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A series involving Harmonic numbers

Posted: Wed Jul 26, 2017 8:05 pm
by mathofusva
Show that
$$\sum_{n=1}^\infty\,\frac{(-1)^{n+1}}{(n+1)^2}H_nH_{n+1} = \frac{\pi^4}{480},$$
where $H_n$ is the $n$-th Harmonic number.

Re: A series involving Harmonic numbers

Posted: Wed Jul 26, 2017 10:50 pm
by Tolaso J Kos
Let $\mathcal{H}_n$ denote the $n$-th harmonic number and consider the power series

$$\sum_{n=1}^{\infty} \mathcal{H}_n \mathcal{H}_{n+1} x^n \quad , \quad -1 \leq x <1$$

Since $\mathcal{H}_{n+1} = \mathcal{H}_n + \frac{1}{n+1}$ then we have that

\begin{align*}
\sum_{n=1}^{\infty} \mathcal{H}_n \mathcal{H}_{n+1} x^n &= \sum_{n=1}^{\infty} \mathcal{H}_n \left ( \mathcal{H}_n + \frac{1}{n+1} \right ) x^n \\
&=\sum_{n=1}^{\infty} \mathcal{H}_n^2 x^n + \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n+1}x^n \\
&=\frac{\log^2 (1-x) +{\rm Li}_2(x)}{1-x} + \frac{\log^2(1-x)}{2x}
\end{align*}

Thus mapping $x \mapsto -x$ we get that

$$\sum_{n=1}^{\infty} (-1)^{n+1} \mathcal{H}_n \mathcal{H}_{n+1} x^n = - \frac{\log^2 (1+x)+{\rm Li}_2(-x)}{1+x} + \frac{\log^2(1+x)}{2x}$$

Integrating we get that

\begin{align*}
\int \sum_{n=1}^{\infty} (-1)^{n+1} \mathcal{H}_n \mathcal{H}_{n+1} x^n \, {\rm d}x&= \sum_{n=1}^{\infty} (-1)^{n+1} \mathcal{H}_n \mathcal{H}_{n+1} \int x^n \, {\rm d}x \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n \mathcal{H}_{n+1} x^{n+1}}{n+1}\\
&=\int \left ( \frac{\log^2 (1+x)+{\rm Li}_2(-x)}{1+x} + \frac{\log^2(1+x)}{2x} \right ) \, {\rm d}x\\
&= -3 {\rm Li}_3 (1+x) + {\rm Li}_2(-x) \log(1+x)+ 3{\rm Li}_2 (1+x) \log(1+x) \\
&\quad \quad + \frac{\log^3(1+x)}{3} + \frac{3}{2} \log(-x) \log^2 \left ( 1+x \right )
\end{align*}

Hence

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n \mathcal{H}_{n+1} x^{n-1}}{n+1} = \frac{1}{x} \bigg[-3 {\rm Li}_3 (1+x) + {\rm Li}_2(-x) \log(1+x)+$$
$$+3{\rm Li}_2 (1+x) \log(1+x)+ \frac{\log^3(1+x)}{3} + \frac{3}{2} \log(-x) \log^2 \left ( 1+x \right ) \bigg] $$

Integrating from $0$ to $1$ we must get the result .... :roll: :roll: There must be something more sufficient and clever here , no?

Re: A series involving Harmonic numbers

Posted: Fri Jul 28, 2017 9:01 pm
by r9m
This is closely related to problem 11993 from American Mathematical Monthly Journal.

Now the problem presented in that integral form can be dealt with rather easily and one avoids having to calculate the last Euler Sum I left off .. :) There's an old blog post of mine with spoilers for this problem, :mrgreen: but honestly the problem is much simpler than I ever imagined.