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Suppose we have an alternating series of the form $$\sum_{n=2}^{\infty} \frac{(-1)^{f(n)}}{n^2} \text{ where } f(n) = \begin{cases}1 & \text{n is prime} \\ 0 & \text{otherwise}\end{cases}$$.

How would one go about proving that this series converges or not? Does this series have a closed form solution?

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    $\begingroup$ It converges absolutely, so there is no problem with convergence. I doubt there is a convenient form for the sum, however. $\endgroup$ – lulu Feb 23 '18 at 14:26
  • $\begingroup$ Can you see that $\left|\dfrac{(-1)^{f(n)}}{n^2}\right|\le\dfrac{1}{n^2}$? $\endgroup$ – egreg Feb 23 '18 at 14:27
  • $\begingroup$ Yes this seems true and because of this equation the function converges absolutely and so has a finite limit less then $\frac{1}{n^2}$ Thank you for the response. $\endgroup$ – Razvan Feb 23 '18 at 14:34
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    $\begingroup$ Just a little thing: pay attention to the domain of your terms. You can't evaluate $\frac{(-1)^m}{n^2}$ in $n=0$, and I don't know how you would define $f(n)$ for $n=0,1$, so I suggest to start from $n=2$. $\endgroup$ – Ottavio Bartenor Feb 23 '18 at 14:36
  • $\begingroup$ I added the question to reflect this observation $\endgroup$ – Razvan Feb 23 '18 at 14:40

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