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Does $\sum_{n=1}^{\infty}\sin(n\pi)/n^{2}$ in $\mathbb{C}$ converge or diverge?

My guess is that the series does not converge due to the periodicity of trigonometric functions but I can't be sure without figuring it out more formally.

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$\sin(n\pi)=0$. Has the question been typed correctly? If so, the seris converges, and has sum $0$, since every term is $0$. – André Nicolas Jan 23 '12 at 15:30


$$\begin{array} &\sin(\pi) = 0 \\ \sin(2 \pi) = 0 \\ \sin(3\pi) = 0 \\ \qquad\qquad\vdots \end{array}$$

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I'm a little embarrassed -- the question is trivially easy. Should've taken another look at it. Thanks though! – Emir Jan 23 '12 at 15:33

Let's suppose you meant $\displaystyle \sum \frac{\sin n}{n^2}$ instead. According to your logic, since the sine function is periodic, this sum can't converge.

But it does converge. How does one show this? Perhaps it's easiest to show that the partial sums are Cauchy. This is true because the tail of the $k$th partial sum is contained within a distance at most $\sum_{k+1}^\infty \frac{1}{n^2}$ away, which decreases as $k$ increases (and converges, of course).

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