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Harmonic series

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

    \[\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n(n+1)(n+2)} = \frac{\pi^2}{12} - \frac{1}{2}\]

Solution

We recall the series \displaystyle \sum_{n=1}^{\infty} \mathcal{H}_n x^n = - \frac{\ln (1-x)}{1-x}. Integrating we get

    \[\sum_{n=1}^\infty \frac{\mathcal{H}_n}{n+1}x^{n+1} = \frac{1}{2}\ln^2(1-x)\]

The last takes the form

    \[\sum_{n=1}^\infty \frac{\mathcal{H}_n}{n}x^{n} = \frac{1}{2}\,\ln^2(1-x) + \mathrm{Li}_2(x)\]

where \mathrm{Li}_2 denotes the dilogarithm function. Thus,

    \begin{align*} \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n \left ( n+1 \right ) \left ( n+2 \right )} &= \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n \left ( n+1 \right )} \int_{0}^{1} x^{n+1} \, \mathrm{d}x \\ &= \int_{0}^{1} \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n \left ( n+1 \right )} x^{n+1} \,\mathrm{d}x \\ &= \int_{0}^{1} \sum_{n=1}^{\infty} \mathcal{H}_n \left ( \frac{1}{n} - \frac{1}{n+1} \right ) x^{n+1} \, \mathrm{d}x \\ &= \frac{1}{2}\int_{0}^{1} x \ln^2 (1-x)\, \mathrm{d}x +\int_0^1 x \mathrm{Li}_2(x) \, \mathrm{d}x \\ &\quad \quad \quad \quad -\frac{1}{2} \int_0^1 \ln^2(1-x) \, \mathrm{d}x \\ &=\frac{1}{2} \cdot \frac{7}{4} + \frac{\pi^2}{12} - \frac{9}{24} - 2 \cdot \frac{1}{2} \\ &= \frac{\pi^2}{12} - \frac{1}{2} \end{align*}

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