Show that $\sum_{n=1}^{\infty} \frac{\binom{2n}{n} (H_{2n} - H_n)}{4^n (2n - 1)^2} = 2 + \frac{3\pi}{2} \log(2) - 2G - \pi$

Question

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 $$J = \int \frac{1}{t^2 \sqrt{1 - t^2}} \log\left(\frac{1 + \sqrt{1 - t^2}}{2}\right) \, dt$$

$$J = \int \frac{1}{\sin^2 \theta \cos \theta} \log\left(\frac{1 + \cos \theta}{2}\right) \cos \theta \, d\theta$$

Answer 1

We have: $$ \begin{aligned} \sum_{n\ge 1} \binom{2n}{n} \frac{(H_{2n} - H_{n})}{4^n (2n - 1)} x^{2n - 1} &= - \int_{0}^{x} \frac{1}{t^2 \sqrt{1 - t^2}} \log\left(\frac{1 + \sqrt{1 - t^2}}{2}\right) \, dt \\ \sum_{n\ge 1} \binom{2n}{n} \frac{(H_{2n} - H_{n})}{4^n (2n - 1)} x^{2n - 2} &= - \frac 1x \int_{0}^{x} \frac{1}{t^2 \sqrt{1 - t^2}} \log\left(\frac{1 + \sqrt{1 - t^2}}{2}\right) \, dt \\ \\ \sum_{n\ge 1} \binom{2n}{n} \frac{(H_{2n} - H_{n})}{4^n (2n - 1)^2} \phantom{x^{2n - 2} } &= - \int_0^1 \frac {dx}x \int_{0}^{x} \frac{1}{t^2 \sqrt{1 - t^2}} \log\left(\frac{1 + \sqrt{1 - t^2}}{2}\right) \, dt \\ &= - \iint_{0\le t\le x\le 1} \frac 1x \frac{1}{t^2 \sqrt{1 - t^2}} \log\left(\frac{1 + \sqrt{1 - t^2}}{2}\right) \, dx\, dt \\ &= - \int_0^1 \frac{1}{t^2 \sqrt{1 - t^2}} \log\left(\frac{1 + \sqrt{1 - t^2}}{2}\right)\, dt \int_t^1\frac{dx}x \\ &= \int_0^1 \frac{1}{t^2 \sqrt{1 - t^2}} \log\left(\frac{1 + \sqrt{1 - t^2}}{2}\right) \log t \; dt \\ &\qquad\text{Substitution: $t=\sin 2u$} \\ &= \int_0^{\pi/4} \frac{1}{\sin^2 2u \color{gray}{\cos 2u}} \log\left(\frac{1 + \cos 2u}{2}\right) \log (\sin 2u) \cdot 2\color{gray}{\cos 2u}\; du \\ &= \underbrace{ \int_0^{\pi/4} \frac 1{4\sin^2 u \cos^2u} \log\left(\cos ^2 u\right) \log (2\sin u\cos u) \; 2\; du}_{:=J} \ . \end{aligned} $$ Let us compute this last integral $J$.

