Limit of the series $\sum{a_n}$ where $a_n= \frac{1}{n}$ if $n$ is a square and $\frac{1}{n^2}$ otherwise

Question

Recently I have stumbled upon a question like this-

$$a_n = \begin{cases} \frac{1}{n} \ \mbox{ if } n \mbox{ is a perfect square} \\ \frac{1}{n^2} \ \mbox{ otherwise}, \end{cases}$$ Comment on the convergence or divergence of the series $\sum{a_n}$.

I am not sure but I thought like this- $$\sum{a_n} =\\(1+\frac{1}{4} +\frac{1}{9}+\frac{1}{4}+\frac{1}{25}+...)<2(1+\frac{1}{4}+\frac{1}{9}+...)<\infty$$ Thus $\sum{a_n}$ converges. But I am not sure if the steps are correct or not. Any help would be very much appreciated!

Answer 1

To make your idea rigorous, we can let $k$ be the largest integer so that $k^2\leq N$. We have $$ \sum_{n=1}^Na_n<\sum_{n=1}^k\frac1{n^2}+\sum_{n=1}^N\frac1{n^2}<2\sum_{n=1}^N\frac1{n^2}. $$ Now, use that the RHS converges.

Answer 2

The set of perfect squares greater than $1$, $S$, is enumerated by $\{ k^2, k \in \mathbb{N}^* \}$ (not difficult to prove if required), then :

$$\sum_{n \in S} \frac 1 n = \sum_{k\ge 1} \frac 1 {k^2}.$$

Then :

$$\sum_{n \ge 1} a_n = \sum_{n \in S} a_n + \sum_{n \notin S} a_n < \sum_{k \ge 1 } \frac 1 { k^2} + \sum_{k \ge 1} \frac 1 {k^2} = 2 \sum_{k \ge 1} \frac 1 {k^2} <\infty$$

Answer 3

As an alternative, consider \begin{align*} \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty a_n &= \sum_{n=1} \left(\frac{1}{n^2} - a_n\right) \\ &= \sum_{n=1}^\infty \begin{cases} \frac{1}{n^2} - \frac{1}{n} & \text{if $n$ is a perfect square} \\ 0 & \text{otherwise} \end{cases} \\ &= \sum_{m=1}^\infty \left(\frac{1}{m^4} - \frac{1}{m^2}\right) \\ &= \zeta(4) - \zeta(2) = \frac{\pi^4}{90} - \frac{\pi^2}{6}. \end{align*} Therefore, $$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{n^2} - \left(\frac{\pi^4}{90} - \frac{\pi^2}{6}\right) = \frac{\pi^2}{3} - \frac{\pi^4}{90}.$$