TITLE: Formula for partial sum of Riemann zeta function
QUESTION [6 upvotes]: Possible Duplicate:
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!

REPLY [4 votes]: Perhaps that you are thinking at the formula proposed by Woon 'A New Representation of the Riemann Zeta Function zeta(s)'.

REPLY [3 votes]: 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.
$$H^{(s)}_k=\frac{(-1)^{-s}B_{-s+1}+B_{-s+1}(k+1)}{-s+1}$$

REPLY [2 votes]: 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.