The Comparison Test for Positive Series Examples 2

# The Comparison Test for Positive Series Examples 2

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 more examples of applying the comparison test.

## Example 1

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

Note that for all $n \in \mathbb{N}$ that $\cos^2 n ≥ 0$. Therefore $n^2 - \cos^2 n ≤ n^2$, which implies that $\frac{1}{n^2 - \cos^2 n} ≥ \frac{1}{n^2}$. Therefore we have that:

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

We note that $\sum_{n=1}^{\infty} \frac{1}{n}$ is the divergent harmonic series, and so by the comparison test, we have that $\sum_{n=1}^{\infty} \frac{1}{n^2 - \cos^2 n}$ also diverges.

## Example 2

Using the comparison test, determine whether the series $\sum_{n=1}^{\infty} \frac{1}{e^n + n^e}$ converges or diverges.

Note that for all $n \in \mathbb{N}$ that $e^n + n^e ≥ e^n$. Therefore we have that $\frac{1}{e^n + n^e} ≤ \frac{1}{e^n}$.

We note that the following series converges:

(2)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{1}{e^n} = \sum_{n=1}^{\infty} \left ( \frac{1}{e} \right )^n \end{align}

This series converges as a geometric series with $r = \frac{1}{e}$ and $\mid r \mid < 1$. Therefore by the comparison test, we have that $\sum_{n=1}^{\infty} \frac{1}{e^n + n^e}$ also converges.

## Example 3

Suppose that $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ are series such that $a_n ≥ 0$ and $b_n > 0$ for all $n \in \mathbb{N}$. Suppose that $L = \lim_{n \to \infty} \frac{a_n}{b_n}$. Prove that if $L > 0$ and $L$ is finite, then either both series converge or both series diverge.

This example is the major case of The Limit Comparison Test for Positive Series that we will look at next.

Now suppose that $L = \lim_{n \to \infty} \frac{a_n}{b_n}$ and such that $0 < L < \infty$. Since $0 < L < \infty$, then there exists two positive numbers $m, M \in \mathbb{R}$ such that:

(3)
\begin{align} \quad m < L < M \end{align}

Since $\lim_{n \to \infty} \frac{a_n}{b_n} = L$, then for some $N \in \mathbb{N}$, if $n ≥ N$ then:

(4)
\begin{align} \quad m < \frac{a_n}{b_n} < M \\ \quad m b_n < a_n < M b_n \end{align}

From above, if $\sum_{n=1}^{\infty} a_n$ converges, then by the comparison test we must then have that $\sum_{n=1}^{\infty} b_n$ converges as well. Similarly, if $\sum_{n=1}^{\infty} b_n$ diverges then we must have that $\sum_{n=1}^{\infty} a_n$ diverges.