A beautiful sum with inverse central binomial coefficients

[akotronis]

Evaluate \(\displaystyle\sum_{n\geq1}\frac{4^n}{\binom{2n}{n}(4n^2-1)}\).

[Tolaso J Kos]

Evaluate \(\displaystyle\sum_{n\geq1}\frac{4^n}{\binom{2n}{n}(4n^2-1)}\).

A nice way of attacking sums containing inverse binomial coefficients is using the Beta function. Indeed, one can easily verify the relation:

$$\frac{1}{\binom{n}{r}}= (n+1)\int_{0}^{1}t^r \left ( 1-t \right )^{n-r}\, {\rm d}t$$

Hence for $n \mapsto 2n$ and $r \mapsto n$ we get that:

$$\begin{equation}\frac{1}{\binom{2n}{n}}= (2n+1)\int_{0}^{1}t^n \left ( 1-t \right )^n \, {\rm d}t \end{equation}$$

Hence for our given sum we have that:

\begin{align*}
\sum_{n=1}^{\infty}\frac{4^n}{\binom{2n}{n}\left(4n^2-1 \right)} &\overset{(1)}{=}\sum_{n=1}^{\infty}\frac{4^n}{\left ( 2n-1 \right )\cancel{\left ( 2n+1 \right )}} \cancel{\left(2n+1\right)}\int_{0}^{1}t^n \left ( 1-t \right )^n\, {\rm d}t \\
&=\sum_{n=1}^{\infty}\frac{1}{2n-1}\int_{0}^{1}\left [ 4 t\left ( 1-t \right ) \right ]^n \, {\rm d}t \\
&=\sum_{n=1}^{\infty}\frac{1}{2n-1}\int_{0}^{1}\left [ 4t-t^2 \right ]^n \, {\rm d}t \\
&=\sum_{n=1}^{\infty}\frac{1}{2n-1}\int_{0}^{1}\left [ {\color{red}{1}}+4t-4t^2 -{\color{red}{1}} \right ]^n \, {\rm d}t \\
&=\sum_{n=1}^{\infty}\frac{1}{2n-1}\int_{0}^{1}\left [ 1-\left ( 2t-1 \right )^2 \right ]^n \, {\rm d}t \\
&\overset{u=2t-1}{=\! =\! =\! =\! =\!} \; \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{2n-1}\int_{-1}^{1}\left ( 1-u^2 \right )^n \, {\rm d}u = \sum_{n=1}^{\infty}\frac{1}{2n-1}\int_{0}^{1}\left ( 1-u^2 \right )^n \, {\rm d}u \qquad \qquad \text{(due to parity)} \\
&\overset{u= \sin y}{=\! =\! =\! =\!} \sum_{n=1}^{\infty}\frac{1}{2n-1}\int_{0}^{\pi/2}\cos^{2n+1}y \, {\rm d}y \\
&=\int_{0}^{\pi/2}\cos^2 y \sum_{n=1}^{\infty}\frac{\cos^{2n-1} y}{2n-1}\, {\rm d}y \\
&=\int_{0}^{\pi/2}\cos^2 y \sum_{n=1}^{\infty}\frac{\cos^n y}{n}\, {\rm d}y - \frac{1}{2}\int_{0}^{\pi/2}\cos^2 x \sum_{n=1}^{\infty}\frac{\cos^{2n}y}{n}\, {\rm d}y \\
&= \frac{1}{2}\int_{0}^{\pi/2}\cos^2 y\ln \left ( \frac{1+\cos y}{1-\cos y} \right )\, {\rm d}y \\
&=\frac{1}{2}\int_{0}^{\pi/2}\left ( \frac{y}{2}+ \frac{\sin 2y}{4} \right )' \ln \left ( \frac{1+\cos y}{1-\cos y} \right )\, {\rm d}y \\
&=\int_{0}^{\pi/2}\left ( \frac{y}{2}+ \frac{\sin 2y}{4} \right )\cdot \frac{1}{\sin y}\, {\rm d}y \\
&=\frac{1}{2}\int_{0}^{\pi/2}\cos y \, {\rm d}y + \frac{1}{2}\int_{0}^{\pi/2}\frac{y}{\sin y}\, {\rm d}y \\
&=\mathcal{G}+ \frac{1}{2}
\end{align*}

where $\mathcal{G}$ stands for the Catalan's constant. In the evaluations of the series we used the formula $\ln (1-y)=-\sum \limits_{n=1}^{\infty} \frac{y^n}{n}, \; |y|<1$. The last integral is evaluated because:

$$\begin{equation} \mathcal{G}= \frac{1}{4}\int_{-\pi/2}^{\pi/2}\frac{t}{\sin t}\, {\rm d}t \end{equation}$$

holds. One way to prove $(2)$ is by parts in combination with the Fourier cosine series identity $\displaystyle \ln \sin x=-\sum_{n=1}^\infty\frac{\cos 2nx}{n}-\ln 2$.

... by Seraphim ...