---
title: Absolute Convergence, the Ratio and Root Tests, and Rearrangements
module: Sequences and Series
moduleNumber: 2
lessonNumber: 6
order: 206
summary: >
  Absolute convergence is the strong form of convergence that permits free
  manipulation; conditional convergence is fragile. Absolute convergence implies
  convergence, and the ratio and root tests detect it by comparison with the
  geometric series. The alternating series test supplies conditionally
  convergent series, Riemann's theorem rearranges any of them to any sum, and
  Mertens' theorem multiplies series when at least one converges absolutely.
topics: [Sequences and Series]
sources:
  - book: Lebl
    ref: "§2.5 Series; §2.6 More on series"
  - book: Rosenlicht
    ref: "Ch. 2 — The real number system"
draft: false
---

The [comparison test](/real-analysis/sequences-series/series-convergence)
handles nonnegative series. For series with mixed signs there are two grades of
convergence. A series that converges even after every term is made positive may
be reordered, multiplied, and combined like a finite sum. A series that
converges only because of cancellation between positive and negative terms
admits no such manipulation: reordering it can change the sum or make it
diverge.

## Absolute and conditional convergence

> **Definition (Absolute and conditional convergence).** A series
> $\sum_{n=1}^\infty x_n$ **converges absolutely** if $\sum_{n=1}^\infty |x_n|$
> converges. If $\sum x_n$ converges but $\sum |x_n|$ diverges, it **converges
> conditionally**.

Absolute convergence is the stronger condition, and it implies ordinary
convergence.

> **Proposition.** If $\sum_{n=1}^\infty x_n$ converges absolutely, then it
> converges.

> **Proof.** Use the Cauchy criterion for series. Given $\epsilon > 0$, absolute
> convergence gives $M$ with $\sum_{i=k+1}^n |x_i| < \epsilon$ for all $n > k
> \ge M$. The triangle inequality for finite sums then bounds the block of the
> original series:
> $$
> \left| \sum_{i=k+1}^n x_i \right| \le \sum_{i=k+1}^n |x_i| < \epsilon.
> $$
> So the partial sums of $\sum x_n$ are Cauchy, and the series
> converges.[^lebl-abs] $\square$

Passing to the limit in the finite triangle inequality gives its infinite
version for an absolutely convergent series:

$$
\left| \sum_{n=1}^\infty x_n \right| \le \sum_{n=1}^\infty |x_n|.
$$

The two sums are generally different numbers; the inequality only bounds one by
the other. The alternating harmonic series $\sum_{n=1}^\infty
\frac{(-1)^n}{n}$ converges (proved below) while $\sum \frac{1}{n}$ diverges, so
it is the standard example of conditional convergence.

## The ratio and root tests

Both standard tests for absolute convergence compare the tail of a series to a
geometric series, generalizing the constant ratio $r$ of $\sum r^n$.

> **Proposition (Ratio test).** Let $x_n \ne 0$ for all $n$ and suppose
> $L = \lim_{n\to\infty} \frac{|x_{n+1}|}{|x_n|}$ exists. If $L < 1$ the series
> $\sum x_n$ converges absolutely; if $L > 1$ it diverges.

> **Proof.** If $L > 1$, the [ratio test for
> sequences](/real-analysis/sequences-series/limit-laws-monotone) shows
> $\{x_n\}$ is unbounded, so the terms miss zero and the series diverges. If
> $L < 1$, pick $r$ with $L < r < 1$; past some $M$ the ratios stay below $r$,
> so $|x_n| \le |x_M| r^{-M} r^n$. The partial sums of $\sum |x_n|$ are then
> bounded by a constant plus a convergent geometric series, hence bounded, so
> $\sum |x_n|$ converges.[^lebl-ratio] $\square$

> **Proposition (Root test).** For a series $\sum x_n$, let $L = \limsup_{n\to
> \infty} |x_n|^{1/n}$. If $L < 1$ the series converges absolutely; if $L > 1$
> it diverges.

