Convergence tests for series without fog

An infinite series is not a long plus sign. It is a sequence of partial sums. The series $\sum a_n$ converges when $s_N=a_1+\cdots+a_N$ approaches a finite number. It diverges when that sequence of sums does not. Tests exist because you often cannot write $s_N$ in closed form. A good test answers one question only. Do those partial sums settle?

This page treats three workhorses: the comparison test, the ratio test, and the root test. They are cousins, not rivals. Comparison borrows a series you already understand. Ratio and root ask whether your terms eventually look geometric. Read this after sequences, series, and when infinite sums make sense, not instead of it. The topic map is calculus foundations. If you want the habit of writing the hypothesis before the verdict, use real analysis and proof habits and the intro to proofs pathway. A course shaped order that includes this page is the AP Calculus AB bridge.

The term test, which never proves convergence

If $a_n$ does not go to $0$, the series diverges. Partial sums cannot settle if each new term stays a fixed distance away from zero. That is the term test, sometimes called the divergence test. It never proves convergence. Terms can go to zero slowly, and the sum can still run off.

The harmonic series $\sum 1/n$ is the famous example. The terms vanish. The partial sums do not. A picture that does not need a named integral test: on each interval $[n,n+1]$, the function $1/x$ sits under the constant $1/n$. So

$$\sum_{n=1}^{N}\frac{1}{n} > \int_1^{N+1}\frac{1}{x}\,dx=\ln(N+1).$$

The natural log grows without a bound, so the partial sums do too. Terms going to zero were required and not enough. Any later test that "succeeds" on a series whose terms refuse to go to zero has been misapplied.

Positive term series are the cleanest setting for the three tests below. Partial sums of nonnegative terms only increase, so they either settle or run to infinity. There is no oscillation to hide the answer. If terms change sign, ask first about absolute convergence, meaning convergence of $\sum |a_n|$. Absolute convergence implies ordinary convergence, because the partial sums of $a_n$ are then squeezed by a convergent series of absolute values. The converse can fail. A short note on that failure is later on this page. It is not a reason to skip the absolute question.

Comparison: sandwich against a known series

Suppose $0\le a_n\le b_n$ for all large $n$. If $\sum b_n$ converges, then $\sum a_n$ converges. The partial sums of $a_n$ are increasing, and they stay below a convergent pile plus a finite head, so they settle. If instead $a_n\ge c_n\ge 0$ and $\sum c_n$ diverges, then $\sum a_n$ diverges. You cannot stay above a divergent stack of nonnegative blocks and still have a finite sum.

The usual comparison partners are geometric series and $p$ series. A geometric series $\sum ar^{n}$ with $a\neq 0$ converges when $|r|<1$ and diverges when $|r|\ge 1$. A $p$ series $\sum 1/n^p$ converges when $p>1$ and diverges when $p\le 1$. The harmonic series is the case $p=1$. You must actually check an inequality, or a limit form of it. "It looks like $1/n^2$" is a hint about which partner to try. It is not yet a proof.

Example For $a_n=1/(n^2+3n+1)$, note that $n^2+3n+1>n^2$ when $n>0$, so $0<a_n<1/n^2$. The $p$ series with $p=2$ converges, so this series converges. The inequality is doing the work. A looser bound such as $a_n<1$ is true and useless, because $\sum 1$ diverges and an upper bound by a divergent series proves nothing.

Limit comparison is the same idea with less inequality pain. If $a_n>0$, $b_n>0$, and $a_n/b_n$ approaches a limit $L$ with $0<L<\infty$, then $\sum a_n$ and $\sum b_n$ converge or diverge together. The terms are eventually the same size up to a constant factor, and a constant factor does not change convergence for positive terms.

Example Let $a_n=(2n+1)/(n^2+5)$. Compare with $b_n=1/n$. Then

$$\frac{a_n}{b_n}=\frac{n(2n+1)}{n^2+5}=\frac{2n^2+n}{n^2+5}\to 2.$$

The limit is positive and finite, and $\sum 1/n$ diverges, so $\sum a_n$ diverges. A direct inequality reaches the same verdict. For $n\ge 1$, the numerator $2n+1$ is at least $2n$, and the denominator $n^2+5$ is at most $6n^2$, so $a_n\ge (2n)/(6n^2)=1/(3n)$. A constant times the harmonic series still diverges.

Ratio: ask if the terms become geometric

For $a_n\neq 0$, form $L=\lim |a_{n+1}/a_n|$ when the limit exists. The ratio test says:

• If $L<1$, the series converges absolutely.
• If $L>1$, or if the ratio tends to infinity, the series diverges.
• If $L=1$, the test is silent.

Why $L<1$ wins: eventually $|a_{n+1}|\le r|a_n|$ for some fixed $r$ with $L<r<1$. From there on, the tail is dominated by a geometric series with ratio $r$. Comparison finishes the argument. If $L>1$, the terms eventually grow, so they cannot go to zero, and the term test already says the series diverges.

The silent case $L=1$ includes both $\sum 1/n$ and $\sum 1/n^2$. For $1/n$ the successive ratio is $n/(n+1)$, which approaches $1$. For $1/n^2$ the successive ratio is $n^2/(n+1)^2$, which also approaches $1$. The ratio test cannot see the difference between $p=1$ and $p=2$. That is not a defect in your algebra. It is a limit of the test. Switch to comparison. Staring at the same limit will not make it drop below $1$.

Example For $a_n=n^2/3^n$,

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{n^2}\cdot\frac{1}{3}\to\frac{1}{3}<1.$$

The polynomial in the numerator is helpless against the exponential in the denominator. The series converges absolutely. Factorials in a denominator produce the same kind of verdict for many Taylor tails, because a factorial eventually beats any fixed base to a power. Compute the ratio before you announce that. The slogan is not the test.

