The Alternating Series Test Examples 2

# The Alternating Series Test Examples 2

Let $\{ a_n \}$ be a sequence. Recall from The Alternating Series Test page that if for $n$ sufficiently large we have that:

• $a_na_{n+1} < 0$
• $\mid a_{n+1} \mid ≤ \mid a_n \mid$
• $\lim_{n \to \infty} a_n = 0$

Then the series $\sum_{n=1}^{\infty} a_n$ is convergent.

We will now look at some examples applying this test.

## Example 1

Using the alternating series test, determine whether the series $\sum_{n=1}^{\infty} \frac{200 \cos (n \pi)}{2n + 3}$ converges or diverges.

Note that $\cos (n \pi ) = -1$ if $n$ is odd, and $\cos (n \pi) = 1$ if $n$ is even. Therefore:

(1)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{200 \cos (n \pi)}{2n + 3} = \sum_{n=1}^{\infty} \frac{200 (-1)^n}{2n + 3} \end{align}

Clearly $a_na_{n+1} < 0$. Furthermore, $\mid a_{n+1} \mid ≤ \mid a_n \mid$ for all $n \in \mathbb{N}$. Lastly, since $\lim_{n \to \infty} \frac{200}{2n + 3} = 0$, we have that:

(2)
\begin{align} \quad \lim_{n \to \infty} \frac{200 (-1)^n}{2n + 3} = 0 \end{align}

Therefore by the alternating series test, $\sum_{n=1}^{\infty} \frac{200 \cos (n \pi)}{2n + 3}$ converges.

## Example 2

Using the alternating series test, determine whether the series $\sum_{n=1}^{\infty} (-1)^n \frac{20n^2 -n - 1}{n^3 + n^2 + 33}$ converges or diverges.

Notice that the $\frac{20n^2 -n - 1}{n^3 + n^2 + 33} > 0$ for all $n \in \mathbb{N}$. Therefore we have that $a_na_{n+1} < 0$. Furthermore, it's not hard to see that $\mid a_{n+1} \mid ≤ \mid a_n \mid$. Lastly, we note that:

(3)
\begin{align} \quad \lim_{n \to \infty} \frac{20n^2 -n - 1}{n^3 + n^2 + 33} = 0 \end{align}

Therefore:

(4)
\begin{align} \quad \lim_{n \to \infty} (-1)^n \frac{20n^2 -n - 1}{n^3 + n^2 + 33} = 0 \end{align}

Therefore by the alternating series test, $\sum_{n=1}^{\infty} (-1)^n \frac{20n^2 -n - 1}{n^3 + n^2 + 33}$ converges.