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Squared harmonic sum

Let \mathcal{H}_n denote the n-th harmonic sum. Prove that

    \[\sum_{n=1}^{\infty} \frac{\mathcal{H}_n^2}{n(n+1)}  = 3\zeta(3)\]

where \zeta denotes the Riemann zeta function.

Solution

Nothing more than a series manipulation. Successively we have:

    \begin{align*} \sum_{n=1}^{\infty} \frac{\mathcal{H}_n^2}{n\left ( n+1 \right )} &= \sum_{n=1}^{\infty}\mathcal{H}_n^2 \left [ \frac{1}{n} - \frac{1}{n+1} \right ] \\ &=\sum_{n=1}^{\infty} \left[ \frac{\mathcal{H}_n^2}{n} - \frac{\mathcal{H}_n^2}{n+1} \right] \\ &=\sum_{n=1}^{\infty} \left[ \frac{\mathcal{H}_n^2}{n} - \frac{\left ( \mathcal{H}_{n+1} - \frac{1}{n+1} \right )^2}{n+1} \right] \\ &=\sum_{n=1}^{\infty} \left [ \frac{\mathcal{H}_n^2}{n} -\frac{\mathcal{H}_{n+1}^2}{n+1} \right ] + 2 \sum_{n=1}^{\infty} \frac{\mathcal{H}_{n+1}}{\left ( n+1 \right )^2} - \\ & \quad \quad - \sum_{n=1}^{\infty} \frac{1}{(n+1)^3} \\ &= 1 - \cancelto{0}{\lim_{n \rightarrow +\infty} \frac{\mathcal{H}_{n+1}^2}{n+1}} + 2 \left ( 2 \zeta(3) - 1 \right ) -\\ & \quad \quad - \left ( \zeta(3) - 1 \right ) \\ &=1 + 4 \zeta(3) - 2 - \zeta(3) + 1 \\ &=3\zeta(3) \end{align*}

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