Series Convergence and Divergence Practice Examples 2

# Series Convergence and Divergence Practice Examples 2

We will now look at some more examples of applying the various convergence/divergence tests we have looked at so far to some series without being given what test to apply specifically.

More examples of evaluating series can be found on the following page:

## Example 1

Does the series $\sum_{n=1}^{\infty} \frac{100^n}{n!}$ converge or diverge?

Our first though is to apply the ratio test, that is:

(1)
\begin{align} \quad \lim_{n \to \infty} \frac{\frac{100^{n+1}}{(n+1)!}}{\frac{100^n}{n!}} = \lim_{n \to \infty} \frac{100^{n+1}n!}{100^n (n+1)!} = \lim_{n \to \infty} \frac{100}{n+1} = 0 = L \end{align}

Since $0 ≤ L < 1$, we have that $\sum_{n=1}^{\infty} \frac{100^n}{n!}$ converges by the ratio test.

## Example 2

Does the series $\sum_{n=1}^{\infty} \frac{1+n}{2+n}$ converge or diverge?

Notice that $\lim_{n \to \infty} \frac{1+n}{2+n} = \lim_{n \to \infty} \frac{\frac{1}{n} + 1}{\frac{2}{n} + 1} = 1$, so $\sum_{n=1}^{\infty} \frac{1+n}{2+n}$ diverges by the divergence test.

## Example 3

Does the series $\sum_{n=1}^{\infty} \frac{1 + \sin n}{n^3}$ converge or diverge?

We note that $0 ≤ 1 + \sin n ≤ 2$ for all $n \in \mathbb{N}$. So we have that $\frac{1 + \sin n}{n^3} ≤ \frac{2}{n^3}$. By the comparison test, we know that $\sum_{n=1}^{\infty} \frac{2}{n^3}$ is convergent, and since $\sum_{n=1}^{\infty} \frac{1 + \sin n}{n^3} ≤ \sum_{n=1}^{\infty} \frac{2}{n^3}$, then $\sum_{n=1}^{\infty} \frac{1 + \sin n}{n^3}$ is also convergent.

## Example 4

Does the series $\sum_{n=1}^{\infty} \frac{1}{2^n(n+1)}$ converge or diverge?

Applying the ratio test we get that:

(2)
\begin{align} \quad \lim_{n \to \infty} \frac{\frac{1}{2^{n+1}(n+2)}}{\frac{1}{2^n(n+1)}} = \lim_{n \to \infty} \frac{2^n(n+1)}{2^{n+1}(n+2)} = \lim_{n \to \infty} \frac{n+1}{2n + 4} = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2 + \frac{4}{n}} = \frac{1}{2} = L \end{align}

Since $0 ≤ L < 1$, by the ratio test we get that $\sum_{n=1}^{\infty} \frac{1}{2^n(n+1)}$ converges.

Note that we could have also applied the root test as follows (omitting to show that $\lim_{n \to \infty} (1 + n)^{1/n} = 1$):

(3)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{1}{2^n(n+1)} \biggr \rvert^{1/n} = \lim_{n \to \infty} \left ( \frac{1}{2^n(n+1)} \right)^{1/n} = \lim_{n \to \infty} \frac{1}{2(n+1)^{1/n}} = \frac{1}{2} \cdot \frac{1}{\lim_{n \to \infty} (n+1)^{1/n}} = \frac{1}{2} \cdot 1 = \frac{1}{2} = L \end{align}

Once again since $0 ≤ L < 1$, by the root test we get that $\sum_{n=1}^{\infty} \frac{1}{2^n(n+1)}$ converges.

## Example 5

Does the series $\sum_{n=1}^{\infty} \frac{1}{\pi^n + 5}$ converge or diverge?

We note that $\frac{1}{\pi^n + 5} < \frac{1}{\pi^n} = \left ( \frac{1}{\pi} \right)^n$. We note that $\sum_{n=1}^{\infty} \left ( \frac{1}{\pi} \right)^n$ is a convergent geometric series since the common ratio $\mid r \mid = \biggr \rvert \frac{1}{\pi} \biggr \rvert < 1$, and so by the comparison test since $\sum_{n=1}^{\infty} \frac{1}{\pi^n + 5} < \sum_{n=1}^{\infty} \left ( \frac{1}{\pi} \right)^n$ we have that $\sum_{n=1}^{\infty} \frac{1}{\pi^n + 5}$ converges.