[CALCULUS] Tests for series convergence

Choose and apply the main convergence tests for positive, alternating and power-shaped series.

Most series have no closed-form sum, so this page collects tests that decide whether the partial sums converge without finding their limit. Begin with the $n$ -th term test, then look for a special shape:

geometric or telescoping;
positive terms suited to an integral or comparison;
factorials or exponentials suited to the ratio test;
large powers suited to the root test;
alternating signs suited first to absolute convergence and then to the alternating-series test.

Each test has its own hypotheses and conclusions. An inconclusive limit proves neither convergence nor divergence.

Benchmark positive series with integrals

When the terms are positive and come from a decreasing function, $a_n=f(n)$ , an improper integral measures the same tail behaviour. The resulting $p$ -series test is the benchmark for algebraic decay.

[DEFINITION]

Infinite series

The series $\sum_{n=1}^{\infty}a_n$ converges to $L$ precisely when its partial sums satisfy

\lim_{k\to\infty}s_k=L, \qquad s_k=\sum_{n=1}^{k}a_n.

In that case,

\sum_{n=1}^{\infty}a_n=L.

[THEOREM]

Integral test

Suppose $f$ is positive, continuous and decreasing for $x\ge N$ , where $N$ is a positive integer and $a_n=f(n)$ . Then

\sum_{n=N}^{\infty}a_n

and

\int_N^\infty f(x)\,dx

either both converge or both diverge.

Positivity and monotonicity give, for $m\ge N$ ,

\int_N^{m+1}f(x)\,dx \le\sum_{n=N}^{m}f(n) \le f(N)+\int_N^{m}f(x)\,dx.

The integral and partial sums therefore have the same convergence behaviour.

[THEOREM]

The $p$ -series test

For real $p$ ,

\sum_{n=1}^{\infty}\frac1{n^p} \begin{cases} \text{converges},&p>1,\\ \text{diverges},&p\le1. \end{cases}

When $p\le0$ , the terms do not tend to zero. When $p>0$ , the integral test uses

\int_1^\infty\frac{dx}{x^p},

which is finite exactly when $p>1$ . The special case $p=2$ is evaluated exactly in Euler's Basel sum.

[1]:Integral test

[EXAMPLE]

A logarithmic borderline

For

\sum_{n=2}^{\infty}\frac1{n\ln n},

the corresponding integral is

\int_2^\infty\frac{dx}{x\ln x} =\lim_{b\to\infty}\bigl(\ln(\ln b)-\ln(\ln2)\bigr) =\infty.

Therefore the series diverges.

[exam]

[1]:Integral test

Compare with a known tail

Comparison tests suit terms that can be bounded by, or shown to have the same dominant behaviour as, the terms of a familiar series.

[THEOREM]

Direct comparison test

Suppose $0\le a_n\le b_n$ for all sufficiently large $n$ . Then:

if $\sum b_n$ converges, $\sum a_n$ converges;
if $\sum a_n$ diverges, $\sum b_n$ diverges.

The inequality must point towards the conclusion. A smaller positive series is controlled by a convergent larger series; a larger positive series inherits divergence from a divergent smaller series.

[1]:Infinite series
[2]:Direct comparisons

[EXAMPLE]

Direct comparisons

For

\sum_{n=1}^{\infty}\frac n{n^3+2},

we have

0<\frac n{n^3+2}\le\frac n{n^3}=\frac1{n^2}.

The $p$ -series test proves convergence.

For

\sum_{n=2}^{\infty}\frac1{\sqrt{n^2-2}},

we have $\sqrt{n^2-2}<n$ , hence

\frac1{\sqrt{n^2-2}}>\frac1n.

The given series dominates the divergent harmonic series and therefore diverges.

For

\sum_{n=1}^{\infty}\frac1{\sqrt{n^2+3}},

the direct comparison with $1/n$ points in the wrong direction. Limit comparison gives the right conclusion.

[exam]

[1]:Direct comparison test

Direct comparison needs an inequality with the useful direction. Limit comparison replaces that inequality with a finite positive ratio, which is often easier to calculate.

[THEOREM]

Limit comparison test

If $a_n>0$ and $b_n>0$ for all sufficiently large $n$ and

\lim_{n\to\infty}\frac{a_n}{b_n}=K

with $0<K<\infty$ , then

\sum a_n\text{ converges} \quad\Longleftrightarrow\quad \sum b_n\text{ converges}.

The finite positive ratio means that $a_n$ and $b_n$ are eventually bounded above and below by positive multiples of one another. If the ratio is $0$ , only one-way conclusions are available: convergence of $\sum b_n$ implies convergence of $\sum a_n$ and divergence of $\sum a_n$ implies divergence of $\sum b_n$ .

[EXAMPLE]

Dominant-term comparisons

For

\frac{(2n+1)/(n^3-2)}{1/n^2}\longrightarrow2,

the series

\sum\frac{2n+1}{n^3-2}

converges like $\sum1/n^2$ .

For

\frac{(n-1)/(n^4+1)^{1/3}}{1/n^{1/3}} \longrightarrow1,

the series

\sum\frac{n-1}{(n^4+1)^{1/3}}

diverges like $\sum1/n^{1/3}$ .

Also,

\frac{\dfrac{3n^2+5n}{2^n(n^2+1)}}{3/2^n} \longrightarrow1,

so the series has the same convergence behaviour as a geometric series.

Finally, for

\sum_{n=2}^{\infty}\frac{\ln n}{\sqrt{n^3+1}},

use $\ln n\le n^{1/4}$ for sufficiently large $n$ :

\frac{\ln n}{\sqrt{n^3+1}} \le\frac{n^{1/4}}{n^{3/2}} =\frac1{n^{5/4}}.

