# The Comparison Test for Positive Series

Recall the Comparison Test for Improper Integral Convergence/Divergence that says:

Theorem (Comparison Test for Integral Convergence/Divergence): Suppose that $0 ≤ g(x) ≤ f(x)$. If $\int_a^{\infty} f(x) \: dx$ also converges, then $\int_a^{\infty} g(x) \: dx$ converges. Furthermore, if $\int_a^{\infty} g(x) \: dx$ diverges, then $\int_a^{\infty} f(x) \: dx$ also diverges. |

We will now develop an analogous test which we will call The Comparison Test for Positive Series as follows.

Theorem (The Comparison Test for Positive Series): Let $\{ a_n \}$ and $\{ b_n \}$ be ultimately positive sequences such that for some $N \in \mathbb{N}$, if $n ≥ N$ then $0 ≤ a_n ≤ b_n$. If the series $\sum_{n=1}^{\infty} b_n$ is convergent, then the series $\sum_{n=1}^{\infty} a_n$ is also convergent. Similarly if the series $\sum_{n=1}^{\infty} a_n$ is divergent, then the series $\sum_{n=1}^{\infty} b_n$ is divergent. |

**Proof of Theorem:**We know that for $n ≥ N$, then $0 ≤ a_n ≤ b_n$. Let $s_n = a_1 + a_2 + ... + a_n$ and let $s_n' = b_1 + b_2 + ... + b_n$.

- We note that $s_n ≤ s_n'$. Suppose that $\sum_{n=1}^{\infty} b_n$ is convergent. Then $s_n'$ is bounded by some $M \in \mathbb{R}$ such that $M > 0$ and so we get that $s_n ≤ M$ and thus it follows that $\{ s_n \}$ is bounded above. We also note that $\{ a_n \}$ is ultimately positive and so $\sum_{n=1}^{\infty} a_n$ must be convergent.

- Now suppose that $\sum_{n=1}^{\infty} a_n$ is divergent, namely to infinity. We still have that $s_n ≤ s_n'$ and so as $n \to \infty$, $s_n \to \infty$, so it follows that as $n \to \infty$ $s_n' \to \infty$ as well. Therefore $\sum_{n=1}^{\infty} b_n$ is divergent. $\blacksquare$

Before we look at some examples of applying the comparison theorem for positive series, let's first recall some series that we know are in fact convergent or divergent to use as a comparison. The list below is helpful to memorize.

- The geometric series $\sum_{n=1}^{\infty} ar^{n-1}$ is convergent if $\mid r \mid < 1$.

- The geometric series $\sum_{n=1}^{\infty} ar^{n-1}$ is divergent to infinity $r ≥ 1$.

- By the p-Series test, if $p > 1$ then $\sum_{n=1}^{\infty} \frac{1}{x^p}$ is convergent.

- By the p-Series test, if $p ≤ 1$ then $\sum_{n=1}^{\infty} \frac{1}{x^p}$ is divergent to infinity.

## Example 1

**Using the comparison test, determine whether the series $\sum_{n=1}^{\infty} \frac{1}{3^n + 1}$ is convergent or divergent.**

We suspect that this series may be convergent and will compare it to the series $\sum_{n=1}^{\infty} \frac{1}{3^n}$.

We note that for all $n ≥ 1$, that $\frac{1}{3^n + 1} < \frac{1}{3^n}$. We also know that the series $\sum_{n=1}^{\infty} \frac{1}{3^n} = \sum_{n=1}^{\infty} \left ( \frac{1}{3} \right)^n$ is convergent since the common ratio $r = \frac{1}{3} < 1$.

Therefore by the comparison theorem it follows that $\sum_{n=1}^{\infty} \frac{1}{3^n + 1}$ is convergent as well.

## Example 2

**Using the comparison test, determine whether the series $\sum_{n=1}^{\infty} \frac{n^2 + 1}{\sqrt{n}}$ is convergent or divergent.**

We suspect that this series may be divergent and will compare it to the series $\sum_{n=1}^{\infty} \frac{1}{n^3}$. We note that for all $n ≥ 1$, $\frac{1}{\sqrt{n}} < \frac{n^2 + 1}{\sqrt{n}}$. We know by the p-Series test since $p = \frac{1}{2} ≤ 1$ then the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ diverges, and thus by the comparison theorem it follows that $\sum_{n=1}^{\infty} \frac{n^2 + 1}{\sqrt{n}}$ also diverges.