Determine the convergence of the series with an = 1/n or an = 1/n2

Consider the series

$\sum_{n=1}^{\infty} a_n$

where

$a_n = \begin{dcases} \frac{1}{n} & \text{if } n \text{ is a square},\\ \frac{1}{n^2} &\text{otherwise}. \end{dcases}$

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series converges absolutely.

Proof. For each positive integer $n$ the term $\frac{1}{n^2}$ appears at most twice (appearing twice in the case that $n$ is a perfect square), and no other terms are in the series. Hence, we apply the comparison test,

$\sum_{n=1}^{\infty} a_n < \sum_{n=1}^{\infty} \frac{2}{n^2} = 2 \sum_{n=1}^{\infty} \frac{1}{n^2}.$

Since this last series converges, we conclude that the $\sum a_n$ converges by the comparison test $. \qquad \blacksquare$

Stumbling Robot

Determine the convergence of the series with a_n = 1/n or a_n = 1/n²

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