Integration and series · Definition

Comparison test

real analysisseries

Before you start you need to know

It turns an unknown series into a solved problem: if it is trapped below a convergent series, it converges too.

Definition

A test that decides whether a series \(\sum a_n\) of nonnegative terms converges by comparing it with another series \(\sum b_n\) such that \(a_n\le b_n\).

\[0\le a_n\le b_n\ \Longrightarrow\ \Big(\sum b_n\text{ converges}\Rightarrow\sum a_n\text{ converges}\Big)\]

The idea is that, with nonnegative terms, partial sums can only grow. If \(0\le a_n\le b_n\) and \(\sum b_n\) converges, the partial sums of \(\sum a_n\) are increasing and bounded by the sum of \(\sum b_n\); by the monotone convergence theorem they converge. That is the whole proof.

The contrapositive is just as useful: if \(\sum a_n\) diverges, then \(\sum b_n\) diverges. Thus \(\frac1{n^2+1}\le\frac1{n^2}\) gives convergence, and \(\frac1{2n}\le\frac1{n+1}\) gives divergence. The usual reference series are the geometric series, the harmonic series \(\sum\frac1n\) (divergent) and \(\sum\frac1{n^2}\) (convergent).

When the inequality is awkward, use the limit version: if \(a_n,b_n>0\) and \(\frac{a_n}{b_n}\to L\) with \(0<L<\infty\), both series converge or both diverge. For instance, \(\frac{2n+1}{n^3+4}\) behaves like \(\frac1{n^2}\), with \(L=2\).

Typical mistakes are comparing in the wrong direction (lying above a convergent series says nothing) and forgetting that the terms must be nonnegative. This test is the starting tool for absolute convergence, where \(\sum|a_n|\) is compared with a known series.

Think about it

If \(a_n\ge b_n\ge0\) and \(\sum b_n\) converges, what follows about \(\sum a_n\)?

Nothing: lying above a convergent series does not force convergence.

Why is \(a_n\ge0\) required?

With nonnegative terms the partial sums are increasing, so only a bound is needed; with mixed signs they could oscillate.

Exercises 10

  1. Knowing that \(\sum\frac1{n^2}\) converges, which series converges by direct comparison with it?
  2. To prove that \(\sum\frac1{n+1}\) diverges, which \(b_n\) of a divergent series gives \(0\le b_n\le\frac1{n+1}\) for \(n\ge1\)?
  3. It is known that \(\sum_{n=1}^{\infty}\frac1{2^n}=1\). Since \(\frac1{2^n+n}\le\frac1{2^n}\), what upper bound for \(\sum_{n=1}^{\infty}\frac1{2^n+n}\) does the comparison give directly?
  4. If \(0\le b_n\le a_n\) for all \(n\) and \(\sum b_n\) diverges, what happens to \(\sum a_n\)?
  5. Since \(-\frac1n\le\frac1{n^2}\) and \(\sum\frac1{n^2}\) converges, can we conclude that \(\sum\left(-\frac1n\right)\) converges?
  6. Let \(a_n=\dfrac{2n+1}{n^3+4}\) and \(b_n=\dfrac1{n^2}\). Compute \(\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}\). (If the limit is finite and positive, both series behave alike.)
  7. For \(a_n=\dfrac{2+\cos n}{3^n}\) we have \(0\le a_n\le b_n\) with \(b_n=\dfrac{3}{3^n}\). What is \(\displaystyle\sum_{n=1}^{\infty}b_n\), an upper bound for \(\sum a_n\)?
  8. Which comparison correctly proves that \(\sum_{n=1}^{\infty}\frac1{3^n-1}\) converges?
  9. Let \(a_n\ge0\) with \(\sum a_n\) convergent. What can be said about \(\sum a_n^2\)?
  10. Using \(n!\ge2^{n-1}\) for \(n\ge1\), what upper bound for \(\sum_{n=1}^{\infty}\frac1{n!}\) comes from summing the geometric series \(\sum_{n=1}^{\infty}\frac1{2^{n-1}}\)?

References

  • Stephen Abbott, Understanding Analysis, 2nd edition, Springer, 2015. §2.7, Teorema 2.7.4
  • Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976. Cap. 3, Teorema 3.25
  • Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, 4th edition, Wiley, 2011. §3.7