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: 
The Laurent series expansion of the Riemann Zeta function around $s=1$ is
$$\zeta(s) = \dfrac1{s-1} + \sum\_{n=0}^{\infty} \dfrac{(-1)^n}{n!} \gamma\_n (s-1)^n$$
Further, recall that since $\zeta(2) = \dfrac{\pi^2}6 < 50$ and since $\zeta(s)$ is a rapidly decreasing function close to $1$, we have $s \in (1,2)$. Approximating $\zeta(s)$ with the first three terms gives us a quadratic in $s$, i.e.,
$$\dfrac1{s-1} + \gamma\_0 - \gamma\_1(s-1) \approx 50$$
Solving the quadratic gives us $s \approx 1.020234186$. Better approximations can be obtained by considering more terms and solving the resulting polynomial equation using numerical root finding techniques.
Since I have seen this graph for only real numbers ([source](https://pl.wikipedia.org/wiki/Funkcja_dzeta_Riemanna#/media/File:Zeta_plot.gif))
[![enter image description here](https://i.stack.imgur.com/1TH6i.jpg)](https://i.stack.imgur.com/1TH6i.jpg)
- I have start thinking about to approximate the zeta function into a more elementary function: *Hyperbola*:

> 
> $\zeta(x)=\displaystyle\sum\_{n=1}^{\infty}\frac{1}{n^x} \rightarrow
>  H(x)=\frac{\alpha x+\beta}{\lambda x-x\_0}+y\_0$ for x>1
> 
> 
> 

To fit our hyperbola we gonna need [particular values of the Riemann zeta function](https://en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function), like: 

$(2;\frac{\pi^2}{6})$ and $(4;\frac{\pi^4}{90})$

One of the *variables* for $H(x)$ is $x\_0=1$

Of course $y\_0$ should be $1$ but we gonna set $y\_0=0$ and $\alpha=0$ as well.

Hense:
$h(x)=\frac{\beta}{\lambda x-1}$

> 
> (For X variables we need (at least) X points to find our formula)
> 
> 
> 

Result: $h(x)=\frac{\pi^4}{6}\frac{1}{30-\pi^2+\frac{4(\pi^2-15)}{x}}$ for $1<x<5.2$ 

Solving for $h(x)=50$ we get $x=\frac{4(\pi^2-15)}{\pi^2(\frac{\pi^2}{300}+1)-30}\approx1.036$ not really close but by this simplification is good.

> 
> Epilog: A $H(x)$ can be fit for $x<1$ as well- here an example:
> 
> 
> 

Via $(-1;-\frac{1}{12})$,$(0;-\frac{1}{2})$ and $(\frac{1}{2};\zeta(\frac{1}{2}))$ I have fit:

$h\_2(x)=\frac{5(b+1)|x|}{12(b|x|+1)}-\frac{1}{2}$;where $b=\frac{17+24\zeta(\frac{1}{2})}{1+12\zeta(\frac{1}{2})}$;$x<1$