Series Convergence and Divergence Practice Examples 4

# Series Convergence and Divergence Practice Examples 4

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 $\sum_{n=1}^{\infty} \frac{\cos (n \pi)}{(n+1) \ln (n+1)}$ converge (absolutely), converge (conditionally) or diverge?

We note that this is an alternating series since $\cos (n \pi) = 1$ if $n$ even, and $\cos (n \pi) = -1$ if $n$ is odd. Let's first check the divergence test to see if this series diverges. We first note that

(1)
\begin{align} 0 ≤ \biggr \rvert \frac{\cos (n \pi)}{(n+1) \ln (n+1)} \biggr \rvert ≤ \frac{1}{(n+1)\ln (n+1)} \end{align}

We note that $\lim_{n \to \infty} \frac{1}{(n+1)\ln (n+1)} = 0$, and so by the squeeze theorem $\lim_{n \to \infty} \frac{\cos (n \pi)}{(n+1) \ln (n+1)} = 0$. So the divergence test FAILS in this case. We now will check to see if this series is absolutely convergent as follows by determining whether $\sum_{n=1}^{\infty} \biggr \rvert \frac{\cos (n \pi)}{(n+1) \ln (n+1)} \biggr \rvert = \sum_{n=1}^{\infty} \frac{1}{(n+1) \ln (n+1)}$ converges or diverges. If our original series is absolutely convergent then we are done. We will use the integral test to determine the convergence of this absolute-valued series. Let $f(x) = \frac{1}{(x+1)\ln (x+1)}$

(2)
\begin{align} \int_{1}^{\infty} \frac{1}{(x+1)\ln (x+1)} \: dx \end{align}

We first want to evaluate the follow indefinite integral by letting $u = \ln (x +1)$ so that $du = \frac{1}{x+1} \: dx$ and so:

(3)
\begin{align} \quad \int \frac{1}{(x+1)\ln (x+1)} \: dx = \frac{1}{u} \: du = \ln u + C = \ln(\ln(x+1)) + C \end{align}

Therefore we have that:

(4)
\begin{align} \quad \int \frac{1}{(x+1)\ln (x+1)} \: dx = \lim_{b \to \infty} \int_1^b \frac{1}{(x+1)\ln (x+1)} \: dx = \lim_{b \to \infty} [\ln(\ln(x+1))]_1^b = \ln(\ln(b+1)) - \ln(\ln(2)) = \infty \end{align}

Therefore $\sum_{n=1}^{\infty} \biggr \rvert \frac{\cos (n \pi)}{(n+1) \ln (n+1)} \biggr \rvert = \sum_{n=1}^{\infty} \frac{1}{(n+1) \ln (n+1)}$ is divergent, and so $\sum_{n=1}^{\infty} \frac{\cos (n \pi)}{(n+1) \ln (n+1)}$ is not absolutely convergent.

Our last chance is to check using the alternating series test. We note that $\lim_{n \to \infty} \frac{1}{(n+1)\ln (n+1)} = 0$. We also know this series is ultimately decreasing, and that $a_na_{n+1} < 0$. Therefore by the alternating series test $\sum_{n=1}^{\infty} \frac{\cos (n \pi)}{(n+1) \ln (n+1)}$ is convergent (conditionally).

## Example 2

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

Applying the root test we have that:

(5)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{2^n}{n^n} \biggr \rvert^{1/n} = \lim_{n \to \infty} \left ( \frac{2^n}{n^n} \right )^{1/n} = \lim_{n \to \infty} \frac{2}{n} = 0 \end{align}

Therefore by the root test we have that $\sum_{n=1}^{\infty} \frac{2^n}{n^n}$ converges.

## Example 3

Does the series $\sum_{n=1}^{\infty} \frac{(2n)!6^n}{(3n)!}$ converge or diverge?

Applying the ratio test we get that:

(6)
\begin{align} \quad \lim_{n \to \infty} \frac{\frac{(2(n+1))!6^{n+1}}{(3(n+1))!}}{\frac{(2n)!6^n}{(3n)!}} = \lim_{n \to \infty} \frac{(2n+2)!(3n)!6^{n+1}}{(3n+3)!(2n!)6^n} = \lim_{n \to \infty} \frac{(2n+2)!(3n)!6}{(3n+3)!(2n!)} = \lim_{n \to \infty} \frac{6(2n+2)(2n+1)}{(3n+3)(3n+2)(3n+1)} = 0 \end{align}

So by the ratio test we have that $\sum_{n=2}^{\infty} \frac{(2n)!6^n}{(3n)!}$ converges.

## Example 4

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

We should note to NOT make the mistake of assuming this is an alternating series. In fact, notice that:

(7)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{1 + (-1)^n}{\sqrt{n}} = 0 + \frac{2}{\sqrt{2}} + 0 + \frac{2}{\sqrt{4}} + 0 + \frac{2}{\sqrt{6}} + ... \end{align}

This series can be compressed such that $\sum_{n=1}^{\infty} \frac{1 + (-1)^n}{\sqrt{n}} = 2 \sum_{k=1}^{\infty} \frac{1}{\sqrt{2k}}$. We note that $2 \sum_{n=1}^{\infty} \frac{1}{\sqrt{2k}} = \sqrt{2} \sum_{n=1}^{\infty} \frac{1}{\sqrt{k}}$, and this series is a divergent p-series, and so $\sum_{n=1}^{\infty} \frac{1 + (-1)^n}{\sqrt{n}}$ diverges.

## Example 5

Let $f(x) = x^3$. Does the series $\sum_{n=1}^{\infty} \frac{f'(n) }{f''(n) + n^2}$ converge or diverge?

We first note that $f'(x) = 3x^2$ and that $f''(x) = 6x$. Making these substitutions we get that:

(8)
\begin{align} \sum_{n=1}^{\infty} \frac{f'(n) }{f''(n) + n^2} = \sum_{n=1}^{\infty} \frac{3n^2}{6n + n^2} \end{align}

Now we have that:

(9)
\begin{align} \quad \lim_{n \to \infty} \frac{3n^2}{6n + n^2} = \lim_{n \to \infty} \frac{3}{\frac{6}{n} + 1} = 3 \end{align}

Therefore by the divergence test the series $\sum_{n=1}^{\infty} \frac{f'(n) }{f''(n) + n^2}$ diverges.