Comparison tests

Most series you meet in the wild are too tangled to sum in closed form — there is no neat formula for $\sum \frac{1}{n^2+3}$ the way there is for a geometric series. What you can do is compare the tangled series, term by term or in the limit, against a benchmark series whose fate you already know. If your series is squeezed below a convergent benchmark, it converges; if it dominates a divergent one, it diverges too. This article builds that toolkit — the direct comparison test, the limit comparison test, and the integral test — around the two benchmarks every engineering student should know cold: the $p$ -series and the geometric series. Master these and the ratio and root tests that follow will feel like natural extensions of the same idea.

The core insight

A series whose terms are trapped below (or above) a benchmark series of known behaviour must converge (or diverge) the same way, because partial sums that are bounded and increasing are forced to settle down.

The core idea

The benchmarks. Everything here rests on knowing two families of series cold. The $p$ -series,

\sum_{n=1}^{\infty} \frac{1}{n^p},

where: $p$ is a fixed real exponent, converges if $p>1$ and diverges if $p \le 1$ (the case $p=1$ is the harmonic series, the classic slow-diverging example). The geometric series $\sum ar^n$ converges iff $|r|<1$ . Between these two families you can benchmark almost anything that shows up in a first course.

Direct comparison test. Suppose $0 \le a_n \le b_n$ for all $n$ past some point $N$ . If $\sum b_n$ converges, so does $\sum a_n$ — a series trapped underneath a convergent series cannot itself blow up. If $\sum a_n$ diverges, then $\sum b_n$ must diverge too, since $b_n$ is even bigger. The test says nothing if the inequality points the wrong way: knowing $a_n \le b_n$ with $b_n$ divergent tells you nothing about $a_n$ .

Limit comparison test. Suppose $a_n, b_n > 0$ and the ratio settles to a finite, positive number:

\lim_{n\to\infty} \frac{a_n}{b_n} = L, \qquad 0 < L < \infty.

where: $L$ is the limiting ratio of the two term sequences. Then $\sum a_n$ and $\sum b_n$ share the same fate — both converge or both diverge. This sidesteps the fiddly job of proving an inequality; you only need to spot the right order of growth. If $L=0$ or $L=\infty$ , the test is inconclusive in general (it still gives one-directional information in special cases, but for this course treat $0<L<\infty$ as the safe regime to rely on).

Integral test. If $f(x)$ is positive, continuous, and decreasing for $x \ge N$ , with $a_n = f(n)$ , then

\sum_{n=N}^{\infty} a_n \ \text{converges} \iff \int_{N}^{\infty} f(x)\,dx \ \text{converges (is finite)}.

This is precisely how the $p$ -series rule is proved in the first place: $\int_1^\infty x^{-p}\,dx$ converges exactly when $p>1$ , matching the series result. The integral test converts a discrete summing question into a continuous area question you already know how to evaluate.

Why any of this works. All three tests are dressed-up versions of one fact about real numbers: a sequence of partial sums $S_N = \sum_{n\le N} a_n$ (with $a_n \ge 0$ ) is monotonically increasing, and a monotonically increasing sequence that is bounded above must converge — the Monotone Convergence Theorem. Trap $a_n$ term-by-term below a convergent series and you have bounded the partial sums, so they must settle. Trap $a_n$ above a divergent series and the partial sums are forced to climb without limit. The comparison-type tests are just different ways of arranging that trap.

Figure

Two curves on a graph showing a bounded series squeezed under a convergent p series curve and an unbounded series pushed above a divergent harmonic series curve

Derivation ADirect comparison for a rational series

The goal is to prove that $\sum_{n=1}^{\infty} \dfrac{1}{n^2+3}$ converges, showing in full generality how a direct-comparison argument is built from an algebraic inequality plus a known benchmark, so that the same three-step pattern can be reused on any series whose terms are dominated by a recognizable polynomial or power expression.

Step 1. Choose a benchmark by inspecting the dominant behaviour of the terms for large $n$ . Here $n^2+3$ behaves like $n^2$ once $n$ is large, since the constant $3$ becomes negligible next to $n^2$ . This suggests comparing against the $p$ -series $\sum 1/n^2$ , which is known to converge because $p=2>1$ .

Step 2. Establish the inequality that will do the trapping. For every $n \ge 1$ , $n^2+3 > n^2$ , and since both sides are positive, taking reciprocals reverses the inequality: $\dfrac{1}{n^2+3} < \dfrac{1}{n^2}$ . Writing $a_n = \dfrac{1}{n^2+3}$ and $b_n = \dfrac{1}{n^2}$ , this gives $0 < a_n < b_n$ for all $n \ge 1$ , with no exceptions to worry about.

