Summations of $\frac{1}{n^p}$ I wasn't able to find any information on this, so I'm just curious and am asking here. 
We know the summation (n sums from 1 to infinity) of $\frac{1}{n^2}$, $\frac{1}{n^4}$, and I'm sure finding $\frac{1}{n^k}$ where k is a multiple of 2 shouldn't be an issue. 
My question is, do we know the value of $\frac{1}{n^p}$ for any prime p? As far as I know, we know approximately what $\frac{1}{n^3}$ sums to but we don't have a precise value as we do, for instance, for $\frac{1}{n^2}$.
Again, I'm just curious, and would love to learn more. 

If you are concerned by the patrial sums$$S_n=\sum_{k=1}^n\frac{1}{k^s}=H_n^{(s)}$$ where appear the generalized harmonic numbers.
Usig their asymptotics, we have
$$S_n=n^{-s} \left(\frac{n}{1-s}+\frac{1}{2}-\frac{s}{12 n}+\frac{s (s+1) (s+2)}{720
   n^3}+O\left(\frac{1}{n^5}\right)\right)+\zeta (s)$$
Just for the fun of it, use it for $s=\pi$ and $n=10$. The "exact" value is $1.1732134857$ while the above truncated expression would give $1.1732134864$
