---
title: The Convergence Tests
module: Infinite Sequences and Series
moduleNumber: 9
lessonNumber: 3
order: 903
summary: >
  The comparison, alternating-series, ratio, and root tests decide convergence
  without a closed-form partial sum. Absolute convergence is stronger than
  conditional convergence and is preserved under rearrangement; an alternating
  series errs by less than its first omitted term. A test is chosen from the
  shape of the general term.
topics: [Infinite Sequences and Series]
sources:
  - book: Stewart
    ref: "Ch. 11; §11.4 The Comparison Tests; §11.5 Alternating Series"
  - book: Stewart
    ref: "§11.6 Absolute Convergence and the Ratio and Root Tests; §11.7 Strategy for Testing Series"
---

The [Integral Test](/calculus/sequences-and-series/series-and-the-integral-test)
needs an antiderivative, which most series do not offer. The comparison,
alternating, ratio, and root tests need none: each reads convergence off the
general term directly, by comparison with a known series, by the sign pattern, or
by the ratio of consecutive terms.

## The Comparison Test

If a series of positive terms is dominated term by term by a convergent series,
its partial sums are bounded, so it converges too. The reverse comparison forces
divergence.

> **Theorem (Comparison Test).** Let $\sum a_n$ and $\sum b_n$ have positive
> terms.
> - If $\sum b_n$ converges and $a_n \le b_n$ for all $n$, then $\sum a_n$
>   converges.
> - If $\sum b_n$ diverges and $a_n \ge b_n$ for all $n$, then $\sum a_n$
>   diverges.

The proof is the Monotonic Sequence Theorem again. With $s_n = \sum_{i\le n} a_i$
and $t_n = \sum_{i\le n} b_i$, positivity makes both increasing. If $\sum b_n = t$
then $s_n \le t_n \le t$, so $\{s_n\}$ is bounded above and converges. The
divergent case is the contrapositive.

$$
% caption: Term-by-term comparison: the tested series (dark) sits under a
% convergent series (light) whose terms cap it at every index, so its partial
% sums stay bounded.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (0,0) -- (7.4,0) node[anchor=north east] {$n$};
\draw[black, ->] (0,0) -- (0,3.2) node[anchor=north east] {};
% b_n = 1/2^n dominating (light), a_n = 1/(2^n+1) tested (dark), scaled
\foreach \n/\yb/\ya in {1/1.5/1.0, 2/0.75/0.6, 3/0.375/0.33, 4/0.19/0.17, 5/0.094/0.09} {
  \fill[black!10] (\n-0.34,0) rectangle (\n-0.02, \yb);
  \draw[black] (\n-0.34,0) rectangle (\n-0.02, \yb);
  \fill[acc!55] (\n-0.34,0) rectangle (\n-0.02, \ya);
}
\node[black!70, anchor=west] at (2.2,1.9) {$b_n = 1/2^n$ (convergent)};
\node[acc, anchor=west] at (2.2,1.4) {$a_n = 1/(2^n+1) \le b_n$};
\end{tikzpicture}
$$

For comparison the two standard families supply the reference series:

- a **$p$-series** $\sum 1/n^p$ converges iff $p > 1$;
- a **geometric series** $\sum ar^{n-1}$ converges iff $|r| < 1$.

> **Worked example.** Test $\displaystyle\sum_{n=1}^{\infty}
> \frac{5}{2n^2 + 4n + 3}$ for convergence.
>
> Dropping the two smaller denominator terms only enlarges the fraction:
>
> $$
> \frac{5}{2n^2 + 4n + 3} < \frac{5}{2n^2}.
> $$
>
> The dominating series $\sum 5/(2n^2) = \tfrac{5}{2}\sum 1/n^2$ is a constant
> times a convergent $p$-series ($p = 2$). By the first part of the Comparison
> Test, the given series converges.

The direction matters. To conclude convergence the terms must be
_smaller_ than a convergent series; being smaller than a divergent one says
nothing. The next test removes that restriction.

## The Limit Comparison Test

The inequality version fails on series like $\sum 1/(2^n - 1)$: the natural
comparison $1/(2^n - 1) > 1/2^n$ points the wrong way. Comparing _limits_ of the
ratio sidesteps the issue.

> **Theorem (Limit Comparison Test).** Let $\sum a_n$ and $\sum b_n$ have
> positive terms. If
> $$
> \lim_{n\to\infty} \frac{a_n}{b_n} = c
> $$
> with $0 < c < \infty$, then both series converge or both diverge.