> **Proof.** If $L < 1$, choose $r$ with $L < r < 1$; by definition of the limit
> superior, $|x_n|^{1/n} < r$ for all large $n$, so $|x_n| < r^n$, and
> comparison with the geometric series gives absolute convergence. If $L > 1$, a
> subsequence has $|x_{n_k}|^{1/n_k} > 1$, so $|x_{n_k}| > 1$ and the terms
> cannot go to zero.[^lebl-root] $\square$

Neither test gives information when its limit equals $1$. The root test uses a $\limsup$
and so applies to every series, making it slightly stronger than the ratio test,
but the ratio test is usually easier to compute — for factorials it is
immediate.

$$
% caption: The ratio and root tests both read a single number L and give the same
% verdict; the boundary value one is uninformative for either test.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize,
  box/.style={draw, minimum width=32mm, minimum height=10mm, align=center, font=\footnotesize}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box, draw=acc, text=acc] (r) at (0,0) {$L$ from ratio or root};
\node[box] (lt) at (5.4,1.8) {$L < 1$: absolute};
\node[box] (eq) at (5.4,0) {$L = 1$: no verdict};
\node[box] (gt) at (5.4,-1.8) {$L > 1$: diverges};
\draw[->, black, thick] (r) -- (lt);
\draw[->, black, thick] (r) -- (eq);
\draw[->, black, thick] (r) -- (gt);
\end{tikzpicture}
$$

> **Worked example.** $\sum_{n=1}^\infty \frac{2^n}{n!}$ converges absolutely by
> the ratio test:
> $$
> \frac{2^{n+1}/(n+1)!}{2^n/n!} = \frac{2^{n+1}}{2^n} \cdot \frac{n!}{(n+1)!} = \frac{2}{n+1} \longrightarrow 0 < 1.
> $$

> **Worked example.** $\sum_{n=1}^\infty \frac{n}{2^n}$ converges absolutely.
> The ratio of consecutive terms is
> $$
> \frac{(n+1)/2^{n+1}}{n/2^n} = \frac{n+1}{2n} = \frac{1}{2}\left(1 + \frac{1}{n}\right) \longrightarrow \frac{1}{2} < 1.
> $$
> The root test reaches the same verdict: $\left(\frac{n}{2^n}\right)^{1/n} =
> \frac{n^{1/n}}{2} \to \frac{1}{2}$, using the
> [standard limit](/real-analysis/sequences-series/limit-laws-monotone)
> $n^{1/n} \to 1$.

## The alternating series test

Conditional convergence has its own sufficient condition. A series whose terms
alternate in sign and decrease to zero converges, even when the absolute series
diverges.

> **Proposition (Alternating series test).** Let $\{x_n\}$ be a monotone
> decreasing sequence of positive numbers with $\lim_{n\to\infty} x_n = 0$. Then
> $\sum_{n=1}^\infty (-1)^n x_n$ converges.

> **Proof.** Group the even partial sums in pairs:
> $$
> s_{2k} = (-x_1 + x_2) + (-x_3 + x_4) + \cdots + (-x_{2k-1} + x_{2k}) = \sum_{\ell=1}^k (-x_{2\ell-1} + x_{2\ell}).
> $$
> Each pair is $\le 0$ since $\{x_n\}$ decreases, so $\{s_{2k}\}$ is decreasing;
> regrouping the other way shows $s_{2k} \ge -x_1$, so $\{s_{2k}\}$ is bounded
> below. A bounded monotone sequence converges; call its limit $a$. Because
> $s_{2k+1} = s_{2k} - x_{2k+1}$ and $x_{2k+1} \to 0$, the odd partial sums
> share the limit $a$, so $s_m \to a$.[^lebl-alt] $\square$

$$
% caption: Partial sums of an alternating series. Odd sums fall and even sums
% rise; each pair brackets the limit more tightly, so the sums zig-zag inward.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (9.4,0) node[right, black] {$m$};
\draw[black] (0,0) -- (0,3.4);
\draw[acc, thick] (0,2.08) -- (9.4,2.08) node[right, black] {limit};
\draw[black]
  (1.05,3.0) -- (2.1,1.5) -- (3.15,2.5) -- (4.2,1.75) -- (5.25,2.35) -- (6.3,1.85) -- (7.35,2.28) -- (8.4,1.9);
\foreach \m/\y in {1/3.0, 2/1.5, 3/2.5, 4/1.75, 5/2.35, 6/1.85, 7/2.28, 8/1.9}
  \fill[black] (\m*1.05,\y) circle (2.2pt);
\node[black, anchor=south] at (1.05,3.05) {$s_1$};
\node[black, anchor=north] at (2.1,1.45) {$s_2$};
\end{tikzpicture}
$$

> **Worked example.** The alternating harmonic series
> $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ converges conditionally. The sequence
> $x_n = \frac{1}{n}$ is positive, monotone decreasing
> ($\frac{1}{n+1} \le \frac{1}{n}$), and tends to $0$, so the alternating series
> test gives convergence. The absolute series is
> $\sum \left| \frac{(-1)^n}{n} \right| = \sum \frac{1}{n}$, the
> [harmonic series](/real-analysis/sequences-series/series-convergence), which
> diverges. Convergent but not absolutely convergent is precisely conditional
> convergence.

