Comparison Test for Series: Convergence & Examples

What Is The Comparison Test?

The Comparison Test is a rule for deciding whether an infinite series of positive terms converges (adds up to a finite number) or diverges (grows without bound), by comparing its terms to those of a second series whose behaviour is already known. It only applies when the terms are non-negative, and it comes in two directions.

Let $\sum a_n$ be the series you want to judge, and let $\sum b_n$ be a benchmark series you already understand.

$$\text{If } 0 \le a_n \le b_n \text{ for all large } n \text{ and } \sum b_n \text{ converges, then } \sum a_n \text{ converges.}$$

$$\text{If } a_n \ge b_n \ge 0 \text{ for all large } n \text{ and } \sum b_n \text{ diverges, then } \sum a_n \text{ diverges.}$$

Read plainly: a series trapped underneath a convergent one must also converge, and a series sitting above a divergent one must also diverge. The phrase "for all large $n$" matters, because the first few terms never change whether a series converges; only the tail does. For the wider map of when infinite sums settle to a value, see convergence and divergence of series, and for the object itself, the infinite series overview.

Which Series Do You Compare Against?

The Comparison Test is only as good as the benchmark you bring to it. Two families cover almost every first-year problem, because their convergence is settled by a single glance at one number.

Table: The standard benchmark series used with the Comparison Test.

The harmonic series is just the $p = 1$ edge of the p-series family, and it is the benchmark that proves most divergences. The geometric series, whose closed form comes straight from geometric progressions, is the benchmark for anything with a constant base raised to $n$. Learn these three cold, because the whole test rests on recognising which one your series resembles.

How Do You Prove Convergence With The Comparison Test?

The idea is the same picture as the hook: a bounded staircase. When every term is positive, the partial sums $S_N = a_1 + a_2 + \dots + a_N$ climb as $N$ grows, so they form an increasing sequence. An increasing sequence that never rises above a fixed ceiling has to settle to a limit. The benchmark supplies the ceiling.

If $0 \le a_n \le b_n$ and $\sum b_n$ converges to some finite total $B$, then every partial sum of $\sum a_n$ is trapped:

$$S_N = \sum_{n=1}^{N} a_n ; \le ; \sum_{n=1}^{N} b_n ; \le ; B.$$

The partial sums of $\sum a_n$ are increasing and bounded above by $B$, so they converge. That is the entire argument. For divergence, flip it: if $a_n \ge b_n \ge 0$ and the benchmark partial sums grow without bound, then the larger partial sums of $\sum a_n$ are dragged up with them, so $\sum a_n$ diverges as well.

How Do You Use The Comparison Test? Worked Examples

Each example states the benchmark, checks the inequality direction, and names the conclusion. Every benchmark used is a p-series or geometric series from the table above.

Example 1: A series that converges (compare to a p-series).

Test $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2 + n}$.

Since $n^2 + n > n^2 > 0$ for every $n \ge 1$, making the denominator larger makes the fraction smaller:

$$0 < \frac{1}{n^2 + n} < \frac{1}{n^2}.$$

The benchmark $\sum \frac{1}{n^2}$ is a p-series with $p = 2 > 1$, so it converges. The given series is trapped underneath it, so by the Comparison Test it converges too. As a bonus check, $\frac{1}{n^2+n} = \frac{1}{n} - \frac{1}{n+1}$ telescopes, and the total is exactly $1$, confirming a finite sum.

Final answer: $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2 + n}$ converges.

Example 2: A series that converges (compare to a geometric series).

Test $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^n + 1}$.

Because $2^n + 1 > 2^n > 0$, again the larger denominator gives a smaller fraction:

$$0 < \frac{1}{2^n + 1} < \frac{1}{2^n} = \left(\frac{1}{2}\right)^{n}.$$

The benchmark $\sum \left(\tfrac{1}{2}\right)^{n}$ is geometric with ratio $r = \tfrac{1}{2}$, and $\lvert r\rvert < 1$, so it converges (its total is $1$). The given series lies below it, so it converges by the Comparison Test.

Final answer: $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^n + 1}$ converges.

Example 3: A series that diverges (compare to the harmonic series).

Test $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n - 1}$.

