# Formula for partial sum of Riemann zeta function [duplicate]

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Finite Sum of Power?

Suppose $f(s,k) = \sum_{n=1}^k n^{-s}$ is the Riemann zeta function truncated at the k-th term. I read on mathoverflow that there is a formula for $f(s,k)$ in terms of Bernoulli numbers, but I can't find it on the web. Would someone happen to know it or could point to a link? I am primarily interested in the case when $s$ is a negative real number.

Thanks!

## marked as duplicate by J. M. is a poor mathematician, Argon, user5783, Rudy the Reindeer, Asaf Karagila♦Jul 23 '12 at 12:22

• Thank you all so much! These are wonderful answers and links! – Thad Jul 12 '12 at 21:45

Perhaps that you are thinking at the formula proposed by Woon 'A New Representation of the Riemann Zeta Function zeta(s)'.

• Thanks for the Edit @Ali Caglayan, – Raymond Manzoni Apr 4 '16 at 23:16

Note that

$$\sum_{n=1}^k n^{-s}=H^{(s)}_k$$

where $H^{(s)}_k$ is the generalized harmonic number.

Some identities are mentioned here, e.g.

$$H^{(s)}_k=\frac{(-1)^{-s}B_{-s+1}+B_{-s+1}(k+1)}{-s+1}$$

Presumably mathoverflow was talking about Faulhaber's formula $$\sum_{k=1}^n k^p = \frac{B_{p+1}(n+1)-B_{p+1}(0)}{p+1}$$ in terms of Bernoulli polynomials. If $p$ is a positive integer, then the coefficients of the Bernoulli polynomials are essentially Bernoulli numbers.

Faulhaber's polynomial provides the answer when $s \in \mathbb{Z}^{-}$. If $s \in \mathbb{R}^-$, a good approximation can be obtained using Euler-Maclaurin which also contain the Bernoulli numbers.