Calculus · Grade 12 and undergraduate

The Comparison Tests for Series

Quick answer

A series of positive terms that is smaller, term by term, than a convergent series converges, and one that is larger than a divergent series diverges. That is the direct comparison test. When the inequality points the wrong way or is hard to prove, the limit comparison test compares the ratio of the terms instead: if aₙ/bₙ approaches a positive finite number, the two series share their fate. The benchmarks are geometric series and p-series, found by keeping only the dominant powers.

What you'll learn

  • Apply the direct comparison test in the right direction
  • Apply the limit comparison test
  • Choose a benchmark series from the dominant terms
  • Recognize when a comparison proves nothing

Comparing term by term

Most series are neither geometric nor a pp-series, but many resemble one. ∑1n2+5\sum \tfrac{1}{n^2 + 5} looks like ∑1n2\sum \tfrac{1}{n^2}, and it is smaller term by term. Since the partial sums of a positive series only go up, a series held below a convergent one cannot climb forever.

The direct comparison test

Suppose 0≤an≤bn0 \le a_n \le b_n for all nn from some point on.

  • If ∑bn\sum b_n converges, then ∑an\sum a_n converges.
  • If ∑an\sum a_n diverges, then ∑bn\sum b_n diverges.

Why the direction matters

A comparison works like a ceiling or a floor. A ceiling that stays finite holds up everything under it, so smaller than convergent means convergent. A floor that rises forever lifts everything above it, so larger than divergent means divergent. The other two cases give no information: being smaller than a divergent series, or larger than a convergent one, is consistent with anything. A comparison proves convergence from above and divergence from below, and nothing the other way round.

Limit comparison

Sometimes the inequality points the useless way. 1n2−10\tfrac{1}{n^2 - 10} is larger than 1n2\tfrac{1}{n^2}, yet for large nn the two are nearly equal. The limit comparison test captures “nearly equal”:

The limit comparison test

Suppose an>0a_n > 0 and bn>0b_n > 0, and lim⁡n→∞anbn=c\displaystyle\lim_{n \to \infty} \frac{a_n}{b_n} = c with 0<c<∞0 < c < \infty. Then ∑an\sum a_n and ∑bn\sum b_n both converge or both diverge.

If the ratio approaches cc, then far enough out ana_n lies between c2bn\tfrac{c}{2}b_n and 2c bn2c\,b_n, and direct comparison works in both directions.

To choose bnb_n, keep the dominant term of the numerator and of the denominator. For 2n+1n3−n+4\tfrac{2n + 1}{n^3 - n + 4} that leaves 2nn3=2n2\tfrac{2n}{n^3} = \tfrac{2}{n^2}, so compare with 1n2\tfrac{1}{n^2}.

Worked examples

Common mistakes

Practice problems

  1. Does ∑n=1∞1n3+n\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^3 + n} converge?

    Answer

    Yes

    Full solution

    1n3+n<1n3\tfrac{1}{n^3 + n} < \tfrac{1}{n^3}, and ∑1n3\sum \tfrac{1}{n^3} converges.

  2. Does ∑n=1∞1n−0.5\displaystyle\sum_{n=1}^{\infty} \frac{1}{n - 0.5} converge?

    Answer

    No

    Full solution

    1n−0.5>1n\tfrac{1}{n - 0.5} > \tfrac{1}{n}, and the harmonic series diverges.

  3. Does ∑n=1∞3n4n+1\displaystyle\sum_{n=1}^{\infty} \frac{3^n}{4^n + 1} converge?

    Answer

    Yes

    Full solution

    3n4n+1<(34)n\tfrac{3^n}{4^n + 1} < \left(\tfrac{3}{4}\right)^n, a convergent geometric series.

