$\sum\limits _{n=1}^{\infty}\frac{1}{n^s}=50$ Riemann-Zeta 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)
 A: 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, 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.
A: 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.
A: $$s\simeq1.0202341852181139861332276132729412593842116437344617189788796190\ldots$$
A: Mathematica yields 
$$s \approx 1.020234185218113986133227613272941259384211643734461718978879619075931...$$
I don't think you are going to get a good answer without just numerically approximating it. 
A: Since I have seen this graph for only real numbers (source)

- 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, 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 $b=\frac{17+24\zeta(\frac{1}{2})}{1+12\zeta(\frac{1}{2})}$;$x<1$
