Series Convergence and Divergence Practice Examples 3

# Series Convergence and Divergence Practice Examples 3

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 $\sum_{n=1}^{\infty} \frac{1 - (-1)^n}{n^4}$ converge or diverge?

We first might be tempted to use the alternating series test since we notice the term $(-1)^n$ is the numerator, however, this series is NOT alternating signs. Notice that every odd term is positive, and every even term is zero. Instead, let's use the comparison test.

We note that $\frac{1 - (-1)^n}{n^4} ≤ \frac{2}{n^4}$, and we know that $\sum_{n=1}^{\infty} \frac{2}{n^4}$ converges by the p-series test. Since $\sum_{n=1}^{\infty} \frac{1 - (-1)^n}{n^4} ≤ \sum_{n=1}^{\infty} \frac{2}{n^4}$, we have that $\sum_{n=1}^{\infty} \frac{1 - (-1)^n}{n^4}$ converges by the comparison test.

## Example 2

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

This example will be difficult to evaluate using some of the simpler tests, so let's try the integral test. Let $f(x) = \frac{1}{x \ln x \sqrt{ \ln \ln x}}$ and evaluate the following integral:

(1)
\begin{align} \int_{3}^{\infty} \frac{1}{x \ln x \sqrt{ \ln \ln x}} \: dx \end{align}

Let's first figure out what this indefinite integral is. Let $u = \ln \ln x$. Therefore $du = \frac{1}{x} \cdot \frac{1}{\ln x} \: dx = \frac{1}{x \ln x} \: dx$, and so:

(2)
\begin{align} \quad \int \frac{1}{x \ln x \sqrt{ \ln \ln x}} \: dx = \int \frac{1}{\sqrt{u}} \: du = \int u^{-1/2} \: du = 2\sqrt{u} + C = 2\sqrt{\ln \ln x} + C \end{align}

Therefore we have that:

(3)
\begin{align} \quad \int_{3}^{\infty} \frac{1}{x \ln x \sqrt{ \ln \ln x}} \: dx = \lim_{b \to \infty} \int_{3}^{b} \frac{1}{x \ln x \sqrt{ \ln \ln x}} \: dx = \lim_{b \to \infty} \left [ 2\sqrt{\ln \ln x} \right ]_{3}^{b} = \lim_{b \to \infty} 2\ln \ln b - 2 \ln \ln 3 = \infty \end{align}

By the integral test $\sum_{n=3}^{\infty} \frac{1}{n \ln n \sqrt{ \ln \ln n}}$ diverges.

## Example 3

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

Applying the ratio test we get that:

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

So by the ratio test $\sum_{n=1}^{\infty} \frac{n!}{n^2e^n}$ diverges.

## Example 4

Does the series $\sum_{n=1}^{\infty} \frac{n + 4}{n^3 - 2n + 3}$ converge or diverge?

We note that for large $n$ that $\frac{n + 4}{n^3 - 2n + 3}$ behaves like $\frac{n}{n^3} = \frac{1}{n^2}$, and using the limit comparison test:

(5)
\begin{align} \lim_{n \to \infty} \frac{\frac{n + 4}{n^3 - 2n + 3}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^3 + 4n^2}{n^3 - 2n + 3} = \lim_{n \to \infty} \frac{1 + \frac{4}{n}}{1 - \frac{2}{n^2} + \frac{3}{n^3}} = 1 \end{align}

So since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges as a p-series, it follows that $\sum_{n=1}^{\infty} \frac{n + 4}{n^3 - 2n + 3}$ also converges.

## Example 5

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

Using the trigonometric identity that $\sin ^2 n + \cos ^2 n = 1$ we have that $\sum_{n=1}^{\infty} \frac{n}{\sin ^2 n + \cos ^2 n} = \sum_{n=1}^{\infty} n$, and we note that $\lim_{n \to \infty} n = \infty$, so by the divergence test we have that $\sum_{n=1}^{\infty} \frac{n}{\sin ^2 n + \cos ^2 n}$ diverges.