# How to find the sum of this cos series

$$S = \sum_{k=1}^{\infty} \frac{cos(\theta\log(k))}{k^a}$$

How do I go about finding the value of S, given that $\theta \to \infty$ and $0 < a < 1$.

Any special techniques that might be helpful in calculating this sum?

EDIT: Just to give some background,

I was actually trying to figure out $$\sum_{k=1}^{\infty} \frac{cos(\theta\log(k))}{k^a} - \sum_{k=1}^{\infty} \frac{cos(\theta\log(k + 0.5))}{(k+0.5)^a}$$

Since that expression was a bit complicated, I decided to write the common version...

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## 1 Answer

Your value is basically the value of the Riemann zeta function: $$S = \sum_{k=1}^{\infty} \frac{Re (e^{i \theta \log(k)})}{k^a} = Re( \sum_{k=1}^{\infty} k^{i \theta - a} ) = Re( \zeta ( a - i \theta ))$$ You want to evaluate this on the critical strip $0 < a < 1$.

The good news is that there is an enormous amount of literatue on the Riemann zeta. The bad news is that this function is nasty on the critical strip.

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More bad(?) news: The sum defining the zeta function is only convergent for $a>1$. To get into the critical strip one needs to use analytic continuation. –  Hans Lundmark Oct 18 '10 at 13:02
Spot on!!! +1. Yea, I was playing around with properties of the zeta function... More precisely the Dirichlet eta function... I got stuck while evaluating this... Maybe I should edit this question and give some details on what I was doing... –  Roupam Ghosh Oct 18 '10 at 14:21
@rpg16: I presume you've seen the Riemann-Siegel formula, which I believe was what Michael and Hans were alluding to? –  Guess who it is. Oct 18 '10 at 14:42
Thanks, I'll check out that link J.M. –  Roupam Ghosh Oct 18 '10 at 14:57