Here you need the divergence direction, so bound the series from below by something known to diverge. Since $n - 1 < n$ for $n \ge 2$, the smaller denominator gives a larger fraction:

$$\frac{1}{n - 1} \ge \frac{1}{n} > 0.$$

The benchmark $\sum \frac{1}{n}$ is the harmonic series, which diverges. The given series sits above a divergent series, so by the Comparison Test it diverges.

Final answer: $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n - 1}$ diverges.

Example 4: A series that diverges (a logarithm over $n$).

Test $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n}$.

For every $n \ge 3$ the logarithm satisfies $\ln n > 1$, so the numerator only helps the series grow:

$$\frac{\ln n}{n} \ge \frac{1}{n} > 0 \qquad (n \ge 3).$$

The benchmark $\sum \frac{1}{n}$ diverges, and the given terms are eventually at least as large, so by the Comparison Test the series diverges. The two starting terms at $n = 1, 2$ do not matter, because only the tail of the series decides convergence.

Final answer: $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n}$ diverges.

When Should You Switch To The Limit Comparison Test?

The Comparison Test needs a clean inequality pointing the right way, and sometimes the obvious benchmark points the wrong way. Consider

$$\sum_{n=2}^{\infty}\frac{1}{n^2 - 1}.$$

The natural benchmark is $\frac{1}{n^2}$, but $n^2 - 1 < n^2$ makes $\frac{1}{n^2 - 1} > \frac{1}{n^2}$. That is the divergence-direction inequality, yet the benchmark converges, so the Comparison Test gives no conclusion here. Rather than hunt for a cleverer bound, switch to the limit comparison test, which only asks that the terms behave alike in the limit:

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

The limit is a positive finite number, so the two series share a fate. Since $\sum \frac{1}{n^2}$ converges, $\sum \frac{1}{n^2 - 1}$ converges as well. Use the direct Comparison Test when the inequality is obvious, and reach for the Limit Comparison Test when the algebra of the inequality is awkward but the dominant behaviour is clear.

Why Does The Comparison Test Work?

The test is not a trick; it is the monotone bounded convergence principle for sequences wearing a series costume.

• Positive terms make partial sums climb. When every $a_n \ge 0$, adding another term can only push the running total up or leave it flat. The partial sums form an increasing sequence, so they either rise to a finite limit or run off to infinity, with nothing in between.

• A finite benchmark is a ceiling. Bounding $a_n$ above by a convergent $b_n$ caps every partial sum by the benchmark's finite total. An increasing sequence with a ceiling must settle, which is convergence.

• A divergent benchmark is a floor that rises forever. Bounding $a_n$ below by a divergent $b_n$ forces the partial sums past every level the benchmark passes, so they cannot settle.

This is why the positivity condition is not optional. If terms can be negative, the partial sums may rise and fall instead of climbing steadily, and a ceiling no longer forces a limit. For series that alternate in sign, a different tool such as the divergence test or an alternating-series argument is needed first.

Who Shaped The Comparison Test?

Deciding whether an infinite sum is finite is older than calculus itself, and the first great result was a comparison argument in disguise.

The idea became a formal test much later. Augustin-Louis Cauchy (1789–1857, France), in his 1821 Cours d'Analyse, gave series convergence its first rigorous treatment, defining convergence through partial sums and stating comparison and ratio criteria as theorems rather than intuitions. From Cauchy onward, the Comparison Test sat beside the ratio test, the root test, and the integral test as one of the standard tools for settling an infinite sum.

Where Is The Comparison Test Used In The Real World?

The test is a background check: before anyone trusts an infinite sum, they confirm it is finite by comparing it to something known.

• Numerical analysis and computing: before summing a series on a computer, an analyst checks convergence by comparison, because a divergent sum would produce a meaningless answer no matter how many terms are added.

• Physics: perturbation series in quantum mechanics and orbital mechanics are checked against known convergent bounds to decide whether adding more correction terms actually improves the estimate.

• Engineering signals: the energy of a discrete signal is an infinite sum of squared samples, and comparing it to a convergent benchmark confirms the signal has finite energy before it is filtered or transmitted.

