# 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?

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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:

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