A series involving Harmonic numbers
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A series involving Harmonic numbers
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.
$$\sum_{n=1}^\infty\,\frac{(-1)^{n+1}}{(n+1)^2}H_nH_{n+1} = \frac{\pi^4}{480},$$
where $H_n$ is the $n$-th Harmonic number.
- Tolaso J Kos
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Re: A series involving Harmonic numbers
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 .... There must be something more sufficient and clever here , no?
$$\sum_{n=1}^{\infty} \mathcal{H}_n \mathcal{H}_{n+1} x^n \quad , \quad -1 \leq x <1$$
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 .... There must be something more sufficient and clever here , no?
Imagination is much more important than knowledge.
Re: A series involving Harmonic numbers
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, but honestly the problem is much simpler than I ever imagined.
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, but honestly the problem is much simpler than I ever imagined.
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