Conditional convergence can be arbitrarily slow: $\sum \frac{(-1)^n}{n^p}$
converges for every $p > 0$ yet fails to converge absolutely whenever
$p \le 1$.

## Rearrangements

A **rearrangement** of $\sum x_n$ is the series $\sum x_{\sigma(k)}$ for a
bijection $\sigma \colon \mathbb{N} \to \mathbb{N}$ — the same terms summed in a
different order. Order becomes irrelevant precisely when the series converges
absolutely.

> **Theorem (Rearrangement of absolutely convergent series).** If
> $\sum_{n=1}^\infty x_n$ converges absolutely to $x$, then every rearrangement
> $\sum_{n=1}^\infty x_{\sigma(n)}$ converges absolutely to the same $x$.

> **Proof (idea).** Given $\epsilon > 0$, choose $M$ with
> $\left| \sum_{n=1}^M x_n - x \right| < \epsilon/2$ and
> $\sum_{n=M+1}^\infty |x_n| < \epsilon/2$. Since $\sigma$ hits
> $\{1, \dots, M\}$ within its first $K$ values for some $K$, any partial sum
> $\sum_{n=1}^N x_{\sigma(n)}$ with $N \ge K$ contains all of
> $x_1, \dots, x_M$; the remaining terms have indices exceeding $M$, so their
> total is bounded by the tail $\sum_{n>M} |x_n| < \epsilon/2$. Combining the
> two estimates keeps the rearranged partial sum within $\epsilon$ of
> $x$.[^lebl-rearr] $\square$

Conditional convergence has no such stability: reordering a conditionally
convergent series can produce any sum.

> **Theorem (Riemann rearrangement).** A conditionally convergent series can be
> rearranged to converge to any prescribed real number $L$ (and also to diverge
> to $\pm\infty$).

The mechanism is that the positive terms alone sum to $+\infty$ and the negative
terms alone to $-\infty$; only their interleaving produces a finite sum. For a
target $L$, add positive terms in order until the running sum first exceeds $L$,
then add negative terms until it drops below $L$, and repeat. Because the terms
tend to zero, each overshoot shrinks, and the running partial sum is eventually
trapped within any tolerance of $L$.[^lebl-riemann]

$$
% caption: Why the steering works: for a conditionally convergent series the
% positive terms alone sum to plus infinity and the negative terms alone to minus
% infinity, so each sign's supply never runs out.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (10.2,0) node[right, black] {$k$};
\draw[black] (0,-2.3) -- (0,2.6);
% positive-part partial sums grow without bound
\foreach \k/\y in {1/0.5, 2/0.85, 3/1.12, 4/1.34, 5/1.53, 6/1.70, 7/1.85, 8/1.99, 9/2.12, 10/2.24}
  \fill[black] (\k*0.9,\y) circle (1.9pt);
% negative-part partial sums fall without bound
\foreach \k/\y in {1/-0.4, 2/-0.68, 3/-0.90, 4/-1.08, 5/-1.24, 6/-1.38, 7/-1.51, 8/-1.63, 9/-1.74, 10/-1.84}
  \fill[black] (\k*0.9,\y) circle (1.9pt);
\node[black, anchor=west] at (7.4,2.35) {positive terms only};
\node[black, anchor=west] at (7.4,-2.1) {negative terms only};
\end{tikzpicture}
$$