If the ratio settles to a positive finite $c$, the two series are the same size
for large $n$, so they converge or diverge together. For $\sum 1/(2^n - 1)$, compare with
$b_n = 1/2^n$:

$$
\lim_{n\to\infty} \frac{1/(2^n - 1)}{1/2^n}
= \lim_{n\to\infty} \frac{2^n}{2^n - 1}
= \lim_{n\to\infty} \frac{1}{1 - 2^{-n}} = 1,
$$

and since $\sum 1/2^n$ converges, so does the given series. In practice, build
$b_n$ by keeping only the highest powers of $n$ in numerator and denominator.

> **Worked example.** Test $\displaystyle\sum_{n=1}^{\infty}
> \frac{2n^2 + 3n}{\sqrt{5 + n^5}}$ for convergence.
>
> Keep the dominant power in each part: the numerator behaves like $2n^2$ and
> $\sqrt{5 + n^5}$ like $n^{5/2}$, so take $b_n = 2n^2/n^{5/2} = 2/n^{1/2}$,
> i.e. $b_n = 2/\sqrt n$. The ratio limit is
>
> $$
> \lim_{n\to\infty}\frac{a_n}{b_n}
> = \lim_{n\to\infty}\frac{2n^2 + 3n}{\sqrt{5 + n^5}}\cdot\frac{\sqrt n}{2}
> = \lim_{n\to\infty}\frac{n^{5/2}(2 + 3/n)}{2n^{5/2}\sqrt{5/n^5 + 1}} = 1,
> $$
>
> a positive finite number. Since $\sum 2/\sqrt n$ is a divergent $p$-series
> ($p = \tfrac{1}{2}$), the given series diverges as well.

## Alternating series

A series whose terms alternate in sign,

$$
\sum_{n=1}^{\infty} (-1)^{n-1} b_n = b_1 - b_2 + b_3 - b_4 + \cdots, \qquad b_n > 0,
$$

converges under a light condition: the magnitudes need only decrease
to zero.

> **Theorem (Alternating Series Test).** If $b_n > 0$ satisfies
> (i) $b_{n+1} \le b_n$ for all $n$ and (ii) $\lim_{n\to\infty} b_n = 0$, then
> $\sum (-1)^{n-1} b_n$ converges.

The partial sums step right by $b_1$, left by $b_2$, right by $b_3$, and so on.
Because the steps shrink, the even partial sums $s_2, s_4, s_6, \ldots$ increase
while the odd ones $s_1, s_3, s_5, \ldots$ decrease, and the two bracket a common
limit $s$ from below and above.

$$
% caption: Partial sums of an alternating series close in on the sum $s$ from
% both sides: even sums rise, odd sums fall, and each lies within one term of
% $s$.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{red}{HTML}{C0392B}
\draw[black, ->] (0,0) -- (10,0) node[anchor=north west] {};
\draw[acc, dashed] (6.3,-0.9) -- (6.3,0.9);
\node[acc, anchor=south] at (6.3,0.9) {$s$};
% odd partial sums (from above), even (from below)
\foreach \x/\lab in {9/{$s_1$}, 5.1/{$s_3$}, 7.1/{$s_5$}} {
  \fill (\x,0) circle (2.4pt);
  \node[anchor=south, font=\scriptsize] at (\x,0.12) {\lab};
}
\foreach \x/\lab in {2.4/{$s_2$}, 5.7/{$s_4$}, 6.0/{$s_6$}} {
  \fill (\x,0) circle (2.4pt);
  \node[anchor=north, font=\scriptsize] at (\x,-0.12) {\lab};
}
\draw[black, |-|] (2.4,-0.55) -- (9,-0.55);
\node[black, anchor=north, font=\scriptsize] at (5.7,-0.55) {$s$ lies between consecutive sums};
\end{tikzpicture}
$$

Formally, the even sums are increasing and bounded above by $b_1$, so they
converge to some $s$; the odd sums converge to the same $s$ because
$s_{2n+1} = s_{2n} + b_{2n+1}$ and $b_{2n+1} \to 0$. The **alternating harmonic
series**

$$
1 - \tfrac{1}{2} + \tfrac{1}{3} - \tfrac{1}{4} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}
$$

converges by this test (with $b_n = 1/n$), even though the harmonic series
itself diverges. Its sum is $\ln 2$.

When the monotonicity of $b_n$ is not obvious, test the related function's
derivative. For $b_n = n^2/(n^3 + 1)$, the function $f(x) = x^2/(x^3+1)$ has
$f'(x) = x(2 - x^3)/(x^3+1)^2 < 0$ once $x > \sqrt[3]{2}$, so $b_n$ eventually
decreases — enough for the test.

### Estimating an alternating sum

The bracketing picture also yields an error bound: since $s$ lies between
$s_n$ and $s_{n+1}$, the error in stopping at $s_n$ is smaller than the first
term left out.

