Bino Harmonic Series 1

Method 1 :

Note that,

\displaystyle \dfrac{1}{n+1} \dbinom{2n}{n} = \dfrac{\Gamma(2n+1)}{\Gamma(n+1) \Gamma(n+2)} \tag{1}

Using Gamma Duplication Formula, we have,

\displaystyle \Gamma \left(\dfrac{2n+1}{2}\right) = \dfrac{2^{-2n} \sqrt{\pi} \ \Gamma(2n+1)}{\Gamma(n+1)} \tag{2}

Using $(1)$ and $(2)$ , we have,

$\displaystyle \dfrac{1}{n+1} \dbinom{2n}{n} = \dfrac{2^{2n}}{\sqrt{\pi}} \dfrac{\Gamma \left(\dfrac{2n+1}{2}\right)}{\Gamma(n+2)}$

$\displaystyle = \dfrac{2^{2n+1}}{\pi} \cdot \operatorname{B} \left( n + \dfrac{1}{2} , \dfrac{3}{2} \right)$

$\displaystyle \implies \text{S} = \dfrac{2}{\pi} \sum_{n=1}^{\infty} H_{n} \cdot \operatorname{B} \left( n + \dfrac{1}{2} , \dfrac{3}{2} \right)$

$\displaystyle = \dfrac{4}{\pi} \int_{0}^{\frac{\pi}{2}} \sum_{n=1}^{\infty} H_{n} \sin^{2n} \theta \cos^2 \theta \ \mathrm{d}\theta$

From the generating function of harmonic number (which can be easily proved by its integral representation), we have,

$\displaystyle (1-x) \sum_{n=1}^{\infty} H_{n} x^{n} = - \ln(1-x) \ ; \ |x| < 1$

Putting $x = \sin^2 \theta$ and integrating from $0$ to $\frac{\pi}{2}$ , we have,

$\displaystyle \text{S} = -\dfrac{8}{\pi} \int_{0}^{\frac{\pi}{2}} \ln (\cos \theta) \ \mathrm{d} \theta$

$= 4 \ln 2$

$\implies A + B + C = \boxed{6}$

Method 2 :

Lemma : $\sum_{n=0}^{\infty} \dbinom{2n}{n} H_{n} x^{n} = \dfrac{2}{\sqrt{1-4x}} \ln\left(\dfrac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\right)$

Proof :

Note that,

\begin{aligned} \sum_{n=1}^\infty \binom{2n}{n} x^n &= \frac{1}{\sqrt{1-4x}} \ ; \ |x|<\frac{1}{4} \tag{1}\\ H_n &= \int_0^1 \frac{1-t^n}{1-t} \mathrm{d}t \tag{2} \end{aligned}

From $(1)$ and $(2)$ ,

$\displaystyle \begin{aligned} \sum_{n=1}^\infty \binom{2n}{n} H_n x^n &= \sum_{n=1}^\infty \binom{2n}{n} x^n \int_0^1 \frac{1-t^n}{1-t} \mathrm{d}t \\ &= \int_0^1 \frac{\frac{1}{\sqrt{1-4x}}-\frac{1}{\sqrt{1-4xt}}}{1-t} \mathrm{d}t \\ &= \frac{1}{\sqrt{1-4x}}\int_0^1 \frac{\sqrt{1-4xt}-\sqrt{1-4x}}{(1-t)\sqrt{1-4xt}} \mathrm{d}t \\ &= \frac{1}{\sqrt{1-4x}} \Bigg[ -\log(1-t)+\log \Bigg|\frac{\sqrt{1-4xt}-\sqrt{1-4x}}{\sqrt{1-4xt}+\sqrt{1-4x}} \Bigg|\Bigg]_{t=0}^{t=1} \\ &= \frac{2}{\sqrt{1-4x}}\log \left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}} \right) \end{aligned}$

$\therefore \displaystyle \text{S} = \int_{0}^{\frac{1}{4}} \sum_{n=1}^\infty \binom{2n}{n} H_{n} x^n \ \mathrm{d}x = \int_{0}^{\frac{1}{4}} \frac{2}{\sqrt{1-4x}}\log \left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}} \right) \ \mathrm{d}x$

$= 4 \ln 2$

$\implies A+B+C = \boxed{6}$