In it, it is easy to pass from $\log(\cos u)$ to $\log(\cos^2 u)$ back and forth, similarly for the passage between $\log(\sin u)$ and $\log(\sin^2 u)$, if needed. We split the last integral in three corresponding integrals $J_1,J_2,J_3$ to the relation $$ \log (2\sin u\cos u) =\log(2) +\log(\sin u) + \log(\cos u)\ . $$ Two of the resulted integrals are more or less straightforward: $$ \begin{aligned} J_1 &:=\int_0^{\pi/4} \frac 1{4\sin^2 u \cos^2u} \log\left(\cos ^2 u\right) \log (2) \; 2\; du \\ &=-2\log 2\int_0^{\pi/4} (\cot 2u)' \log\left(\cos u\right)\; du \\ &= -2\log 2 \Bigg[ \cot 2u\cdot \log\left(\cos u\right)\Bigg]_0^{\pi/4} -2\log 2 \int_0^{\pi/4} \cot 2u\cdot \tan u\; du \\ &= -\log 2 \int_0^{\pi/4} (1-\tan^2 u)\; du =-\log2\left(\frac \pi2-1\right)\ , \\[2mm] J_2 &:=\int_0^{\pi/4} \frac 1{4\sin^2 u \cos^2u} \log\left(\cos ^2 u\right) \log (\cos u) \; 2\; du \\ &=-2\int_0^{\pi/4} (\cot 2u)' \log^2(\cos u)\; du \\ &= -2 \Bigg[ \cot 2u\cdot \log^2(\cos u)\Bigg]_0^{\pi/4} -2 \int_0^{\pi/4} \cot 2u\cdot \tan u\cdot2\log(\cos u)\; du \\ &= -\int_0^{\pi/4} 2(1-\tan^2 u)\log(\cos u)\; du = -\int_0^{\pi/4} 2(2u-\tan u)'\log(\cos u)\; du \\ &=\left(\frac \pi2-1\right)\log 2 - 2\int_0^{\pi/4} (2u-\tan u)\tan u\; du \\ &=\left(\frac \pi2-1\right)\log 2 -\left(2G -\frac\pi2\log 2\right) +\left(2-\frac \pi2\right) \\ &=(\pi-1)\log 2- 2G + 2-\frac \pi2 \ , \\[2mm] J_3 &:=\int_0^{\pi/4} \frac 1{4\sin^2 u \cos^2u} \log\left(\cos ^2 u\right) \log (\sin u) \; 2\; du \\ &=-2\int_0^{\pi/4} (\cot 2u)' \log(\cos u) \log(\sin u) \; du \\ &= -2\Big[\cot 2u \log(\cos u) \log(\sin u) \Big]_0^{\pi/4} \\ &\qquad - 2\int_0^{\pi/4} \cot 2u \cdot \tan u \cdot\log(\sin u) \; du + 2\int_0^{\pi/4} \cot 2u \cdot\log(\cos u) \cdot\cot u \; du \\ &= \underbrace{ \int_0^{\pi/4} (\tan^2 u - 1) \log(\sin u) \; du}_{J_{31}} + \underbrace{ \int_0^{\pi/4} (\cot^2 u - 1) \log(\cos u) \; du}_{J_{32}} \\ &=J_{31} + J_{32} \ , \\ &\qquad\text{ Substitution: $t=\tan u$, $u=\arctan t$, $du=dt/(1+t^2)$} \\ J_{31} &= \frac 12 \int_0^1(t^2-1)\log\frac {t^2}{1+t^2}\;\frac{dt}{1+t^2} \\ &= \frac 12 \int_1^\infty \frac {1-t^2}{t^2}\log\frac 1{1+t^2} \;\frac{dt}{1+t^2} \ , \\ J_{32} &= \frac 12 \int_0^1 \frac {1-t^2}{t^2}\log\frac 1{1+t^2} \;\frac{dt}{1+t^2}\ ,\qquad\text{ so adding the two expressions:} \\ J_3 &= \frac 12 \int_0^\infty \frac {t^2-1}{t^2(1+t^2)} \log(1+t^2) \;dt \\ &= \int_0^\infty \frac 1{1+t^2} \log(1+t^2) \;dt - \frac 12 \underbrace{ \int_0^\infty \frac 1{t^2} \log(1+t^2) \;dt}_{=\pi}\ , \text{ so let us consider} \\ K(a) &= \int_0^\infty \frac 1{1+t^2} \log(a^2+t^2) \;dt\ , \\ K'(a) &= \int_0^\infty \frac 1{1+t^2}\cdot \frac {2a}{a^2+t^2} \;dt = \frac {2a}{a^2-1} \int_0^\infty \left(\frac 1{1+t^2}- \frac 1{a^2+t^2}\right) \;dt \\ &=\frac \pi{a+1}\ ,\\ K(a)&=\pi\log(a+1)\ ,\\ J_3&=K(1) -\frac \pi 2=\pi\log 2 -\frac \pi 2\ . \\[5mm] &\qquad\text{ Putting all together:} \\[5mm] J&=J_1+J_2+J_3 \\ &= -\log2\left(\frac \pi2-1\right) +\left((\pi-1)\log 2- 2G + 2-\frac \pi2\right) +\left(\pi\log 2 -\frac \pi 2\right) \\ &=\bbox[yellow]{\qquad\frac{3\pi}2\log 2 - 2G + 2-\pi\qquad}\ . \end{aligned} $$

Answer 2

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.