> **Theorem (Alternating Series Estimation).** If $s = \sum (-1)^{n-1} b_n$ meets
> the two conditions of the Alternating Series Test, then
> $$
> |R_n| = |s - s_n| \le b_{n+1}.
> $$

> **Worked example.** Approximate $\displaystyle\sum_{n=0}^{\infty}
> \frac{(-1)^n}{n!}$ to three decimal places.
>
> The magnitudes $b_n = 1/n!$ decrease to $0$, so the estimation theorem applies.
> The first omitted term after $s_6$ is $b_7 = 1/5040 < 0.0002$, which sets the
> error, so the partial sum through $n = 6$ already fixes three decimals:
>
> $$
> s_6 = 1 - 1 + \tfrac{1}{2} - \tfrac{1}{6} + \tfrac{1}{24} - \tfrac{1}{120}
> + \tfrac{1}{720} \approx 0.368056,
> $$
>
> so $s \approx 0.368$. The exact sum is $e^{-1}$.

The rule "error below the first omitted term" is special to alternating series
meeting these two conditions; it does not apply to series in general.

## Absolute convergence, the Ratio Test, and the Root Test

For series with irregular signs, test the absolute values.

> **Definition (Absolute and conditional convergence).** $\sum a_n$ is
> **absolutely convergent** if $\sum |a_n|$ converges. It is **conditionally
> convergent** if it converges but $\sum |a_n|$ diverges.

> **Theorem.** Absolute convergence implies convergence.

The proof uses $0 \le a_n + |a_n| \le 2|a_n|$: if $\sum |a_n|$ converges then
$\sum(a_n + |a_n|)$ converges by comparison, and $\sum a_n = \sum(a_n + |a_n|) -
\sum |a_n|$ is a difference of convergent series. The alternating harmonic series
converges but $\sum 1/n$ diverges, so it is conditionally convergent. Absolute
convergence is the stronger property: it is preserved under rearrangement,
whereas a conditionally convergent series can be reordered to sum to _any_ real
number (Riemann's rearrangement theorem).

$$
% caption: Absolute convergence is a strict subset of convergence. The outer
% region holds every convergent series; the inner disk holds the absolutely
% convergent ones; the ring between them holds the conditionally convergent
% series, such as the alternating harmonic series.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) ellipse (3.6 and 2.3);
\fill[acc!12] (-0.7,0) ellipse (1.9 and 1.5);
\draw[acc] (-0.7,0) ellipse (1.9 and 1.5);
\node[acc, align=center] at (-0.7,0) {absolutely\\convergent};
\node[black!70, anchor=east] at (3.5,1.55) {convergent};
\node[black, align=center, font=\scriptsize] at (2.15,-0.35) {conditionally\\convergent};
\fill[black] (2.05,0.35) circle (1.6pt);
\node[black, anchor=west, font=\scriptsize] at (2.2,0.55) {alt. harmonic};
\end{tikzpicture}
$$

The next two tests detect absolute convergence by measuring how fast $|a_n|$
decays.

> **Theorem (Ratio Test).** Let $L = \lim_{n\to\infty} \left|
> \dfrac{a_{n+1}}{a_n} \right|$.
> - If $L < 1$, $\sum a_n$ converges absolutely.
> - If $L > 1$ (or $L = \infty$), $\sum a_n$ diverges.
> - If $L = 1$, the test is inconclusive.

The idea is comparison with a geometric series. If $L < 1$, pick $r$ with
$L < r < 1$; past some $N$ the ratios stay below $r$, so $|a_{N+k}| \le |a_N|
r^k$, and the tail is dominated by a convergent geometric series. If $L > 1$ the
terms eventually grow, so $a_n \not\to 0$ and the Test for Divergence applies.

$$
% caption: The Ratio-Test limit $L$ decides on a number line: $L < 1$ forces
% absolute convergence, $L > 1$ forces divergence, and $L = 1$ leaves the
% question open.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{red}{HTML}{C0392B}
\draw[acc, line width=2pt] (0,0) -- (5,0);
\draw[red!70, line width=2pt] (5,0) -- (9,0);
\foreach \x in {0,5} \draw[black] (\x,0.12) -- (\x,-0.12);
\node[anchor=north] at (0,-0.15) {$0$};
\node[anchor=north] at (5,-0.15) {$L = 1$};
\node[anchor=north] at (9,-0.15) {$L$};
\node[acc, anchor=south] at (2.5,0.2) {converges absolutely};
\node[red!80, anchor=south] at (7,0.2) {diverges};
\fill[black] (5,0) circle (2.6pt);
\node[black, anchor=south] at (5,0.55) {inconclusive};
\end{tikzpicture}
$$

The Ratio Test is most effective when $a_n$ contains factorials or constants
raised to the $n$th power, where successive terms cancel cleanly. For
$\sum (-1)^n n^3/3^n$,

$$
\left|\frac{a_{n+1}}{a_n}\right|
= \frac{(n+1)^3}{3^{n+1}} \cdot \frac{3^n}{n^3}
= \frac{1}{3}\Big(1 + \frac{1}{n}\Big)^3 \to \frac{1}{3} < 1,
$$

so the series converges absolutely.

