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)?

For any decreasing $f(x)$, we have that:
$$ \int_{a}^{b+1} f(i) \; \mathrm d i \le \sum_{i=a}^b f(i) \le \int_{a-1}^b f(i) \; \mathrm d i $$
So in our case, with $b = \infty$, and $a = k$:
$$ \int_{k}^{\infty} i^{-3} \; \mathrm d i \le \sum_{i=k}^\infty i^{-3} \le \int_{k-1}^\infty i^{-3} \; \mathrm d i $$
Giving:
$$ \int_{k}^{\infty} i^{-3} \; \mathrm d i \le \sum_{i=k}^\infty i^{-3} \le \int_{k-1}^\infty i^{-3} \; \mathrm d i $$
$$ \frac{1}{2k^2} \le \sum_{i=k}^\infty i^{-3} \le \frac{1}{2(k-1)^2} $$
