Help in evaluating $\sum\limits_{k=1}^\infty \frac1{k}\sin (\frac{a}{k})$ I would like to try to evaluate
$$\sum\limits_{k=1}^\infty \frac{\sin (\frac{a}{k})}{k}$$
However, all of my attempts have been fruitless.  Even Wolfram Alpha cannot evaluate this sum.  Can someone help me evaluate this interesting sum?

This is the Hardy-Littlewood function (see this and this as well), which is known to be very slowly convergent. Walter Gautschi, in this article, shows that
$$\sum_{k=1}^\infty \frac1{k}\sin\frac{x}{k}=\int_0^\infty \frac{\operatorname{bei}(2\sqrt{xu})}{\exp\,u-1}\mathrm du$$
where $\operatorname{bei}(x)=\operatorname{bei}_0(x)$ is a Kelvin function, through Laplace transform techniques (i.e., $\mathcal{L} \{\operatorname{bei}(2\sqrt{xt})\}=\sin(x/s)/s$), and gives a few methods for efficiently evaluating the integral.

Here's a plot of the Hardy-Littlewood function:

