The comparison test

Infinite series: Convergence tests

The comparison test

We can estimate series just like we can estimate normal limits. The following two propositions are together called the comparison test.

The comparison test (1) Let\(\textstyle\sum a_n\) and \(\textstyle\sum b_n\) be series such that \(a_n\geq 0\) for all \(n\). If \(\textstyle\sum a_n\) converges and \(|b_n|\leq a_n\) for all \(n\) then \(\textstyle\sum b_n\) converges as well.

Let \(\epsilon>0\) be given. Since \(\textstyle\sum a_n\) meets the Cauchy criterion, there exists \(N\) such that \(\left|\sum_{k=m}^n a_k\right|<\epsilon\) for \(n,m\geq N\). It now follows that \[ \left| \sum_{k=n}^m b_k \right| \leq \sum_{k=n}^m \left| b_k \right| \leq \sum_{k=n}^m a_k\text,\] so the series \(\sum b_n\) also satisfies the Cauchy criterion. This means that the series converges.

The comparison test (2) Let \(\textstyle\sum a_n\) and \(\textstyle\sum b_n\) be series. If \(\textstyle\sum a_n = \infty\) and \(b_n\geq a_n\) for all \(n\) then \(\textstyle\sum b_n = \infty\).

If \(b_n\geq a_n\) for all \(n\) then for the partial sums it holds that \[ \sum_{k=1}^n b_k \geq \sum_{k=1}^n a_k \text, \] so for the limits we obtain \[ \sum b_n = \lim_{n\to\infty}\sum_{k=1}^n b_k \geq \lim_{n\to\infty}\sum_{k=1}^n a_k = \sum a_n = \infty\text. \]

The first part of the comparison test can be used to prove that a series converges, while the second part can be used to show that a series diverges.

The series \(\sum_{n=1}^\infty \frac{1}{2n+1}\) diverges. This is because \[ \frac{1}{2n+1} > \frac{1}{2n+n} = \frac{1}{3n} \] and \(\textstyle\sum \frac{1}{3n} = \frac{1}{3} \sum \frac{1}{n}\) diverges.