Is there a formula for $\sum_{n=1}^{k} \frac1{n^3}$? I am searching for the value of $$\sum_{n=k+1}^{\infty} \frac1{n^3} \stackrel{?}{=} \sum_{n = 1}^{\infty} \frac1{n^3} - \sum_{n=1}^{k} \frac1{n^3} = \zeta(3) - \sum_{n=1}^{k} \frac1{n^3}$$
For which I think I need to know the value of $$\sum_{n=1}^{k} \frac1{n^3}$$
Does anyone know of a formula (and a reference if it is complicated)?

Not to pile on, but there is a neat expression for finite Riemann zetas, based on the following fact:
$$\int_0^{\infty} dt \frac{t^2 e^{-N t}}{e^t-1} = 2 \sum_{k=N+1}^{\infty} \frac{1}{k^3}$$
You can prove this using a geometric sum-type expansion of the denominator and evaluation of the subsequent integrals.  Applying integration by parts, you get the following expansion for large $N$:
$$\int_0^{\infty} dt \frac{t^2 e^{-N t}}{e^t-1} =  \frac{1}{N^2} - \frac{1}{N^3} + \frac{1}{2 N^4} + O \left ( \frac{1}{N^5} \right )$$
Therefore a good approximation for the sum is
$$\sum_{k=1}^N \frac{1}{k^3} \approx \zeta(3)-\left[\frac{1}{2 N^2} - \frac{1}{2 N^3} + \frac{1}{4 N^4}\right]$$
Even for small $N$, say, $N=5$, the relative error in the the above approximation is vanishingly small, i.e., less than $0.03\%$.  For larger $N \sim 1000$, the error is swamped by machine precision.  For even smaller errors, one may use Richardson extrapolation as discussed in Bender & Orszag, p. 375.
