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)?

We have
$$S_n = \sum_{k=n+1}^{\infty} \dfrac1{k^3} = \int_{n^+}^{\infty} \dfrac{d \lfloor t \rfloor}{t^3} = \left. \dfrac{\lfloor t \rfloor}{t^3} \right \vert_{t=n^+}^{\infty} + 3 \int_{n^+}^{\infty} \dfrac{\lfloor t\rfloor dt}{t^4} = -\dfrac1{n^2}+3 \int_{n^{+}}^{\infty} \dfrac{t-\{t\}}{t^4} dt$$
Hence, we get that
$$S_n = -\dfrac1{n^2} + 3 \int_{n^+}^{\infty} \dfrac{dt}{t^3} - 3 \int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt = \dfrac1{2n^2} - 3 \underbrace{\int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt}_{\mathcal{O}(1/n^3)}$$
You can get better approximations by repeating the above procedure, and this is called as Euler-Maclaurin Summation. To get a higher order approximation, we need to get a good approximation of $\displaystyle \int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt$. This is done as follows.
$$\int_{n^+}^{\infty} \dfrac{\{t\}}{t^4}dt = \int_{n^+}^{\infty} \dfrac{1/2}{t^4}dt+ \int_{n^+}^{\infty} \dfrac{\{t\}-1/2}{t^4}dt = \dfrac1{6n^3}+\int_{n^+}^{\infty} \dfrac{B_1(\{t\})dt}{t^4}\\ = \dfrac1{6n^3} + \overbrace{\left.\dfrac{B_2(\{t\})}{2t^4} \right \vert_{t=n^+}^{\infty}}^0 + 4 \underbrace{\int_{n^+}^{\infty} \dfrac{B_2(\{t\})}{t^5} dt}_{\mathcal{O}(1/n^4)}$$
where $B_n(x)$ are Bernoulli polynomials of order $n$.
Hence, a better asymptotic for $S_n$ is
$$S_n = \dfrac1{2n^2} - \dfrac1{2n^3} + \mathcal{O}(1/n^4)$$
Crank this repeatedly to get higher order estimates.

The Euler MacLaurin for infinite sums, where the integrand is well-behaved (by which I mean all the derivatives of the function vanish as $x \to \infty$) is
$$\sum_{k=n}^{\infty} f(k) = \int_{n}^{\infty} f(x)dx + \dfrac{f(n)}2 - \sum_{k=1}^{\infty} \dfrac{B_{2k}}{(2k)!} f^{(2k-1)'}(n)$$ where $B_{2k}$ are the Bernoulli numbers.
In your case, $f(x) = 1/x^3$ and $f^{(2k-1)'}(x) = -\dfrac{(2k+1)!}{2 \cdot x^{2k+2}}$.
Hence, we get that
$$\sum_{k=n}^{\infty} \dfrac1{k^3} = \dfrac1{2n^2} + \dfrac1{2n^3} + \sum_{k=1}^{\infty} \dfrac{(2k+1)B_{2k}}{2 \cdot n^{2k+2}}$$
$$\sum_{k=n+1}^{\infty} \dfrac1{k^3} = \dfrac1{2n^2} - \dfrac1{2n^3} + \sum_{k=1}^{\infty} \dfrac{(2k+1)B_{2k}}{2 \cdot n^{2k+2}}$$
