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

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

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

Thanks,

• What is $s$ here? – Clement C. Nov 23 '15 at 23:33
• 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. – mattos Nov 23 '15 at 23:36
• @ClementC. It's supposed to be the exponent of the denominator. The excercise is about establishing conditions on s, so the series converges. – Meno11 Nov 23 '15 at 23:40

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 $s>0$.
• @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 – Meno11 Nov 24 '15 at 22:37
• @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. – Olivier Oloa Nov 24 '15 at 23:03