Question: I've to find a value for 's' were the infinit sum gives me the value 50. Is that possible and how do I've to calculate that value.

I've no idea how te begin so, help me!

Solve s for:

$$\sum\limits \_{n=1}^{\infty}\frac{1}{n^s}=50$$

(and I know that it's the Riemann zeta function for real numbers bigger than 1)

Answer: 
We have that the function $\zeta(s)$ is decreasing on the interval $(1,2)$. Since $\lim\_{s\to 1^+}\zeta(s)=+\infty$ while $\zeta(2)=\frac{\pi^2}{6}<50$ we have $\zeta(s\_0)=50$ for some $s\_0\in(1,2)$. Exploiting the asymptotic expansion in a right neighbourhood of $s=1$ (formula $(15)$ [here](http://mathworld.wolfram.com/RiemannZetaFunction.html), it follows from summation by parts) we have:
$$ \zeta(s)\approx \frac{1}{s-1}+\gamma $$
so a reasonable estimate for $s\_0$ is given by:
$$ s\_0\approx\frac{51-\gamma}{50-\gamma} $$
and we can increase the accuracy of such approximation by using Newton's method, for instance.
$$s\simeq1.0202341852181139861332276132729412593842116437344617189788796190\ldots$$