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I'd appreciate any help analyzing the convergence of the following series:

$$\sum_{n=2}^\infty \frac{(-1)^n}{(n+(-1)^n)^s}$$

Thanks,

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  • $\begingroup$ What is $s$ here? $\endgroup$ – Clement C. Nov 23 '15 at 23:33
  • $\begingroup$ First, we need to know what $s$ is. You may want to expand the series and then collect certain terms, you should get something that looks like the Riemann zeta functions. $\endgroup$ – mattos Nov 23 '15 at 23:36
  • $\begingroup$ @ClementC. It's supposed to be the exponent of the denominator. The excercise is about establishing conditions on s, so the series converges. $\endgroup$ – Meno11 Nov 23 '15 at 23:40
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One may observe that, as $n \to \infty$, $$ \begin{align} \frac{(-1)^n}{(n+(-1)^n)^s}&=\frac{(-1)^n}{n^s}\frac1{(1+(-1)^n/n)^s}\\\\ &=\frac{(-1)^n}{n^s}\left(1-s\frac{(-1)^n}{n} +\mathcal{O}\left(\frac1{n^2}\right)\right)\\\\ &=\frac{(-1)^n}{n^s}-\frac{s}{n^{1+s}} +\mathcal{O}\left(\frac1{n^{2+s}}\right) \end{align} $$ giving, for some large $n_0$, $$ \sum_{n\geq n_0}\frac{(-1)^n}{(n+(-1)^n)^s}=\sum_{n\geq n_0}\frac{(-1)^n}{n^s}-s\sum_{n\geq n_0}\frac1{n^{1+s}}+\sum_{n\geq n_0}\mathcal{O}\left(\frac1{n^{2+s}}\right) $$ Thus your series is convergent for all $s>0$.

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    $\begingroup$ Damn, you were fast. $\endgroup$ – Clement C. Nov 23 '15 at 23:42
  • $\begingroup$ @OlivierOloa I can't seem to understand what you did after you factored out $\frac{(-1)^n}{n^s}\frac1{(1+(-1)^n/n)^s}$ how do you get that big o function? Thanks $\endgroup$ – Meno11 Nov 24 '15 at 22:37
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    $\begingroup$ @Meno11 I've used the fact that, as $u\to 0$: $$ \frac1{(1+u)^s}=1-s u +O(u^2)$$ To obtain this, you may just apply the Taylor expansion $f(u)=f(0)+f'(0)u+O(u^2)$ to $f(u):= \frac1{(1+u)^s}$. Thanks. $\endgroup$ – Olivier Oloa Nov 24 '15 at 23:03

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