Summations of $\frac{1}{n^p}$

Question

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.

Answer 1

The Riemann zeta function, commonly given by $\zeta(s)$, can be written like this: $$\zeta(s)=\sum_{k=1}^\infty\frac{1}{k^s}$$. It is known that $\zeta(1)$ is indeterminate, as $$\lim_{\epsilon\rightarrow0}(\zeta(1+\epsilon))=\pm\infty$$ and that $\zeta(2)=\frac{\pi^2}{6}$. As it would seem, the solutions for most even $s$ can be expressed as $\frac{\pi^m}{n}$ ($m,n\in\Bbb{Z}$). However, $\zeta(s)$ for odd $s$ are not as easy to express. For example, probably the nicest way of representing $\zeta(3)$, a.k.a. Apéry's constant, is: $$\zeta(3)=-\frac{1}{3}\int_0^1\frac{\log^3(1-t)^2}{t^3}dt$$ which is actually pretty neat. Except we can't actually express this integral unless we give it in terms of the zeta function. So, the closest we've got so far is approximation by numerical methods. There are other functions capable of representing values for odd $s$, but that's only because the functions were defined for cases like this. Hope this helps.

Answer 2

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 $1.1732134864$