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On a generating function

Let \mathcal{H}_n denote the n-th harmonic number and \mathcal{H}_n^{(2)} the n-th harmonic number of order 2 , namely \mathcal{H}_n^{(2)} = \sum \limits_{k=1}^{n} \frac{1}{k^2}. Prove that

    \[\sum_{n=1}^\infty \left( H_n^2-H_n^{(2)} \right)x^{n}=\frac{\ln^2(1-x)}{1-x}\]

Solution

Lemma: Let \{a_n\}_{n \in \mathbb{N}} be a sequence such tht a_0=0. Then

    \[\sum_{n=1}^\infty a_n x^n=\frac{1}{1-x} \sum_{n=1}^\infty (a_n-a_{n-1}) x^n\]

Proof: We have successively:

    \begin{align*} \sum_{n=0}^\infty a_n x^n&=\left( \frac{1}{1-x}-\frac{x}{1-x} \right) \sum_{n=0}^\infty a_n x^n\\ &=\frac{1}{1-x}\sum_{n=0}^\infty a_n x^n-\frac{1}{1-x} \sum_{n=0}^\infty a_n x^{n+1}\\ &=\frac{1}{1-x}\sum_{n=0}^\infty a_n x^n-\frac{1}{1-x}\sum_{n=1}^\infty a_{n-1}x^{n} \end{align*}

Thus,

    \[\sum_{n=1}^\infty a_n x^n=\frac{1}{1-x} \sum_{n=1}^\infty (a_n-a_{n-1}) x^n\]

Hence for the original problem if we let a_n = \mathcal{H}_n^2 then

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

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