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Harmonic sum with reciprocal central binomial coefficient

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

    \[\sum_{n=1}^{\infty} \frac{2^n \mathcal{H}_n}{\binom{2n}{n}} = \pi - \frac{\pi \log 2}{2} + \mathcal{G}\]

where \mathcal{G} denotes the Catalan’s constant.

Solution

We begin with a lemma:

Lemma: Let \mathrm{Li}_2 denote the dilogarithm function. It holds that

    \[\mathrm{Li}_2 \left ( \frac{1+i}{2} \right ) - \mathrm{Li}_2 \left ( \frac{1-i}{2} \right ) = -\frac{i}{4} \left ( \pi \log 2 - 8 \mathcal{G} \right )\]

Proof: It is well known that

(1)   \begin{equation*} \mathrm{Li}_2(z) + \mathrm{Li}_2 \left ( \frac{z}{z-1} \right ) = - \frac{1}{2}\ln^2 \left ( 1-z \right ) \quad , \quad z \notin [1, +\infty) \end{equation*}

Setting z=\frac{1+i}{2} we have that:

    \begin{align*} \mathrm{Li}_2 \left ( \frac{1+i}{2} \right ) + \mathrm{Li}_2 \left ( -i \right ) &= -\frac{1}{2} \ln^2 \left ( 1 - \frac{1+i}{2} \right ) \\ &=- \frac{1}{2} \ln^2 \left ( \frac{1-i}{2} \right ) \\ &= -\frac{1}{2} \left ( -\frac{\pi^2}{16} + \frac{\log^2 2}{4} + \frac{i \pi \log 2}{4} \right )\\ &=\frac{\pi^2}{32} - \frac{\log^2 2}{8} - \frac{i \pi \log 2}{8} \end{align*}

Setting z=\frac{1-i}{2} we have that:

    \begin{align*} \mathrm{Li}_2 \left ( \frac{1-i}{2} \right ) + \mathrm{Li}_2 \left ( i \right ) &= -\frac{1}{2} \ln^2 \left ( 1 - \frac{1-i}{2} \right ) \\ &=- \frac{1}{2} \ln^2 \left ( \frac{1+i}{2} \right ) \\ &= -\frac{1}{2} \left ( -\frac{\pi^2}{16} + \frac{\log^2 2}{4} - \frac{i \pi \log 2}{4} \right )\\ &=\frac{\pi^2}{32} - \frac{\log^2 2}{8} + \frac{i \pi \log 2}{8} \end{align*}

However,

    \begin{align*} \operatorname{Li}_2(iz) &= -\int_0^{z} \frac{\log (1-ix)}{x}\,dx \\ &= -\int_0^{z} \frac{\log \left((1+x^2)^{1/2}e^{-i\arctan x}\right)}{x}\,dx \\ &= -\frac{1}{2}\int_0^{z} \frac{\log \left(1+x^2\right)}{x}\,dx + i \int_0^{z} \frac{\arctan x}{x}\,dx \\ &= -\frac{1}{4}\int_0^{-z^2} \frac{\log \left(1-x\right)}{x}\,dx + i \int_0^{z} \frac{\arctan x}{x}\,dx \\ &= \frac{1}{4}\operatorname{Li}_2(-z^2) + i\operatorname{Ti}_2(z) \end{align*}

where \mathrm{Ti}_2 is the inverse tangent function. It now follows that

    \[\mathrm{Li}_2(i) = -\frac{\pi^2}{48} + i \mathcal{G} \quad , \quad \mathrm{Li}_2(-i) = - \frac{\pi^2}{48} -i \mathcal{G}\]

in view of the well known series \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} = \mathcal{G}.

Substracting the above relations we get the proof of the lemma.

Theorem: Let t=\frac{z}{z-1} \; , \; |t| \leq 1. It holds that

    \[\sum_{n=0}^{\infty} \frac{\mathcal{H}_n 4^n}{\binom{2n}{n}} z^n = \left ( 1+t \right ) \bigg [ \left (2 + \frac{1}{2} \log \frac{1+t}{4} \right ) \sqrt{t} \arctan \sqrt{t} -\]

    \[\hspace{3em} - \frac{\sqrt{-t}}{2} \left ( \mathrm{Li}_2 \left ( \frac{1+\sqrt{-t}}{2} \right ) - \mathrm{Li}_2 \left ( \frac{1-\sqrt{-t}}{2} \right ) \right ) \bigg ]\]

Setting z=\frac{1}{2} we have that t=1. The result now follows immediately using the lemma above as well as the fact that i=\sqrt{-1}.

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