5.4 Comparison tests

We have seen that the integral test allows us to determine the convergence or divergence of a series by comparing it to a related improper integral. In this section, we show how to use comparison tests to determine the convergence or divergence of a series by comparing it to a series whose convergence or divergence is known. Typically these tests are used to determine convergence of series that are similar to geometric series or p -series.

Comparison test

In the preceding two sections, we discussed two large classes of series: geometric series and p -series. We know exactly when these series converge and when they diverge. Here we show how to use the convergence or divergence of these series to prove convergence or divergence for other series, using a method called the comparison test    .

For example, consider the series

This series looks similar to the convergent series

Since the terms in each of the series are positive, the sequence of partial sums for each series is monotone increasing. Furthermore, since

for all positive integers $n,$ the $k\text{th}$ partial sum $S_k$ of ${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{n^2+1}$ satisfies

(See [link] (a) and [link] .) Since the series on the right converges, the sequence $\left\{S_k\right\}$ is bounded above. We conclude that $\left\{S_k\right\}$ is a monotone increasing sequence that is bounded above. Therefore, by the Monotone Convergence Theorem, $\left\{S_k\right\}$ converges, and thus

converges.

Similarly, consider the series

This series looks similar to the divergent series

The sequence of partial sums for each series is monotone increasing and

for every positive integer $n.$ Therefore, the $k\text{th}$ partial sum $S_k$ of ${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{n-1\text{/}2}$ satisfies

(See [link] (b) and [link] .) Since the series ${\displaystyle \sum }_{n=1}^{\infty }1\text{/}n$ diverges to infinity, the sequence of partial sums ${\displaystyle \sum }_{n=1}^{k}1\text{/}n$ is unbounded. Consequently, $\left\{S_k\right\}$ is an unbounded sequence, and therefore diverges. We conclude that

diverges.

Proof

We prove part i. The proof of part ii. is the contrapositive of part i. Let $\left\{S_k\right\}$ be the sequence of partial sums associated with ${\displaystyle \sum }_{n=1}^{\infty }{a}_n,$ and let $L={\displaystyle \sum }_{n=1}^{\infty }{b}_n.$ Since the terms $a_n\ge 0,$

Therefore, the sequence of partial sums is increasing. Further, since $a_n\le b_n$ for all $n\ge N,$ then