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Let $a_n = \begin{cases}\frac{1}{n}&\text{ if $n$ is a square},\\[6pt] \frac{-1}{n}&\text{ otherwise}.\end{cases}$

Determine if $\displaystyle\sum\limits_{n\geq1}a_n$ is a convergent or divergent series. If is convergent determine its value.

I have no idea how to proceed.

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You know that $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ diverges, and you can write this as $\displaystyle \sum_{n=1}^\infty \frac{1}{n}=\sum_{n=1}^\infty \frac{1}{n^2}+\sum_{n\ne\text{square}}^\infty \frac{1}{n}$, so one of the sums on the right must diverge (which?). Then what can you conclude about your series?

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Hint: Consider a simpler series:

$$ b_n = \begin{cases}1/n&\text{ if $n$ is a square}, \\ 0&\text{ otherwise}.\end{cases} $$

Can you determine if $\sum b_n$ converges? Now, what about $\sum 2b_n-a_n$?

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You're being told that $$a_n=\begin{cases}-n^{-1}&n\neq m^2\\n^{-2}&n=m^2\end{cases}$$ Can you write your series in terms of $\sum n^{-1}$, $\sum n^{-2}$?

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