The Comparison Test for Positive Series Examples 1

# The Comparison Test for Positive Series Examples 1

Let $\{ a_n \}$ and $\{ b_n \}$ be ultimately positive sequences such that for some $N \in \mathbb{N}$, if $n ≥ N$ then $0 ≤ a_n ≤ b_n$. Recall from the The Comparison Test for Positive Series page that 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. ||

We will now look at some examples of applying the comparison test.

## Example 1

Using the comparison test, determine whether the series $\sum_{n=1}^{\infty} \frac{1}{7^n + n}$ is convergent or divergent.

We note that this series is always positive since the numerator is always positive and the denominator is always positive for $n \in \mathbb{N}$. We also note that $\frac{1}{7^n + n} < \frac{1}{7^n}$ for all $n \in \mathbb{N}$ since the denominator of the former is always larger than the denominator of the latter.

Now we note that $\sum_{n=1}^{\infty} \frac{1}{7^n}$ is a geometric series, namely a convergent geometric series since $r = \frac{1}{7}$ and so $\mid r \mid < 1$.

So by the comparison test, since $0 ≤ \sum_{n=1}^{\infty} \frac{1}{7^n + n} ≤\sum_{n=1}^{\infty} \frac{1}{7^n}$ and since $\sum_{n=1}^{\infty} \frac{1}{7^n}$ is convergent, then $\sum_{n=1}^{\infty} \frac{1}{7^n + n}$ is also convergent.

## Example 2

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

We note that this series is always positive. Since $0 ≤ \mid \sin n \mid ≤ 1$ for all $n \in \mathbb{N}$, we have that $\frac{1}{n \mid \sin n \mid} ≥ \frac{1}{n}$ for all $n \in \mathbb{N}$.

We also note that $\sum_{n=1}^{\infty} \frac{1}{n}$ is a divergent p-series since $p = 1$. So by the comparison test since $0 ≤ \sum_{n=1}^{\infty} \frac{1}{n} ≤ \sum_{n=1}^{\infty} \frac{1}{n \mid \sin n \mid}$ and since $\sum_{n=1}^{\infty} \frac{1}{n}$ is divergent, then $\sum_{n=1}^{\infty} \frac{1}{n \mid \sin n \mid}$ is also divergent.

## Example 3

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

We note that this series is always positive. Now we want to find a series to compare with. We note that for large $n$ this series behaves like $\frac{n}{n^2} = \frac{1}{n}$ since $0 ≤ \cos ^2 n ≤ 1$. We also note that since $n^2 ≥ n^2 - \cos ^2 n$ then:

(1)
\begin{align} \frac{1}{n} = \frac{n}{n^2} ≤ \frac{n}{n^2 - \cos ^2 n} \end{align}

Now we know that $0 ≤ \sum_{n=1}^{\infty} \frac{1}{n} ≤ \sum_{n=1}^{\infty} \frac{n}{n^2 - \cos ^2 n}$ and since $\sum_{n=1}^{\infty} \frac{1}{n}$ it follows by the comparison test that $\sum_{n=1}^{\infty} \frac{n}{n^2 - \cos ^2 n}$ also diverges.