Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 11 down vote accepted

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:

plot of Hardy-Littlewood function

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.