# 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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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 $0 < a < 1$.
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