> **Worked example.** Test $\displaystyle\sum_{n=1}^{\infty} \frac{n^n}{n!}$ for
> convergence.
>
> Form the ratio of successive terms and cancel the factorials against the
> powers:
>
> $$
> \frac{a_{n+1}}{a_n}
> = \frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}
> = \frac{(n+1)^{n+1}}{(n+1)\,n^n}
> = \frac{(n+1)^n}{n^n}
> = \left(1 + \frac{1}{n}\right)^n.
> $$
>
> This tends to $e \approx 2.718 > 1$, so the Ratio Test gives divergence. (The
> terms do not tend to $0$, consistent with the verdict.)

The test is useless on rational functions of $n$: for every $p$-series the ratio
tends to $1$.

When the general term is an $n$th power, the **Root Test** is cleaner.

> **Theorem (Root Test).** Let $L = \lim_{n\to\infty} \sqrt[n]{|a_n|}$. The same
> three cases hold: $L < 1$ converges absolutely, $L > 1$ diverges, $L = 1$ is
> inconclusive.

For $\sum \big((2n+3)/(3n+2)\big)^n$, taking the $n$th root gives
$(2n+3)/(3n+2) \to 2/3 < 1$, so the series converges. If the Ratio Test gives
$L = 1$, the Root Test gives $L = 1$ as well, and conversely; the two are
inconclusive on the same series.

## A strategy for testing series

There is no fixed order of tests to try; classify the
series by the _form_ of its general term and pick the matching test.

$$
% caption: Routing a series to a test by the shape of its general term $a_n$,
% from a quick divergence check down to comparison, ratio, root, and integral.
\begin{tikzpicture}[scale=1.0, font=\footnotesize,
  box/.style={draw, minimum width=33mm, minimum height=8mm, align=center, font=\scriptsize}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box, draw=acc, text=acc] (start) at (0,6) {general term $a_n$};
\node[box] (div) at (0,4.7) {$a_n \not\to 0$?  Test for Divergence};
\node[box] (pg) at (0,3.4) {$p$-series or geometric?  known result};
\node[box] (cmp) at (0,2.1) {rational / algebraic in $n$?  comparison};
\node[box] (alt) at (0,0.8) {alternating $(-1)^n b_n$?  Alternating Series};
\node[box] (rat) at (5.4,3.4) {factorials or $c^n$?  Ratio Test};
\node[box] (root) at (5.4,2.1) {$a_n = (b_n)^n$?  Root Test};
\node[box] (int) at (5.4,0.8) {$a_n = f(n)$, $\int f$ easy?  Integral Test};
\draw[->, black] (start) -- (div);
\draw[->, black] (div) -- (pg);
\draw[->, black] (pg) -- (cmp);
\draw[->, black] (cmp) -- (alt);
\draw[->, black] (pg.east) to[bend left=12] (rat.west);
\draw[->, black] (cmp.east) to[bend left=8] (root.west);
\draw[->, black] (alt.east) to[bend left=8] (int.west);
\end{tikzpicture}
$$

The classification in words:

- **$a_n \not\to 0$ at a glance:** Test for Divergence.
- **$\sum 1/n^p$:** $p$-series. **$\sum ar^{n-1}$:** geometric.
- **$a_n$ rational or algebraic in $n$:** comparison or limit comparison with a
  $p$-series, choosing $p$ from the highest powers.
- **Alternating signs:** Alternating Series Test.
- **Factorials or $n$th powers of a constant:** Ratio Test.
- **$a_n = (b_n)^n$:** Root Test.
- **$a_n = f(n)$ with $\int_1^{\infty} f$ tractable:** Integral Test.

| Test | Best for | Key quantity |
| --- | --- | --- |
| Divergence | any, quick screen | $\lim a_n$ |
| Comparison / Limit Comparison | rational, algebraic $a_n$ | ratio to a $p$-series |
| Alternating Series | $(-1)^n b_n$ | $b_n \downarrow 0$; error $\le b_{n+1}$ |
| Ratio | factorials, $c^n$ | $\lim \lvert a_{n+1}/a_n \rvert$ |
| Root | $(b_n)^n$ | $\lim \sqrt[n]{\lvert a_n \rvert}$ |