ALTERNATING INFINITE SUMS WITH RECIPROCAL OF CENTRAL BINOMIAL COEFFICIENT

In today´s post We will prove the following two alternating infinite sums involving the reciprocal of the central binomial coefficient that appear in this Twitter post

$\begin{aligned} &\sum_{n=0}^{\infty} \frac{(-1)^{n} 2^{2 n+1}}{(2 n+1)^{2}\left(\begin{array}{c} 2 n \\ n \end{array}\right)}=\frac{\pi^{2}}{4}-\ln ^{2}(\sqrt{2}-1)\\ &\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)^{2}\left(\begin{array}{c} 2 n \\ n \end{array}\right)}=\frac{\pi^{2}}{6}-3 \ln ^{2} \phi \\ \end{aligned}$

First recall previously proved here:

$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n+1}}{\left(2n+1\right)}$ (1)

Dividing both sides of (1) by x we obtain

$\frac{\arcsin(x)}{x\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{ 4^{n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n}}{\left(2n+1\right)}$ (2)

Now let $x \to ix$ in (2)

$\begin{aligned} \sum_{n=0}^\infty\frac{ 4^{n}(-1)^n}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{x^{2n}}{\left(2n+1\right)}&=\frac{-i\arcsin(ix)}{x\sqrt{1-(ix)^2}}\\ &=\frac{\text{arcsinh}(x)}{x\sqrt{1+x^2}}\\ \end{aligned}$ (3)

If we integrate both sides of the above equation from 0 to 1 we obtain

$\sum_{n=0}^\infty\frac{ 2^{2n}}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{(-1)^n}{\left(2n+1\right)^2}=\int_0^1\frac{\text{arcsinh}(x)}{x\sqrt{1+x^2}}\,dx$ (4)

Let´s now evaluate the integral on the R.H.S.

$\begin{align*} \int_0^1\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}}\,dx&=-\text{arcsinh}(x)\text{arcsinh}\left(\frac{1}{x}\right)\Big|_0^1+\int_0^1\frac{\operatorname{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\,dx \\ &=-\text{arcsinh}^2(1)+\int_1^\infty\frac{\text{arcsinh}\left(x\right)}{x\sqrt{1+x^2}}\,dx \qquad \left(\frac1{x}\to x \right)\\ &=-\frac{\text{arcsinh}^2(1)}{2}+\frac12\int_0^\infty\frac{\text{arcsinh}\left(x\right)}{x\sqrt{1+x^2}}\,dx \\ &=-\frac{\ln^2\left(1+\sqrt{2} \right)}{2}+\frac12\int_0^\infty\frac{x}{\sinh(x)}\,dx \\ &=-\frac{\ln^2\left(1+\sqrt{2} \right)}{2}+\frac12\int_0^\infty\frac{x}{\sinh(x)}\,dx \\ &= \frac{\pi^2}{8}-\frac{\ln^2\left(1+\sqrt{2} \right)}{2} \qquad \blacksquare\\ \end{align*}$

We used the fact that

$\frac{d}{dx}\operatorname{arcsinh}\left(\frac{1}{x}\right)=-\frac{1}{x^2\sqrt{1+\frac{1}{x^2}}}$

and

$\operatorname{arcsinh}(x)=\ln\left(x+\sqrt{1+x^2} \right)$

$\operatorname{arcsinh}(1)=\ln\left(1+\sqrt{2} \right)$

and that

$\begin{aligned} \int_0^\infty\frac{x}{\sinh(x)}\,dx &=2\int_0^\infty\frac{x e^{-x}}{1-e^{-2x}}\,dx \\ &=2\sum_{k=1}^\infty \int_0^\infty x e^{-x(2k-1)}\,dx \\ &=2\sum_{k=1}^\infty \frac{1}{(2k-1)^2}\int_0^\infty x e^{-x}\,dx \\ &=2\sum_{k=1}^\infty \frac{1}{(2k-1)^2}\\ &=2\left(\sum_{k=1}^\infty\frac{1}{k^2} -\sum_{k=1}^\infty \frac{1}{(2k)^2} \right)\\ &=2\left(\zeta(2) -\frac14\zeta(2)\right)\\ &=2 \,\frac{\pi^2}{8}\\ &=\frac{\pi^2}{4} \qquad \blacksquare \end{aligned}$

Now plugging the result obtained back in (4) we conclude that

$\sum_{n=0}^{\infty} \frac{(-1)^{n} 2^{2 n+1}}{(2 n+1)^{2}\left(\begin{array}{c} 2 n \\ n \end{array}\right)}=\frac{\pi^{2}}{4}-\ln ^{2}(\sqrt{2}-1)$

Proving the first series.

