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A generating function involving harmonic number of even index

Let \mathcal{H}_n denote the n-th harmonic number. Prove that forall |x|<1 it holds that

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

Solution

Well we are stating two lemmata.

Lemma 1: For all |x|<1 it holds that

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

Proof: Pretty straight forward calculations show that

    \begin{align*} \frac{\log \left ( 1-x \right )}{1-x} &= \left ( -\sum_{n=1}^{\infty} \frac{x^n}{n} \right )\left ( \sum_{n=0}^{\infty} x^n \right ) \\ &= -\sum_{n=1}^{\infty} \mathcal{H}_n x^n \end{align*}

and Lemma 1 is proved. \blacksquare

Lemma 2: For all |x|<1 it holds that

    \[\sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} = - \frac{1}{2} \left [ \frac{\log \left ( 1-x \right )}{1-x} + \frac{\log\left ( 1+x \right )}{1+x} \right ]\]

Proof: We begin by lemma 1 and successively we have

    \begin{align*} -\frac{\log \left ( 1-x \right )}{1-x} &= \sum_{n=1}^{\infty} \mathcal{H}_{n} x^n \\ &= \sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + \\ &\quad \quad + \sum_{n=0}^{\infty} \mathcal{H}_{2n+1} x^{2n+1} \\ &=\sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + x + \\ & \quad \quad +\sum_{n=1}^{\infty} \left ( \mathcal{H}_{2n} + \frac{1}{2n+1} \right ) x^{2n+1} \\ &=\sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + x + \\ &\quad \quad +x \sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + \sum_{n=1}^{\infty} \frac{x^{2n+1}}{2n+1} \\ &= \sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + x + x \sum_{n=1}^{\infty} \mathcal{H}_{2n} x^{2n} + \\ & \quad \quad +\frac{1}{2} \left [ \log (1+x) - \log \left ( 1-x \right ) \right ] \end{align*}

and Lemma 2 follows. \blacksquare

Now, mapping x \mapsto ix back at Lemma 2 we have that

\displaystyle\sum_{n=1}^{\infty} (-1)^n \mathcal{H}_{2n} x^{2n} = -\frac{1}{2} \left [ \frac{\log(1-ix)}{1-ix} + \frac{\log(1+ix)}{1+ix} \right ]

Integrating we have that

\begin{aligned} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_{2n} x^{2n+1} }{2n+1} &= \frac{i}{4} \left [ \log^2 (1-ix) -\log^2 (1+ix) \right ] \\ &=\frac{i}{4} \bigg [ \log^2 \left ( 1+x^2 \right )-i \arctan x \log \left ( 1+x^2 \right ) \\ & \quad \quad - \log^2 \left ( 1+x^2 \right ) - i \arctan x \log \left ( 1+x^2 \right )\bigg] \\ &= \frac{\arctan x \log \left ( 1+x^2 \right )}{2} \end{aligned}

 

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