• Probability and statistics: confirming that a distribution has a finite mean or variance reduces to showing an infinite sum converges, which is routinely done by comparison to a p-series.

• Computer science: the total cost of a recursive algorithm is often an infinite sum, and comparing it to a geometric series shows at a glance whether the cost stays bounded.

One habit runs through all of them: never trust an infinite total until you have trapped it under a benchmark you already believe.

What Are The Most Common Mistakes With The Comparison Test?

These three errors account for most lost marks, and each matches a question real students ask on r/calculus and r/learnmath and in AP review guides.

Using the wrong inequality direction for the conclusion.

Where it slips in:

A student bounds their series above by a divergent series, or below by a convergent one, and then declares an answer.

Don't do this:

Do not conclude anything from a mismatched pair. Bounding above by a divergent benchmark, or below by a convergent one, tells you nothing at all.

The correct way:

Match the direction to the goal. To prove convergence, bound above by a convergent benchmark: $a_n \le b_n$ with $\sum b_n$ convergent. To prove divergence, bound below by a divergent benchmark: $a_n \ge b_n$ with $\sum b_n$ divergent.

Comparing to a benchmark whose behaviour you never checked.

Where it slips in:

A student compares to a series that looks familiar but does not actually confirm whether that benchmark converges or diverges.

Don't do this:

Do not compare to a mystery series. If you cannot state whether the benchmark converges, the comparison proves nothing.

The correct way:

Always compare to a p-series or geometric series, and say its verdict out loud first. Write "$\sum \frac{1}{n^2}$ converges because $p = 2 > 1$," then run the comparison.

Applying the test to a series that is not positive.

Where it slips in:

A student uses the Comparison Test on a series whose terms change sign, such as one containing $(-1)^n$, and treats it like a positive-term series.

Don't do this:

Do not ignore the sign condition. The test requires $0 \le a_n$; with negative terms the bounded-staircase argument breaks and a ceiling no longer forces a limit.

The correct way:

Confirm the terms are non-negative first. If they alternate, use an alternating-series test, or apply the Comparison Test to the series of absolute values $\sum \lvert a_n\rvert$ to check absolute convergence.

Practice Problems On The Comparison Test

Work each one, then check against the answer. Every benchmark is a p-series, geometric series, or harmonic series.

1. Test $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^3 + 2}$.
   (Answer: $\frac{1}{n^3+2} < \frac{1}{n^3}$, a convergent p-series with $p = 3$, so it converges.)

2. Test $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^3 + 1}$.
   (Answer: $\frac{n}{n^3+1} < \frac{n}{n^3} = \frac{1}{n^2}$, convergent p-series, so it converges.)

3. Test $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^n + n}$.
   (Answer: $\frac{1}{3^n+n} < \frac{1}{3^n} = \left(\tfrac{1}{3}\right)^n$, convergent geometric, so it converges.)

4. Test $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2n + 5}$.
   (Answer: $2n+5 \le 7n$ for $n \ge 1$, so $\frac{1}{2n+5} \ge \frac{1}{7n}$; since $\sum \frac{1}{7n}$ diverges, it diverges.)

5. Test $\displaystyle\sum_{n=1}^{\infty}\frac{2 + \sin n}{n^2}$.
   (Answer: $0 < \frac{2+\sin n}{n^2} \le \frac{3}{n^2}$, convergent p-series, so it converges.)

6. Test $\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n^3}$.
   (Answer: $\ln n \le n$ gives $\frac{\ln n}{n^3} \le \frac{1}{n^2}$, convergent p-series, so it converges.)

Where Should You Go Next After The Comparison Test?

The Comparison Test is one of a family of convergence tests, and several natural doors open from here.

1. Limit comparison test. The flexible cousin of this test, for when the inequality is awkward but the terms behave alike in the limit.

2. Ratio test. The go-to test when terms involve powers or products that grow quickly, where a ratio of consecutive terms settles the question.

3. Integral test. Connects a series to the area under a matching curve, and is the cleanest way to confirm exactly when a p-series converges.

If your child is meeting convergence tests for the first time, a live Bhanzu trainer teaches the Comparison Test from the bounded-staircase picture up, so choosing a benchmark becomes a habit rather than a guess, in the Bhanzu math program.