For the second one, we integrate (3) from 0 to 1/2 to get

$\sum_{n=0}^\infty\frac{ 1}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{(-1)^n}{\left(2n+1\right)^2}=\int_0^{1/2}\frac{\text{arcsinh}(x)}{x\sqrt{1+x^2}}\,dx$

Let´s now evaluate the integral on the R.H.S.

$\begin{aligned} \int_0^{1/2}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}}\,dx&=\int_0^{\ln(\phi)}\frac{x}{\sinh(x)}\,dx \qquad \left(\operatorname{arcsinh}(x) \to x \right)\\ &=2\int_0^{\ln(\phi)}\frac{x e^{-x}}{1-e^{-2x}}\,dx \\ &=2\int_1^{\phi^{-1}}\frac{\ln(x)}{1-x^2}\,dx \qquad e^{-x} \to x \right)\\ &=2\left[\ln(x)\ln(1+x)+\operatorname{Li}_{2}(-x)+\operatorname{Li}_{2}(1-x) \right]_1^{\phi^{-1}}\\ &=2\left[\ln\left(\frac{1}{\phi}\right)\ln\left(1+\frac{1}{\phi}\right)+\operatorname{Li}_{2}\left(-\frac{1}{\phi}\right)+\operatorname{Li}_{2}\left(1-\frac{1}{\phi}\right)-\operatorname{Li}_{2}(-1) \right]\\ &=2\left[-\ln^2\left(\phi\right)+\left(\frac{\ln^2\left(\phi\right)}{2}-\frac{\pi^2}{15} \right)+\operatorname{Li}_{2}\left(\frac{1}{\phi^2}\right)+\eta(2) \right]\\ &=2\left[-\ln^2\left(\phi\right)+\left(\frac{\ln^2\left(\phi\right)}{2}-\frac{\pi^2}{15} \right)+\left(\frac{\pi^2}{15}-\ln^2\left(\phi\right) \right) +\frac{\pi^2}{12}\right]\\ &=2\left[-\frac{3\ln^2\left(\phi\right)}{2}+\frac{\pi^2}{12}\right]\\ &=\frac{\pi^2}{6}-3\ln^2\left(\phi\right) \qquad \blacksqaure\\ \end{aligned}$

And we prove the second series.

$\sum_{n=0}^\infty\frac{ 1}{\left(\begin{array}{c}2n\\ n\end{array}\right)}\frac{(-1)^n}{\left(2n+1\right)^2}=\frac{\pi^2}{6}-3\ln^2\left(\phi\right)$

In the evaluation of the last integral we used that:

$\begin{gathered} \phi=\frac{1}{\phi}+1, \quad \phi^{2}=\phi+1 \\ \phi^{-2}+\phi^{-1}=1 \Rightarrow \frac{1}{\phi^{2}}=1-\frac{1}{\phi} \end{gathered}$

and

$\begin{aligned} \mathrm{Li}_{2}\left(\frac{1}{\phi}\right) &=\frac{\pi^{2}}{10}-\ln ^{2} \phi \\ \mathrm{Li}_{2}\left(-\frac{1}{\phi}\right) &=\frac{\ln ^{2} \phi}{2}-\frac{\pi^{2}}{15} \\ \mathrm{Li}_{2}\left(\frac{1}{\phi^{2}}\right) &=\frac{\pi^{2}}{15}-\ln ^{2} \phi \end{aligned}$

Proved previously in this post.

Also for the integral

$\begin{aligned} \int\frac{\ln(x)}{1-x^2}\,dx&=\frac12\left( \int\frac{\ln(x)}{1+x}+\int\frac{\ln(x)}{1-x}\right)\\ &=\frac12\left( \int\frac{\ln(x)}{1+x}+\operatorname{Li}_{2}(1-x)\right)\\ &=\frac12\left(\ln(x)\ln(1+x)-\int \frac{\ln(1+x)}{x}\,dx+\operatorname{Li}_{2}(1-x)\right)\\ &=\frac12\left(\ln(x)\ln(1+x)+\operatorname{Li}_{2}(-x)+\operatorname{Li}_{2}(1-x)\right) \qquad \blacksquare \end{aligned}$

Appendix:

$\arcsin(ix)=i\,\text{arcsinh}(x)$

Proof:

Let

$x=\sinh(w)$

Then

$\begin{aligned} x=\sinh(w) \, \Rightarrow \, w=\text{arcsinh}(x) \qquad (A.1) \end{aligned}$

On the other hand

$\begin{aligned} \sin(iw)&=i \sinh(w)=ix\\ & \Rightarrow \, iw=\arcsin(ix)\\ & \Rightarrow \, w=-i\arcsin(ix) \qquad \qquad (A.2) \end{aligned}$

Equating (A.1) and (A.2) we obtain the desired result.

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