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A series involving Harmonic numbers

Calculus (Integrals, Series)
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mathofusva
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A series involving Harmonic numbers

#1

Post by mathofusva » Wed Jul 26, 2017 8:05 pm

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.
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Tolaso J Kos
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Re: A series involving Harmonic numbers

#2

Post by Tolaso J Kos » Wed Jul 26, 2017 10:50 pm

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?
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Re: A series involving Harmonic numbers

#3

Post by r9m » Fri Jul 28, 2017 9:01 pm

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.
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