The series converges by direct comparison.

[exam]

[1]:Limit comparison test

Choose ratio or root for factorials and powers

The ratio test is designed for factorials and exponentials because consecutive terms cancel their repeated factors. The root test is designed for large $n$ -th powers because taking an $n$ -th root exposes the base. Both tests reduce the question to whether a number is below or above $1$ .

[THEOREM]

Ratio test

For a series of positive terms, suppose

\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L.

Then:

$L<1$ implies convergence;
$L>1$ implies divergence;
$L=1$ gives no conclusion.

For sign-changing terms, apply the test to $|a_n|$ when testing absolute convergence.

[1]:Infinite series
[2]:Factorials and exponentials

[EXAMPLE]

Factorials and exponentials

For $\sum n^2/2^n$ ,

\frac{(n+1)^2/2^{n+1}}{n^2/2^n} =\frac12\left(1+\frac1n\right)^2 \longrightarrow\frac12.

The series converges.

For $\sum1/n!$ ,

\frac{1/(n+1)!}{1/n!}=\frac1{n+1}\longrightarrow0,

so the series converges.

For $\sum n!/2^n$ ,

\frac{(n+1)!/2^{n+1}}{n!/2^n} =\frac{n+1}{2}\longrightarrow\infty.

The terms do not approach zero, so the series diverges.

[routine]

[1]:Ratio test

[THEOREM]

Root test

\lim_{n\to\infty}\sqrt[n]{|a_n|}=L,

then:

$L<1$ implies absolute convergence;
$L>1$ implies divergence;
$L=1$ gives no conclusion.

If the ratio test gives $1$ , switching to the root test does not automatically help: the root test may also be inconclusive.

[EXAMPLE]

Large powers

For

a_n=\left(\frac n{n+1}\right)^{n^2},

we have

\sqrt[n]{a_n} =\left(\frac n{n+1}\right)^n =\left(1+\frac1n\right)^{-n} \longrightarrow e^{-1}<1.

Therefore $\sum a_n$ converges.

For

a_n=(\arctan n)^n,

\sqrt[n]{a_n}=\arctan n\longrightarrow\frac\pi2>1.

Therefore the series diverges.

[exam]

[1]:Root test

Use cancellation carefully

Positive-term tests cannot see cancellation between positive and negative terms, so first test the series of absolute values. If that fails, the alternating-series test may still prove convergence. Such conditional convergence is fragile: rearranging the terms can change the sum.

[DEFINITION]

Absolute and conditional convergence

A series $\sum a_n$ is absolutely convergent if $\sum|a_n|$ converges. It is conditionally convergent if $\sum a_n$ converges but $\sum|a_n|$ diverges.

Absolute convergence controls the series without relying on cancellation. For example, $\sum(-1)^n/n^2$ is absolutely convergent because $\sum1/n^2$ converges, while $\sum(-1)^n/n$ is a candidate for conditional convergence.

[THEOREM]

Absolute convergence implies convergence

If $\sum|a_n|$ converges, then $\sum a_n$ converges.

If $S_n$ are the partial sums, then for $m>n$

|S_m-S_n| =\left|\sum_{k=n+1}^{m}a_k\right| \le\sum_{k=n+1}^{m}|a_k|.

The tails of the convergent series $\sum|a_k|$ become arbitrarily small, so any two sufficiently late partial sums of $\sum a_k$ are arbitrarily close. Completeness of $\mathbb R$ then guarantees that these partial sums approach a finite limit, even when the signs of the terms vary.

[1]:Absolute and conditional convergence

[THEOREM]

Alternating-series test

If $b_n\ge0$ , $b_{n+1}\le b_n$ and $b_n\to0$ , then

\sum_{n=0}^{\infty}(-1)^n b_n

converges.

The successive partial sums lie on opposite sides of the limiting value and close in on it. The magnitudes need only be eventually decreasing, since finitely many initial terms do not affect convergence.

[EXAMPLE]

Conditional convergence of the alternating harmonic series

The positive magnitudes $1/n$ decrease to zero, so

\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n

converges by the alternating-series test. But its absolute series is the harmonic series, which diverges. The alternating harmonic series is therefore conditionally convergent.

[exam]

[THEOREM]

Rearranging a series

Every rearrangement of an absolutely convergent series has the same sum. A conditionally convergent series can be rearranged to converge to any prescribed real number, or to diverge to $+\infty$ or $-\infty$ .

The second statement is the Riemann rearrangement theorem.

[EXAMPLE]

Rearranging the alternating harmonic series

Let

S=1-\frac12+\frac13-\frac14+\frac15-\cdots.

Take one positive term followed by two negative terms:

1-\frac12-\frac14+\frac13-\frac16-\frac18 +\frac15-\frac1{10}-\frac1{12}+\cdots.

Grouping in blocks gives

\begin{aligned} &\left(1-\frac12-\frac14\right) +\left(\frac13-\frac16-\frac18\right)+\cdots\\ &=\sum_{k=1}^{\infty} \left(\frac1{2k-1}-\frac1{4k-2}-\frac1{4k}\right) =\frac S2. \end{aligned}

The changed sum is a consequence of conditional convergence, not an algebraic contradiction.

[challenge]

[1]:Rearranging a series

The next page uses these tests to decide where a Taylor series converges. For a power series, the ratio and root tests locate the radius, while comparison, integral and alternating tests settle the endpoints.

Tests for series convergence

Infinite series

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