Unit 7, lesson 98

The Comparison Tests

How to settle a series of nonnegative terms by comparing it with a series you already understand, either term by term or through the limit of the ratio of terms.

  • Theory part 195
  • Practice part 196
  • Convergence and Convergence Tests

To find out whether a messy series converges, you rarely need its exact sum. It is enough to set it beside a simpler series whose behavior you know, such as a geometric series or a p-series. If your series is smaller than a convergent one, it converges. If it is bigger than a divergent one, it diverges. These tests make that comparison precise.

Key terms

Series with nonnegative terms

Definition If an≥0a_n \ge 0 for all nn, the partial sums sN=a1+⋯+aNs_N = a_1 + \cdots + a_N form an increasing sequence. By the monotone convergence theorem, ∑an\sum a_n converges if and only if its partial sums are bounded above. Otherwise sN→∞s_N \to \infty.

In plain words When you only ever add nonnegative numbers, the total can only grow. It either levels off below some ceiling or it grows without bound. There is no third option.

Example For ∑n=1∞1n!\sum_{n=1}^{\infty} \frac{1}{n!} note that n!≥2n−1n! \ge 2^{n-1} for n≥1n \ge 1, since n!n! is a product of n−1n - 1 factors that are at least 22 (and a factor 11). So

sN≤1+12+14+⋯+12N−1<2.s_N \le 1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^{N-1}} < 2.

The partial sums are increasing and bounded, so the series converges. Its sum is e−1≈1.718e - 1 \approx 1.718.

Direct comparison test

Definition Suppose 0≤an≤bn0 \le a_n \le b_n for all n≥1n \ge 1.

  1. If ∑bn\sum b_n converges, then ∑an\sum a_n converges, and ∑an≤∑bn\sum a_n \le \sum b_n.
  2. If ∑an\sum a_n diverges, then ∑bn\sum b_n diverges.

In plain words Below a convergent series is safe. Above a divergent series is lost. Being below a divergent series or above a convergent one tells you nothing.

Example For n≥1n \ge 1 we have

0<1n2+n+1<1n2,0 < \frac{1}{n^2 + n + 1} < \frac{1}{n^2},

and ∑1n2\sum \frac{1}{n^2} converges, so ∑1n2+n+1\sum \frac{1}{n^2 + n + 1} converges. In the other direction, ln⁡nn≥1n\frac{\ln n}{n} \ge \frac{1}{n} for n≥3n \ge 3 because ln⁡3>1\ln 3 > 1, and the harmonic series diverges, so ∑ln⁡nn\sum \frac{\ln n}{n} diverges.

Eventual inequalities

Definition In both comparison tests it is enough that the hypotheses hold for all n≥Nn \ge N for some integer NN. This follows because finitely many terms do not affect convergence.

In plain words The inequality only has to be true from some point on. Early terms that break it can be ignored.

Example Consider ∑n=7∞12n−100\sum_{n=7}^{\infty} \frac{1}{2^n - 100}, whose terms are positive since 27=1282^7 = 128. When 2n≥2002^n \ge 200, that is n≥8n \ge 8, we have 2n−100≥2n22^n - 100 \ge \frac{2^n}{2}, so

0<12n−100≤22nfor n≥8.0 < \frac{1}{2^n - 100} \le \frac{2}{2^n} \quad \text{for } n \ge 8.

The geometric series ∑22n\sum \frac{2}{2^n} converges, so the given series converges. The term n=7n = 7 breaks nothing.

Limit comparison test

Definition Suppose an>0a_n > 0 and bn>0b_n > 0 for all n≥Nn \ge N, and

lim⁡n→∞anbn=cwith 0<c<∞.\lim_{n \to \infty} \frac{a_n}{b_n} = c \quad \text{with } 0 < c < \infty.

Then ∑an\sum a_n and ∑bn\sum b_n either both converge or both diverge.

In plain words If two positive sequences are eventually proportional, their series share the same fate. You do not need to fight with exact inequalities, only find the dominant part of ana_n.

Example Consider ∑2n+1n3−n+5\sum \frac{2n + 1}{n^3 - n + 5}. Its terms are positive for n≥1n \ge 1, and for large nn they look like 2nn3=2n2\frac{2n}{n^3} = \frac{2}{n^2}. Compare with bn=1n2b_n = \frac{1}{n^2}.

anbn=2n3+n2n3−n+5→2.\frac{a_n}{b_n} = \frac{2n^3 + n^2}{n^3 - n + 5} \to 2.

Since 0<2<∞0 < 2 < \infty and ∑1n2\sum \frac{1}{n^2} converges, the given series converges.

The zero and infinite ratio cases

Definition Suppose an≥0a_n \ge 0 and bn>0b_n > 0 for all n≥Nn \ge N.

  1. If anbn→0\frac{a_n}{b_n} \to 0 and ∑bn\sum b_n converges, then ∑an\sum a_n converges.
  2. If anbn→∞\frac{a_n}{b_n} \to \infty and ∑bn\sum b_n diverges, then ∑an\sum a_n diverges.

In the other two combinations, ratio 00 with ∑bn\sum b_n divergent or ratio ∞\infty with ∑bn\sum b_n convergent, no conclusion follows.

In plain words A ratio of 00 means ana_n is eventually much smaller than bnb_n, which only helps if ∑bn\sum b_n converges. A ratio of ∞\infty means ana_n is much bigger, which only helps if ∑bn\sum b_n diverges.

Example For ∑ln⁡nn2\sum \frac{\ln n}{n^2} compare with the convergent ∑1n3/2\sum \frac{1}{n^{3/2}}.

(ln⁡n)/n21/n3/2=ln⁡nn→0,\frac{(\ln n)/n^2}{1/n^{3/2}} = \frac{\ln n}{\sqrt{n}} \to 0,

so ∑ln⁡nn2\sum \frac{\ln n}{n^2} converges. For ∑n=2∞1ln⁡n\sum_{n=2}^{\infty} \frac{1}{\ln n} compare with the divergent harmonic series. The ratio is

1/ln⁡n1/n=nln⁡n→∞,\frac{1/\ln n}{1/n} = \frac{n}{\ln n} \to \infty,

so ∑1ln⁡n\sum \frac{1}{\ln n} diverges.

Check yourself

  1. Does ∑n=1∞sin⁡2nn2\sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2} converge?
Show the answer

Yes. Since 0≤sin⁡2n≤10 \le \sin^2 n \le 1, the terms satisfy

0≤sin⁡2nn2≤1n2,0 \le \frac{\sin^2 n}{n^2} \le \frac{1}{n^2},

and ∑1n2\sum \frac{1}{n^2} converges, so direct comparison applies.

  1. Does ∑n=1∞1n2+1\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}} converge?
Show the answer

No. Compare with bn=1nb_n = \frac{1}{n}. The ratio is nn2+1→1\frac{n}{\sqrt{n^2 + 1}} \to 1, and the harmonic series diverges, so this series diverges too.

  1. Why does comparing ∑1n2+1\sum \frac{1}{n^2 + 1} with ∑1n\sum \frac{1}{n} through 1n2+1≤1n\frac{1}{n^2 + 1} \le \frac{1}{n} prove nothing?
Show the answer

Being smaller than a divergent series gives no information. Compare with 1n2\frac{1}{n^2} instead, which shows convergence.

How this lesson connects

It also uses comparison.