General P-series rule This is a p-series:
$$\sum_{n=1}^\infty \frac{1}{n^p}$$
There are 2 p-series (to my knowledge) that somehow reach pi:
$$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
$$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$$
And I noticed that $\frac{\pi^3}{26}$ is really close to $\sum_{n=1}^\infty \frac{1}{n^3}$ and after a minute of digging on the internet I found this series which it's formula is $$
f(n)=(3^{n+1}+(-1)^{n+1})/2-2^{n+1}+1$$
And the series goes $0, 2, 6, 26, 90, 302, 966, 3026...$ and it seemed to line up with the p-series formula pretty well:
$$\sum_{n=1}^\infty \frac{1}{n^2}=1.64453...$$
$$\frac{\pi^{2}}{f(2)}=1.64453...$$
$$$$
$$\sum_{n=1}^\infty \frac{1}{n^3}=1.20205...$$
$$\frac{\pi^{3}}{f(3)}=1.19254...$$
$$$$
$$\sum_{n=1}^\infty \frac{1}{n^4}=1.08232...$$
$$\frac{\pi^{4}}{f(4)}=1.08232...$$
$$$$
$$\sum_{n=1}^\infty \frac{1}{n^5}=1.03692...$$
$$\frac{\pi^{5}}{f(5)}=1.01331...$$
$$$$
$$\sum_{n=1}^\infty \frac{1}{n^6}=1.01734...$$
$$\frac{\pi^{6}}{f(6)}=0.99522...$$
But sadly that certain function only lines up perfectly with $p=2$ and $p=4$ but I was wondering if someone here can modify the function $f(n)$ so: $$\frac{\pi^{x}}{f(x)} = \sum_{n=1}^\infty \frac{1}{n^x}$$
Edit: $f(x)$ starting from 2 should look somthing like: 
$$6, 25.794, 90, 295.122, 945, 2995.285, 9450...$$
 A: What you're calling a p-series is what mathematicians call the Riemann zeta function.  You will find that for any even power, there is a power of $\pi$ in the result:
$\zeta(2n) = (-1)^{n+1}\frac{B_{2n}(2 \pi)^{2n}}{2(2n)!})$
where $B_k$ denotes the $k$th Bernoulli number.
There's much more at https://en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function
For the odd positive integers, much less is known. It is known that $\zeta(3)$ is irrational, that at least one of $\zeta(5),\ \zeta(7),\ \zeta(9),\ \zeta(11)$ is irrational, and that there are infinitely many irrational values for $\zeta(2n+1)$. But no simple closed-form solution of the type you imagine is known.
A: What does happen to be a rational multiple of $\pi^n$ is
$$
\sum_{k=0}^\infty\frac{(-1)^{kn}}{(2k+1)^n}=\left\{\begin{array}{}\beta(n)&\text{if $n$ is odd}\\\zeta(n)\left(1-2^{-n}\right)&\text{if $n$ is even}\end{array}\right.
$$
A recursion for the Dirichlet beta function, $\beta(n)$, is given for odd $n$ in this answer.
A recursion for the Riemann zeta function, $\zeta(n)$, is given for even $n$ in this answer.
