Direct and Limit Comparison Tests

When you can't integrate a series' terms, compare them instead: the direct and limit comparison tests decide convergence of a positive series by lining its terms up against a benchmark series whose fate you already know.

By the end you'll be able to pick a geometric or p-series benchmark, apply direct comparison in a valid direction, and fall back to limit comparison when the inequality points the wrong way.

Predict: if every term of a positive series sits below the matching term of a convergent benchmark, must the series converge? And when the terms poke above the benchmark instead, is all lost? Toggle between the two tests to check.

The blue bars are the terms aₙ of the series under test; the orange bars are the benchmark terms bₙ = (1/3)ⁿ. Slide the probe to read any pair and the ratio aₙ/bₙ; hover or tab to any pair of bars for exact values.

Test & series

Probe n = 3

Direct comparison: every blue bar sits below its orange benchmark bar, and the geometric benchmark converges — so the series converges. At n = 3: aₙ ≈ 0.0333 ≤ bₙ ≈ 0.0370 · ratio aₙ/bₙ = 0.900 → 1 as n grows.

Σ 1/(3ⁿ + n) (series) vs Σ (1/3)ⁿ (convergent geometric benchmark), term by term

aₙ (series under test) bₙ = (1/3)ⁿ (benchmark)

The integral test needs you to integrate $f(x)$, which isn't always feasible. Comparison tests skip the integral entirely: line the terms of your positive series up against a benchmark — term by term (direct) or in the limit (limit comparison) — and inherit the benchmark's verdict. Both tests require $a_n, b_n \ge 0$.

Formal

Direct comparison. If $0 \le a_n \le b_n$ for all sufficiently large $n$: if $\sum b_n$ converges, then $\sum a_n$ converges; if $\sum a_n$ diverges, then $\sum b_n$ diverges. Only these two directions are valid — knowing the larger series converges says nothing about a smaller one you sit above, and a smaller series diverging says nothing about one below it. Limit comparison. If $a_n, b_n > 0$ and $\lim_{n\to\infty} a_n/b_n = c$ for some finite $c > 0$, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge. If the limit is $0$ or $\infty$, the benchmark was a poor choice and the test gives no information.

Procedural

Choosing the benchmark is the whole game. For rational functions of $n$, keep only the leading term of top and bottom: $\frac{2n^2+1}{n^3+5}$ behaves like $\frac{2n^2}{n^3} = \frac{2}{n}$, a divergent multiple of the harmonic series. For terms dominated by an exponential, use a geometric series with the same base. Then: if the inequality against the benchmark is easy and points a valid direction, use direct comparison; if it's awkward (subtraction in a denominator, say), compute $\lim a_n/b_n$ instead — dominant-term benchmarks almost always give a finite positive limit.

Worked example — direct comparison

Test $\displaystyle\sum_{n=1}^{\infty} \frac{1}{3^n + n}$. Since $3^n + n > 3^n$ for $n \ge 1$, $$\frac{1}{3^n+n} < \frac{1}{3^n}.$$ The benchmark $\sum (1/3)^n$ is a convergent geometric series ($r = \tfrac13$, $|r| < 1$), so by direct comparison the original series converges — that's the blue-below-orange picture in the demo's direct mode.

Worked example — limit comparison

Test $\displaystyle\sum_{n=1}^{\infty} \frac{1}{3^n - n}$. Direct comparison is awkward here because $3^n - n < 3^n$ points the wrong way (the terms poke above the benchmark). Instead compare to $b_n = (1/3)^n$: $$\lim_{n\to\infty} \frac{1/(3^n-n)}{1/3^n} = \lim_{n\to\infty} \frac{3^n}{3^n-n} = \lim_{n\to\infty} \frac{1}{1 - n/3^n} = 1,$$ a finite positive number. Since $\sum (1/3)^n$ converges, limit comparison says $\sum \frac{1}{3^n-n}$ converges too.

Your turn

Test $\displaystyle\sum_{n=1}^{\infty} \frac{n+3}{n^3 - 2n + 1}$. Keep only the leading terms: $a_n$ behaves like $\frac{n}{n^3} = \frac{1}{n^2}$, so take $b_n = 1/n^2$. Then $\lim_{n\to\infty} a_n/b_n = $ ____ , and since $\sum 1/n^2$ is a $p$-series with $p = $ ____ , the series ____ .

Reveal the answer

$\dfrac{a_n}{b_n} = \dfrac{(n+3)\,n^2}{n^3-2n+1} = \dfrac{n^3 + 3n^2}{n^3 - 2n + 1} \to $ 1, finite and positive. $\sum 1/n^2$ is a $p$-series with $p = $ 2 $ > 1$, which converges — so by limit comparison the series converges.

More info — why the benchmark's own verdict matters

Both tests only transfer a verdict; the benchmark must arrive with one. That's why the two families from the previous lesson do all the work here: a geometric series $\sum r^n$ converges exactly when $|r| < 1$, and a $p$-series $\sum 1/n^p$ converges exactly when $p > 1$. If you catch yourself comparing against a series whose behavior you'd also have to test, stop and swap in one of these two. And watch the classic traps: both tests need nonnegative terms, direct comparison only works in its two valid directions, and a limit-comparison ratio of $0$ or $\infty$ means "wrong benchmark — pick again", never a conclusion. The Paul's Online Math Notes page in Dive deeper walks several more examples of each trap.

Check your understanding

Question 1 of 4

Test $\sum_{n=1}^{\infty} \frac{1}{2^n + 5}$ for convergence using direct comparison. Which benchmark works, and what is the verdict?

Question 2 of 4

You establish $0 \le a_n \le b_n$ for all $n$, and you know $\sum b_n$ diverges. What does direct comparison tell you about $\sum a_n$?

Question 3 of 4

Applying the limit comparison test to $\sum a_n$ with benchmark $\sum b_n$, you compute $\lim_{n\to\infty} a_n/b_n = 0$. What can you conclude?

Question 4 of 4

Use limit comparison to test $\sum_{n=1}^{\infty} \frac{n}{n^3 + 1}$. Which benchmark and verdict are correct?

Recap

Direct comparison ($0 \le a_n \le b_n$ for large $n$): $\sum b_n$ converges $\Rightarrow \sum a_n$ converges; $\sum a_n$ diverges $\Rightarrow \sum b_n$ diverges. No other direction is valid.
Limit comparison ($a_n, b_n > 0$): if $\lim a_n/b_n = c$ is finite and positive, both series converge or both diverge. A limit of $0$ or $\infty$ gives no information — choose a better benchmark.
Pick benchmarks by dominant behavior: ratio of leading terms $\to$ a $p$-series for rational functions of $n$; a geometric series with matching base for exponential-dominated terms.
Both tests require nonnegative terms ($a_n, b_n \ge 0$).

Dive deeper

Paul's Online Math Notes — Comparison Test / Limit Comparison Test Walk through direct and limit comparison with worked examples.

Sources