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)?
 A: 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.
A: 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:
$$ \int_{k}^{\infty} i^{-3} \; \mathrm d i \le \sum_{i=k}^\infty i^{-3} \le \int_{k-1}^\infty i^{-3} \; \mathrm d i $$
$$ \frac{1}{2k^2} \le \sum_{i=k}^\infty i^{-3} \le \frac{1}{2(k-1)^2} $$
A: 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
$$\sum_{k=n}^{\infty} \dfrac1{k^3} = \dfrac1{2n^2} + \dfrac1{2n^3} + \sum_{k=1}^{\infty} \dfrac{(2k+1)B_{2k}}{2 \cdot n^{2k+2}}$$
$$\sum_{k=n+1}^{\infty} \dfrac1{k^3} = \dfrac1{2n^2} - \dfrac1{2n^3} + \sum_{k=1}^{\infty} \dfrac{(2k+1)B_{2k}}{2 \cdot n^{2k+2}}$$
A: $$\sum _{n=1}^k \frac{1}{n^s}=H_k^{(s)}$$
Where $H_k^{(s)}$ - is the harmonic number of order s.
A: A related problem. No need for the calculations you are doing. Just recall the Herwitz zeta function

$$ \zeta(s,a)=\sum_{n=0}^{\infty}\frac{1}{(n+a)^3}, $$

then you can readily write the sum $\sum_{n=k+1}^{\infty} \frac1{n^3}$ in terms of it as 

$$ \sum_{n=k+1}^{\infty} \frac1{n^3} =  \sum_{n=0}^{\infty} \frac1{(n+k+1)^3}=\zeta(3,k+1). $$

For asymptotic behavior of the Herwitz zeta function with respect to $a$ and $s$, see here. 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Indeed, there is a Zeta Function related identity:
\begin{align}
&\bbox[5px,#ffd]{\sum_{n = k + 1}^{\infty}
{1 \over n^{3}}} =
\zeta\pars{3} - \sum_{n = 1}^{k}{1 \over n^{3}}
\\[5mm] = &\
\zeta\pars{3} -
\underbrace{\bracks{%
\zeta\pars{\color{red}{3}} - {k^{1 - \color{red}{3}} \over \color{red}{3} - 1} + \color{red}{3}\int_{k}^{\infty}{x - \left\lfloor x\right\rfloor \over x^{\color{red}{3} + 1}}\dd x}}_{\ds{\sum_{n = 1}^{k}{1 \over n^{\color{red}{3}}}}}\
\\[5mm] = &\
\bbx{{1 \over 2k^{2}} -
3\int_{k}^{\infty}{x - \left\lfloor x\right\rfloor \over x^{4}}\dd x}
\\[5mm] = &
{1 \over 2k^{2}} - {1 \over 2k^{3}} + {1 \over 4k^{4}} + \mrm{O}\pars{1 \over k^{5}}
\end{align}
