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.

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.

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.