Series Convergence and Divergence Practice Examples 1

Series Convergence and Divergence Practice Examples 1

We will now look at 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 $\sum_{n=1}^{\infty} \frac{1}{n^{e-1}}$ converge or diverge?

We notice that this is a p-series test, and that $p = e - 1 > 1$, and so by the p-series test $\sum_{n=1}^{\infty} \frac{1}{n^{e-1}}$ converges.

Example 2

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

We note that since $0 ≤ \mid \sin n \mid ≤ 1$ for all $n \in \mathbb{N}$:

(1)
\begin{align} \frac{\mid \sin n \mid}{n \sqrt{n}} ≤ \frac{1}{n \sqrt{n}} = \frac{1}{n^{3/2}} \end{align}

We know by the p-series test that $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ is convergent, and since $\sum_{n=1}^{\infty} \frac{\mid \sin n \mid}{n\sqrt{n}} ≤ \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$, by the comparison test $\sum_{n=1}^{\infty} \frac{\mid \sin n \mid}{n\sqrt{n}}$ converges.

Example 3

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

Notice that $\lim_{n \to \infty} \cos \left ( \frac{1}{n^2} \right) = \cos \left ( \lim_{n \to \infty} \frac{1}{n^2} \right ) = \cos (0) = 1$. So by the divergence test, $\sum_{n=1}^{\infty} \cos \left ( \frac{1}{n^2} \right )$ diverges.

Example 4

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

We note that for $n ≥ 2$ that $0 < \ln n < n$ and so $\frac{1}{\ln n} > \frac{1}{n}$. We know that $\sum_{n=2}^{\infty} \frac{1}{n}$ diverges, and since $\sum_{n=2} \frac{1}{\ln n} ≥ \sum_{n=2}^{\infty} \frac{1}{n}$ we have that $\sum_{n=2}^{\infty} \frac{1}{\ln n}$ also converges by the comparison test.

Example 5

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

We note that for large $n$ that $\frac{1}{1 + \sqrt{n}}$ behaves like $\frac{1}{\sqrt{n}}$. We note that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ diverges as a p-series with $p = \frac{1}{2} ≤ 1$. Using the limit comparison test we have:

(2)
\begin{align} \lim_{n \to \infty} \frac{\frac{1}{1 + \sqrt{n }}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{\sqrt{n}}{1 + \sqrt{n}} = \lim_{n \to \infty} \frac{1}{\frac{1}{\sqrt{n}} + 1} = 1 \end{align}

So by the limit comparison test we have that both series diverge, that is $\sum_{n=1}^{\infty} \frac{1}{1 + \sqrt{n}}$ diverges.