\end{tikzpicture}
$$

The terms tend to zero, but not fast enough: their partial sums outgrow every
bound, albeit slowly. Deciding whether a nonnegative series converges is
therefore a question of how fast its terms decay, and the comparison test
answers it.

## Nonnegative series and the comparison test

When every term is nonnegative, the partial sums are monotone increasing, so by
the monotone convergence theorem the series converges exactly when they are
bounded above.

> **Proposition.** If $x_n \ge 0$ for all $n$, then $\sum_{n=1}^\infty x_n$
> converges if and only if its sequence of partial sums is bounded above.

This reduces convergence of a nonnegative series to one inequality, and comparing
against a series already understood settles it.

> **Proposition (Comparison test).** Suppose $0 \le x_n \le y_n$ for all $n$.
> - If $\sum y_n$ converges, then $\sum x_n$ converges.
> - If $\sum x_n$ diverges, then $\sum y_n$ diverges.

> **Proof.** Both partial-sum sequences are increasing, and $\sum_{n=1}^k x_n
> \le \sum_{n=1}^k y_n$. If $\sum y_n$ converges, its partial sums are bounded
> by some $B$, so the smaller partial sums of $\sum x_n$ are bounded by $B$ too,
> hence converge. The second statement is the contrapositive.[^lebl-comp]
> $\square$

$$
% caption: The comparison test. When each term of one nonnegative series sits
% under the matching term of a convergent series, its partial sums stay bounded.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (8.4,0) node[right, black] {$n$};
\draw[black] (0,0) -- (0,3.4);
% y_n bars (outline, dominating)
\foreach \n/\h in {1/3.0, 2/2.1, 3/1.5, 4/1.1, 5/0.82, 6/0.62}
  \draw[black] (\n*1.2-0.22,0) rectangle (\n*1.2+0.22,\h);
% x_n bars (filled, smaller)
\foreach \n/\h in {1/2.2, 2/1.4, 3/0.95, 4/0.62, 5/0.42, 6/0.3}
  \fill[acc!30] (\n*1.2-0.22,0) rectangle (\n*1.2+0.22,\h);
\node[black, anchor=west] at (5.2,3.0) {$y_n$ converges};
\node[acc, anchor=west] at (5.2,2.5) {$x_n$ below $y_n$};
\end{tikzpicture}
$$

> **Worked example.** $\sum_{n=1}^\infty \frac{1}{n^2+1}$ converges. For every
> $n$, $\frac{1}{n^2 + 1} < \frac{1}{n^2}$, and $\sum \frac{1}{n^2}$ converges
> by the $p$-series test below, so the comparison test applies.[^lebl-2518]

> **Worked example.** $\sum_{n=1}^\infty \frac{1}{2n-1}$ diverges. For every
> $n$, $2n - 1 \le 2n$ gives $\frac{1}{2n-1} \ge \frac{1}{2n}$, and
> $$
> \sum_{n=1}^\infty \frac{1}{2n} = \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n}
> $$
> diverges with the harmonic series (a convergent left side would force, by
> linearity, convergence of the right). Comparison then forces
> $\sum \frac{1}{2n-1}$ to diverge.[^lebl-253]

## The p-series

The benchmark family for the comparison test is the $p$-series.

> **Proposition (p-series test).** For $p \in \mathbb{R}$, the series
> $\sum_{n=1}^\infty \frac{1}{n^p}$ converges if and only if $p > 1$.

> **Proof.** For $p \le 1$, $\frac{1}{n^p} \ge \frac{1}{n}$, and since the
> harmonic series diverges, comparison forces divergence. For $p > 1$, group as
> for the harmonic series but bound each block from above by its largest term.
> Blocks ending at $2^{k}-1$ give
> $$
> \sum_{m=2^i}^{2^{i+1}-1} \frac{1}{m^p} < \frac{2^i}{(2^i)^p} = \left( \frac{1}{2^{p-1}} \right)^i,
> $$
> so the partial sums are dominated by the geometric series
> $\sum_i (2^{-(p-1)})^i$, which converges because $2^{-(p-1)} < 1$. Bounded
> partial sums of a nonnegative series converge.[^lebl-p] $\square$

The direction of the block estimate is the whole difference: bounding blocks
from below shows divergence, from above shows convergence. Neither test names
the sum. The value $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ is a separate and much
harder computation, and $\sum \frac{1}{n^p}$ as a function of $p$ is the Riemann
zeta function, the subject of a famous open problem.

| Series | Converges? | Reason |
| --- | --- | --- |
| $\sum r^n$ | iff $\lvert r \rvert < 1$ | geometric, closed form $\frac{1}{1-r}$ |
| $\sum \frac{1}{n(n+1)}$ | yes ($=1$) | telescopes |
| $\sum \frac{1}{n}$ | no | block bound $\ge \tfrac{1}{2}$ |
| $\sum \frac{1}{n^p}$ | iff $p > 1$ | comparison with a geometric series |

## Summary

A series converges by definition when its partial sums do, so sequences supply
all the theory: the Cauchy criterion, linearity, and tail-independence carry
over unchanged. The $n$-th term test rejects any series whose terms miss zero,
but passing it guarantees nothing, as the harmonic series shows. For
nonnegative terms, convergence is boundedness of the partial sums, and the
comparison test against the geometric and $p$-series decides most cases at
sight. The comparison test requires nonnegative terms; series with mixed signs
need [absolute convergence](/real-analysis/sequences-series/absolute-conditional-rearrangement)
and the ratio, root, and alternating-series tests.

[^lebl-254]: Lebl, §2.5, Example 2.5.4.
[^lebl-geo]: Lebl, §2.5, Proposition 2.5.5.
[^lebl-2518]: Lebl, §2.5, Example 2.5.18.
[^lebl-ex251]: Lebl, §2.5, Exercise 2.5.1.
[^lebl-253]: Lebl, §2.5, Exercise 2.5.3.
[^lebl-tail]: Lebl, §2.5, Proposition 2.5.6.
[^lebl-cauchy]: Lebl, §2.5, Definition 2.5.7 and Proposition 2.5.8.
[^lebl-lin]: Lebl, §2.5, Proposition 2.5.12.
[^lebl-nth]: Lebl, §2.5, Proposition 2.5.9.
[^lebl-harm]: Lebl, §2.5, Example 2.5.11; the proof is due to Oresme.
[^lebl-comp]: Lebl, §2.5, Propositions 2.5.13 and 2.5.16.
[^lebl-p]: Lebl, §2.5, Proposition 2.5.17.