What is next step of $\sum\limits_{a=1}^{\infty}\sum\limits_{b=1}^{\infty}\frac{H_{a+b}}{ab(a+b)}$?

Question

$$\sum\limits_{a=1}^{\infty}\sum\limits_{b=1}^{\infty}\sum\limits_{c=1}^{\infty}\frac{1}{abc(a+b+c)}=\sum\limits_{a=1}^{\infty}\sum\limits_{b=1}^{\infty}\frac{H_{a+b}}{ab(a+b)}=\sum\limits_{a=1}^{\infty}?$$

What is the next step?

Answer 1

If you undo the first step like Hurkyl suggested, you are left with $$ \int_{0}^{1}\frac{-\log^3(1-x)}{x}\,dx = \int_{0}^{1}\frac{-\log^3(x)}{1-x}\,dx = \sum_{n\geq 0}\int_{0}^{1}-\log^3(x)x^n\,dx = \sum_{n\geq 0}\frac{6}{(n+1)^4}=6\,\zeta(4)=\color{red}{\frac{\pi^4}{15}}.$$

Answer 2

\begin{eqnarray*} \frac{1}{abc(a+b+c)}= \left( \frac{1}{ab} +\frac{1}{bc}+\frac{1}{ca} \right) \frac{1}{(a+b+c)^2} \end{eqnarray*} so by symmetry \begin{eqnarray*} \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{1}{abc(a+b+c)} = 3\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{1}{ab(a+b+c)^2}. \end{eqnarray*} Similarly \begin{eqnarray*} \frac{1}{ab}= \left( \frac{1}{a} +\frac{1}{b} \right) \frac{1}{(a+b)} \end{eqnarray*} so by symmetry \begin{eqnarray*} \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{1}{abc(a+b+c)} =6 \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{1}{a(a+b)(a+b+c)^2}. \end{eqnarray*} How to evaluate this sum in terms of $ \zeta(4)$ available on request.

Answer 3

Define

$$ f(x) = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{x^{a+b+c}}{abc(a+b+c)} $$

Then, we have

$$ x f'(x) = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{x^{a+b+c}}{abc} = \left( \sum_{a=1}^{\infty} \frac{x^a}{a} \right) \left( \sum_{b=1}^{\infty} \frac{x^b}{b} \right)\left( \sum_{c=1}^{\infty} \frac{x^c}{c} \right) \\ = \mathrm{Li}_1(x)^3 = (-\log(1-x))^3$$

where $\mathrm{Li}_1$ denotes a polylogarithm.

Therefore, you can reduce the problem to the analytic one of determining

$$ \int_0^1 \frac{\mathrm{Li}_1(x)^3}{x} \, \mathrm{d} x $$

Wolframa Alpha computes this integral to be $\frac{\pi^4}{15}$. It can also express indefinite integral in terms of higher polylogarithms.

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Technical note: pay attention to the fact the righthand sums don't converge when $x=1$. However, by analytic continuation, the fact it holds for $|x| < 1$ implies it should also hold on the domain of $f$.