Step 3. Apply the direct comparison test. Since $\sum b_n = \sum 1/n^2$ converges ( $p=2>1$ ) and $0 \le a_n \le b_n$ for every $n$ , the test concludes that $\sum a_n = \sum_{n=1}^{\infty} \dfrac{1}{n^2+3}$ converges. Note the direction: because the smaller series is the one under test and the larger series converges, convergence transfers downward — exactly the pattern to reuse whenever the target series' terms are bounded above by a recognizable $p$ -series or geometric series.

Worked example

Test $\displaystyle\sum_{n=2}^{\infty} \frac{n+1}{n^3-n}$ for convergence using the limit comparison test.

Step 1. Identify the dominant behaviour of numerator and denominator for large $n$ . The numerator $n+1 \approx n$ , and the denominator $n^3 - n \approx n^3$ , so $a_n \approx \dfrac{n}{n^3} = \dfrac{1}{n^2}$ . Choose the benchmark $b_n = \dfrac{1}{n^2}$ , a $p$ -series with $p=2>1$ , known convergent.

Step 2. Compute the limit of the ratio.

\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{n^2(n+1)}{n^3-n} = \lim_{n\to\infty} \frac{n^3+n^2}{n^3-n}.

Dividing numerator and denominator by $n^3$ ,

= \lim_{n\to\infty} \frac{1+1/n}{1-1/n^2} = \frac{1+0}{1-0} = 1.

Step 3. Conclude. Since $L=1$ is finite and positive, $\sum a_n$ and $\sum b_n$ share the same fate. As $\sum 1/n^2$ converges, so does $\displaystyle\sum_{n=2}^{\infty} \frac{n+1}{n^3-n}$ .

Competitive stretch

Now test $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ combined with an integral-test cross-check. The dominant term suggests $p = 1/2 \le 1$ , so this $p$ -series should diverge — but confirm it directly with the integral test rather than just quoting the rule. Take $f(x) = 1/\sqrt{x}$ , positive, continuous, and decreasing for $x \ge 1$ . Then $\displaystyle\int_1^{\infty} x^{-1/2}\,dx = \Big[2x^{1/2}\Big]_1^{\infty}$ , which grows without bound as the upper limit goes to infinity. Since the integral diverges, the integral test confirms $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ diverges, exactly matching the $p$ -series prediction for $p=1/2$ .

Common traps

Comparing in the wrong direction. To prove convergence you need your series term-wise smaller than a known convergent series; to prove divergence you need it larger than a known divergent one. An inequality pointing the wrong way proves nothing at all.
Forgetting the positivity requirement. Both comparison tests, as stated, need nonnegative terms. Applying them straight to a series with mixed signs is invalid — such series call for absolute or alternating convergence tests instead.
Picking a benchmark with the wrong leading power. In limit comparison, choosing $b_n$ from the wrong power of $n$ — say $1/n$ instead of $1/n^2$ in the worked example above — produces $L=0$ or $L=\infty$ , which is inconclusive without extra work. Always match the leading-order growth of numerator and denominator first.
Skipping the hypotheses of the integral test. The test needs $f$ positive, continuous, and genuinely decreasing on the whole interval considered — verify this (for instance via $f'(x)<0$ ) rather than assuming it just because the series terms look like they shrink.

Ask the tutor: Stuck on which benchmark series matches a given $a_n$ , or want more practice orienting the comparison inequality correctly? Ask the AI tutor to work through additional examples with you.

Why does the limit comparison test avoid the inequality hassle of direct comparison?

Direct comparison forces you to prove $a_n \le b_n$ (or $\ge$ ) for every relevant $n$ , which can require algebraic manipulation that is not always obvious, especially when $a_n$ has a messy numerator. Limit comparison only asks for the limiting ratio $a_n/b_n$ as $n \to \infty$ , which is usually a routine algebra-of-limits computation — divide through by the highest power and read off the answer. As long as that limit is finite and positive, the two series are guaranteed to behave identically, so you get the same conclusion with less bookkeeping.

What if the limit comparison ratio comes out to 0 or infinity?

Those cases are not automatically useless, but they need more care than this course expects you to apply blindly. If $L=0$ and $\sum b_n$ converges, you can still conclude $\sum a_n$ converges (since $a_n$ is eventually much smaller than $b_n$ ), but if $\sum b_n$ diverges the test tells you nothing about $a_n$ . Symmetric caveats apply when $L=\infty$ . The safest habit is to choose your benchmark's power so the limit comes out finite and positive in the first place — pick $b_n$ from the exact leading-order term of $a_n$ .

How do I decide whether to use direct comparison, limit comparison, or the integral test?

Use direct comparison when you can spot an obvious, easily provable inequality, such as dropping a small additive term to get a clean $p$ -series bound. Reach for limit comparison when the series is a ratio of polynomials or similar expressions where an exact inequality is awkward to write down but the leading-order behaviour is clear. Save the integral test for cases where the terms come from a function you can actually integrate and where no simple polynomial benchmark applies, such as $1/(n\ln n)$ . In practice, limit comparison covers the widest range of exam-style series with the least algebra.

Comparison tests

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads