Show that $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n} (H_{2n} – H_n)}{4^n (2n – 1)^2} = 2 + \frac{3\pi}{2} \log(2) – 2G – \pi$$
My try :
We know that
$$\sum_{n=1}^{\infty} \binom{2n}{n} (H_{2n} – H_{n}) t^n = – \frac{1}{\sqrt{1 – 4t}} \log\left(\frac{1 + \sqrt{1 – 4t}}{2}\right)$$
In particular we have
$$\sum_{n=1}^{\infty} \binom{2n}{n} \frac{(H_{2n} – H_{n})}{4^n} t^{2n} = – \frac{1}{\sqrt{1 – t^2}} \log\left(\frac{1 + \sqrt{1 – t^2}}{2}\right)$$….(1)
Next, dividing both sides of (1) by $t ^ 2$ and then integrating from 0 to 2, we get
$$\sum_{n=1}^{\infty} \binom{2n}{n} \frac{(H_{2n} – H_{n})}{4^n (2n – 1)} x^{2n – 1} = – \int_{0}^{x} \frac{1}{t^2 \sqrt{1 – t^2}} \log\left(\frac{1 + \sqrt{1 – t^2}}{2}\right) \, dt$$
Making the change of variable $t=sin(\theta)$ in
$$J = \int \frac{1}{t^2 \sqrt{1 – t^2}} \log\left(\frac{1 + \sqrt{1 – t^2}}{2}\right) \, dt$$
$$J = \int \frac{1}{\sin^2 \theta \cos \theta} \log\left(\frac{1 + \cos \theta}{2}\right) \cos \theta \, d\theta$$
Best Answer
Using $$ H_{n} = \int_{0}^{1} \frac{1 - u^n}{1-u} \, du $$ then the series $$ S(t) = \sum_{n=1}^{\infty} \binom{2n}{n} \, \frac{H_{2n} - H_{n}}{4^n \, (2n-1)^2} \, t^n $$ leads to $$ S(t) = \int_{0}^{1} \frac{f(u \, t) - f(u^2 \, t)}{1-u} \, du, $$ where $$ f(t) = \sum_{n=1}^{\infty} \binom{2n}{n} \, \frac{t^n}{4^n \, (2n-1)^2}. $$ It can be shown that $$ f(t) = 1 - \sqrt{1-t} - \sqrt{t} \, \sin^{-1}(\sqrt{t}) $$ and leads to $$ S(t) = \int_{0}^{1} \frac{\sqrt{1-u^2 \, t} - \sqrt{1-u t} + \sqrt{t} \, (u \, \sin^{-1}(\sqrt{u^2 \, t}) - \sqrt{u} \, \sin^{-1}(\sqrt{u t}))}{1-u} \, du. $$ At this point the need to set $t = 1$ makes the calculations of the integrals possible. With this, then \begin{align} \int_{0}^{1} \frac{\sqrt{1-u^2} - \sqrt{1-u}}{1-u} \, du &= \frac{\pi}{2} - 1 \\ \int_{0}^{1} \frac{u \, \sin^{-1}(u) - \sqrt{u} \, \sin^{-1}(\sqrt{u})}{1-u} \, du &= 3 - \frac{3 \pi}{2} + \frac{3 \pi}{2} \, \ln(2) - 2 \bf{G}. \end{align} Using the integrals shows that $$ S(1) = 2 - \pi + \frac{3 \pi}{2} \, \ln(2) - 2 \bf{G}, $$ or $$ \sum_{n=1}^{\infty} \binom{2n}{n} \, \frac{H_{2n} - H_{n}}{4^n \, (2n-1)^2} = 2 - \pi + \frac{3 \pi}{2} \, \ln(2) - 2 \bf{G}, $$ where $\bf{G}$ is Catalan's constant.