Root: a closer cousin of a geometric series

Form $\rho=\limsup\sqrt[n]{|a_n|}$, or an ordinary limit when it exists. The root test uses the same three way split as the ratio test, with $\rho$ in place of $L$. If the root settles below $1$, then $|a_n|\le r^n$ eventually for some $r$ with $\rho<r<1$, and comparison with a geometric series applies again. If the root settles above $1$, the terms do not go to zero.

In a first course the root is kindest when the whole term is already an $n$th power. For $a_n=\bigl(n/(2n+1)\bigr)^n$,

$$\sqrt[n]{a_n}=\frac{n}{2n+1}\to\frac{1}{2}<1.$$

The series converges. The ratio of successive terms is messier and is unnecessary once the root is this clean.

Root can also decide a series whose successive ratios refuse to settle. Let $a_n=2^{-n}$ when $n$ is even, and $a_n=3^{-n}$ when $n$ is odd. Along even indices the next ratio collapses toward $0$. Along odd indices the next ratio blows up. The limit of $|a_{n+1}/a_n|$ does not exist, so the ratio test in the "limit exists" form does not speak. The roots are $1/2$ on even indices and $1/3$ on odd indices, so the largest limit point of the roots is $1/2$, which is below $1$. The root test says the series converges. A direct comparison sees it too, since $a_n\le 2^{-n}$ for every $n$. The example is here so you do not treat a missing ratio limit as a mystery. It is a signal to change tools.

When the ratio limit exists, the root limit exists and they are equal. You may choose whichever expression looks cleaner. Silence at $1$ is still silence for both tests. Neither test is a stronger personality. They fail together on $p$ series.

Signs, absolute convergence, and what these tests do not see

If $\sum |a_n|$ converges, then $\sum a_n$ converges. The proof idea is that the positive part and the negative part are each smaller than the absolute series, so each settles, and their difference settles. You may apply comparison, ratio, or root to $|a_n|$ and then transfer the verdict.

The transfer does not run backward. Let $a_n=(-1)^{n+1}/n$, the alternating harmonic series. The absolute series is the harmonic series, which diverges. The alternating series itself converges. One honest reason: the partial sums with an even number of terms increase, the partial sums with an odd number of terms decrease, and the gap between those two strands is the next term $1/n$, which goes to zero. A course will call this the alternating series test. The hypotheses to remember are that the absolute values decrease to zero. Drop the decrease, and a sign pattern alone does not save you.

For this alternating series the ratio of absolute values is $n/(n+1)$, which approaches $1$. Ratio and root stay silent. Comparison on the absolute values proves divergence of $\sum |a_n|$, which is real information and is not a verdict about $\sum a_n$. The name for "converges, but not absolutely" is conditional convergence. These three tests are built to see absolute convergence. When they are silent and the signs alternate, you need the alternating test, or a different closed form. This page does not pretend to be that later chapter. It does insist that silence not be recorded as convergence.

A small decision map

1. Do the terms fail to go to zero? The series diverges. Stop.
2. Do you see a geometric ratio or a $p$ series after a simple bound? Compare, and check the direction.
3. Do you see a factorial, an exponential, or a rational function times an exponential? Try the ratio.
4. Is the whole term an $n$th power, or do successive ratios fail to settle? Try the root.
5. If the test returns $1$, change tools. Do not stare at the same limit.
6. If the signs alternate and the absolute series diverges, ask the alternating test. Do not force a ratio limit of $1$ to vote.

How to know you understood

1. State the term test in one sentence, and give the harmonic series as a case where terms go to zero and the series still diverges.
2. For $1/(n^2+3n+1)$, name the comparison partner and the direction of the inequality.
3. For $n^2/3^n$, compute the successive ratio limit and say why it is enough.
4. For both $1/n$ and $1/n^2$, show that the ratio limit is $1$, and say which series converges.
5. Say what conditional convergence means, using the alternating harmonic series, and say which of the three tests fails to see it.

A finite sum is a picture, not a verdict

The arithmetic and geometric sequence calculator reports a term $a_n$ and a finite sum $S_n$. It does not test an arbitrary series. Harmonic terms are not an arithmetic or geometric list, so they do not belong in that form. A geometric series does. A Taylor polynomial is another finite object: the Taylor polynomial calculator writes the degree-n truncation and a Lagrange bound. The factorial in those denominators is why the ratio test is the natural test for many Taylor tails. The bound is not itself a convergence proof for the infinite series.

Watching $S_n$ level off can support a guess before you pick a test. A large $S_n$ is not divergence, and a calm $S_n$ is not convergence. Partial sums of the harmonic series grow slowly. A short window looks tame and the series still diverges. Trust the hypothesis of the test you invoked, then the verdict.

If the limit language under "approaches" is the actual gap, read epsilon delta intuition before you stack more tests. The numerical limit estimator can sample a successive ratio such as $n/(n+1)$ and show it nearing $1$. That sample is still not the ratio test. The test is the limit, together with the three way split above. Glossary lines that repay a second look: series, partial sum, convergence.

Citations and further reading

• OpenStax, Calculus Volume 2, comparison tests and ratio and root tests. Comparison · Ratio and root
• Paul’s Online Math Notes, series convergence. tutorial.math.lamar.edu
• Abbott, S. Understanding Analysis, chapters on sequences and series, for epsilon arguments behind the slogans.
• MIT OpenCourseWare, 18.01SC, infinite series unit. ocw.mit.edu