> **Worked example.** Steer the alternating harmonic series to $L = 1.2$. Its
> positive terms are $\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{6}, \dots$ and its
> negative terms $-1, -\tfrac{1}{3}, -\tfrac{1}{5}, \dots$. Adding positive
> terms in order, the running sums are
> $$
> 0.5,\quad 0.75,\quad 0.9167,\quad 1.0417,\quad 1.1417,\quad 1.225,
> $$
> so the sum first exceeds $1.2$ after six terms. One negative term, $-1$, drops
> it to $0.225$, and the rule switches back to positive terms
> $\tfrac{1}{14}, \tfrac{1}{16}, \dots$ until the sum passes $1.2$ again. The
> reordering thus begins
> $$
> \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{6} + \tfrac{1}{8} + \tfrac{1}{10} + \tfrac{1}{12} - 1 + \tfrac{1}{14} + \tfrac{1}{16} + \cdots,
> $$
> and after each switch the sum sits within one term of $L$ — a gap that tends
> to $0$, so the rearranged series converges to $1.2$.[^lebl-riemann]

$$
% caption: Riemann's rearrangement steered to a target. Same-sign runs push the
% running sum past the target line; the switch back overshoots less each time.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) -- (9.8,0) node[right, black] {terms added};
\draw[black] (0,0) -- (0,3.2);
\draw[acc, thick] (0,2.0) -- (9.8,2.0) node[right, black] {target $L$};
\draw[black, thick]
  (0,0) -- (0.7,0.9) -- (1.4,1.5) -- (2.1,2.55) -- (2.8,1.55) -- (3.5,2.05) -- (4.2,2.35) -- (4.9,1.7) -- (5.6,2.18) -- (6.3,2.28) -- (7.0,1.82) -- (7.7,2.1) -- (8.4,1.9) -- (9.1,2.03);
\foreach \x/\y in {2.1/2.55, 2.8/1.55, 4.2/2.35, 4.9/1.7, 6.3/2.28, 7.0/1.82}
  \fill[black] (\x,\y) circle (1.8pt);
\end{tikzpicture}
$$

## Multiplying series: the Cauchy product

Multiplying two series is not term-by-term; the correct product collects all
cross terms of a given total index.

> **Definition (Cauchy product).** The **Cauchy product** of $\sum_{n=0}^\infty
> a_n$ and $\sum_{n=0}^\infty b_n$ is $\sum_{n=0}^\infty c_n$ with
> $$
> c_n = \sum_{i=0}^n a_i b_{n-i} = a_0 b_n + a_1 b_{n-1} + \cdots + a_n b_0.
> $$

> **Theorem (Mertens).** If $\sum a_n \to A$ and $\sum b_n \to B$ both converge
> and at least one converges absolutely, then the Cauchy product converges to
> $AB$.

The absolute-convergence hypothesis cannot be dropped.

> **Worked example.** Take $a_n = b_n = \frac{(-1)^n}{\sqrt{n+1}}$: each series
> converges by the alternating test but neither absolutely (the absolute series
> is the $p$-series with $p = \tfrac{1}{2}$). Since $i + 1 \le n + 1$ and
> $n - i + 1 \le n + 1$ for $0 \le i \le n$, each factor under the root is at
> most $n + 1$, and the product terms satisfy
> $$
> |c_n| = \sum_{i=0}^n \frac{1}{\sqrt{(i+1)(n-i+1)}} \ge \sum_{i=0}^n \frac{1}{\sqrt{(n+1)(n+1)}} = \frac{n+1}{n+1} = 1,
> $$
> so $c_n$ does not tend to zero and the Cauchy product diverges by the $n$-th
> term test.[^lebl-mertens]

## Summary

| Property | Absolute convergence | Conditional convergence |
| --- | --- | --- |
| $\sum \lvert x_n \rvert$ | converges | diverges |
| implies convergence | yes | is convergence |
| rearrangement | same sum, always | any sum (Riemann) |
| Cauchy product | Mertens applies | may diverge |
| typical test | ratio, root, comparison | alternating series |

Absolute convergence is the well-behaved case: the ratio and root tests detect
it, and rearrangement and the Cauchy product preserve the sum. Conditional
convergence, detected by the alternating series test, depends on the exact order
of the terms; by Riemann's theorem, reordering can produce any sum.

[^lebl-abs]: Lebl, §2.5, Definition 2.5.14 and Proposition 2.5.15.
[^lebl-ratio]: Lebl, §2.5, Proposition 2.5.19.
[^lebl-root]: Lebl, §2.6, Proposition 2.6.1.
[^lebl-alt]: Lebl, §2.6, Proposition 2.6.2.
[^lebl-rearr]: Lebl, §2.6, Proposition 2.6.3.
[^lebl-riemann]: Lebl, §2.6, Example 2.6.4.
[^lebl-mertens]: Lebl, §2.6, Theorem 2.6.5 (Mertens) and Example 2.6.6.