# Infinite series (functional) issue

I have a problem solving this series. I'm supposed to resolve it (it could be that I'll have to change the interval).

$$\sum\limits_{n=1}^\infty (-1)^n \frac{\cos(nx)}{n^2 - 1}$$

But I don't even know where to start or how to start. Any help, advices or solutions would be very useful. Thank you very much for your answers.

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The term corresponding to $n = 1$ is undefined. – Michael Albanese Oct 18 '12 at 11:29
What do you mean by "resolving" the series? Do you want to know whether it converges? – Ayman Hourieh Oct 18 '12 at 11:57
Change what interval? You didn't give any interval to change... – rschwieb Oct 18 '12 at 12:16

(i) Let $x \in \mathbb{R} \setminus \{2j\pi: j \in \mathbb{Z}\}$. Note that $$\sum_{k=1}^n \cos kx=-{1 \over 2}+{{\sin\left(n+{1 \over 2}\right)x} \over {2\sin{x \over 2}}}$$ which implies that $$\left|\sum_{k=1}^n \cos kx\right| \leq {1 \over 2}+{1 \over {2\left|\sin{x \over 2}\right|}}.$$ In particular, the sequence $\left(\sum_{k=1}^n \cos kx\right)_{n \in \mathbb{N}}$ is bounded. On the other hand, the series $$\sum_{n=2}^\infty \left|{{(-1)^n} \over {n^2-1}}-{{(-1)^{n+1}} \over {(n+1)^2-1}}\right|$$ converges by the comparison test, since $$\left|{{(-1)^n} \over {n^2-1}}-{{(-1)^{n+1}} \over {(n+1)^2-1}}\right|=\left|{{2n^2+2n-1} \over {n^4+2n^3-n^2-2n}}\right| \leq {{5n^2} \over {n^4}}={5 \over {n^2}} \qquad (n \gg 1).$$ Hence, Dirichlet's test for convergence implies that the series $\sum_{n=2}^\infty (-1)^n {{\cos nx} \over {n^2-1}}$ converges.
(ii) Let $x=2j\pi$ for some $j \in \mathbb{Z}$. Since $\cos nx=1$ for all $n \in \mathbb{N}$, the series $$\sum_{n=2}^\infty (-1)^n {{\cos nx} \over {n^2-1}}=\sum_{n=2}^\infty {{(-1)^n} \over {n^2-1}}$$ converges by the Leibniz's alternating series test.
I'm going to guess that the problem really is, for which values of $x$ does $$\sum_2^{\infty}(-1)^n\cos nx/(n^2-1)$$ converge, and give the hint: what can you prove about $\sum_2^{\infty}(n^2-1)^{-1}$?