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## Series with zetas

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### Series with zetas

Sun May 15, 2016 1:01 pm

Posts: 46
Location: http://www.tolaso.com.gr/
Let $\mathbb{N} \ni m \geq 2$. Evaluate the series

$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{\zeta(2n)+\zeta(3n)+\cdots+\zeta(mn)}{n}$$

in terms of known functions. Here $\zeta$ stands for the Riemann zeta function.

### Re: Series with zetas

Sun May 15, 2016 3:10 pm

Posts: 4
What makes you feel that a closed form exists?

### Re: Series with zetas

Sun May 15, 2016 5:40 pm

Posts: 46
Location: http://www.tolaso.com.gr/
What makes you feel that a closed form exists?

Because there exists. I reached till the very end but I did not actually assemble the pieces to write them in a nicer form. I just left them the way they are.

### Re: Series with zetas

Tue May 17, 2016 5:10 pm

Posts: 4
Can you post the closed form? At least it will help the users to head on to the right path.

### Re: Series with zetas

Thu May 19, 2016 8:49 am

Posts: 46
Location: http://www.tolaso.com.gr/
Can you post the closed form? At least it will help the users to head on to the right path.

Someone needs to know the closed form of this product:

$$\prod_{k=2}^{\infty} \left ( 1- \frac{z^n}{k^n} \right )= \prod_{k=0}^{n-1} \frac{1}{\Gamma \left ( 1-z\exp \left [ \left ( 2\pi k i /n \right ) \right ] \right )}$$

now things are way much simplified. The closed form will be a log and as an arguement it will have products of the above form.

### Re: Series with zetas

Mon Jun 06, 2016 10:35 pm

Posts: 38
VJKey wrote:
Let $\mathbb{N} \ni m \geq 2$. Evaluate the series

$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{\zeta(2n)+\zeta(3n)+\cdots+\zeta(mn)}{n}$$

in terms of known functions. Here $\zeta$ stands for the Riemann zeta function.

Are you sure this is written correctly? This series does not converge for any m.

### Re: Series with zetas

Tue Jun 07, 2016 6:09 pm

Posts: 46
Location: http://www.tolaso.com.gr/
So we need restrictions? I thought that I did not have to impose $m$ in restrictions. But , nonetheless, suppose that $m$ is such a number that the series converges.

### Re: Series with zetas

Tue Jun 07, 2016 6:17 pm

Posts: 38
VJKey wrote:
So we need restrictions? I thought that I did not have to impose $m$ in restrictions. But , nonetheless, suppose that $m$ is such a number that the series converges.

Then no such m exists. $\zeta(an)/n\sim1/n$ as $n\to\infty$ for any positive a.