  4. Does ∑n=1∞n+2n3+1\displaystyle\sum_{n=1}^{\infty} \frac{n + 2}{n^3 + 1} converge?

    Answer

    Yes

    Full solution

    The dominant terms give nn3=1n2\tfrac{n}{n^3} = \tfrac{1}{n^2}. (n+2)/(n3+1)1/n2=n3+2n2n3+1→1\tfrac{(n + 2)/(n^3 + 1)}{1/n^2} = \tfrac{n^3 + 2n^2}{n^3 + 1} \to 1, and ∑1n2\sum \tfrac{1}{n^2} converges.

  5. Does ∑n=1∞nn+4\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n + 4} converge?

    Answer

    No

    Full solution

    Compare with nn=1n\tfrac{\sqrt{n}}{n} = \tfrac{1}{\sqrt{n}}: the ratio nn+4→1\tfrac{n}{n + 4} \to 1, and ∑1n\sum \tfrac{1}{\sqrt{n}} diverges.

  6. Does ∑n=2∞1n2−n\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - n} converge?

    Answer

    Yes

    Full solution

    1/(n2−n)1/n2=n2n2−n→1\tfrac{1/(n^2 - n)}{1/n^2} = \tfrac{n^2}{n^2 - n} \to 1. (It also telescopes: 1n−1−1n\tfrac{1}{n - 1} - \tfrac{1}{n}, with sum 11.)

  7. Does ∑n=1∞ln⁡nn2\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n^2} converge?

    Answer

    Yes

    Full solution

    ln⁡n<n\ln n < \sqrt{n} for every n≥1n \ge 1, so ln⁡nn2<nn2=1n3/2\tfrac{\ln n}{n^2} < \tfrac{\sqrt{n}}{n^2} = \tfrac{1}{n^{3/2}}, a convergent pp-series.

  8. Does ∑n=1∞1n+ln⁡n\displaystyle\sum_{n=1}^{\infty} \frac{1}{n + \ln n} converge?

    Answer

    No

    Full solution

    1/(n+ln⁡n)1/n=nn+ln⁡n→1\tfrac{1/(n + \ln n)}{1/n} = \tfrac{n}{n + \ln n} \to 1, since ln⁡nn→0\tfrac{\ln n}{n} \to 0. The harmonic series diverges.

  9. Does ∑n=1∞2n2+1n4+3n\displaystyle\sum_{n=1}^{\infty} \frac{2n^2 + 1}{n^4 + 3n} converge?

    Answer

    Yes

    Full solution

    The dominant terms give 2n2n4=2n2\tfrac{2n^2}{n^4} = \tfrac{2}{n^2}. Against 1n2\tfrac{1}{n^2} the ratio approaches 22, which is positive and finite.

  10. A student argues that ∑n=1∞1n2+1\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 1} diverges because 1n2+1<1n\tfrac{1}{n^2 + 1} < \tfrac{1}{n} and ∑1n\sum \tfrac{1}{n} diverges. What went wrong?

    Hint

    What can a series smaller than a divergent one do?

    Answer

    Being smaller than a divergent series proves nothing. Compared with 1n2\tfrac{1}{n^2}, the series converges.

    Full solution

    A divergent series lifts only the series above it, and the student’s series lies below. Instead, 1n2+1<1n2\tfrac{1}{n^2 + 1} < \tfrac{1}{n^2}, and ∑1n2\sum \tfrac{1}{n^2} converges, so the series converges.

Frequently asked questions

What does the direct comparison test say?

For positive terms with aₙ ≤ bₙ: if the sum of bₙ converges, so does the sum of aₙ; if the sum of aₙ diverges, so does the sum of bₙ.

What does the limit comparison test say?

For positive terms, if aₙ/bₙ approaches a finite positive number, then the two series either both converge or both diverge.

How do I choose the series to compare with?

Keep only the dominant term in the numerator and the denominator. For (n + 2)/(n³ + 1), that leaves n/n³ = 1/n², a convergent p-series.

Why does being smaller than a divergent series prove nothing?

A smaller series can still diverge, or it can converge. Both 1/(2n) and 1/n² are smaller than 1/n, and one diverges while the other converges.

Do the comparison tests work for negative terms?

No. They need positive terms. For series with mixed signs, test the absolute values or use the alternating series test.

What to learn next

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