I need Help to evaluate series :$\sum_{n=0}^{\infty} \frac{(2n+1)!!}{(2n+2)!!}\frac{1}{n+1}$

Question

I need Help to evaluate series :$$\sum_{n=0}^{\infty} \frac{(2n+1)!!}{(2n+2)!!}\frac{1}{n+1}$$

Let :$u_n=\frac{(2n+1)!!}{(2n+2)!!}\frac{1}{n+1}$

We have $$\lim_{n\to\infty} n\left({\frac{u_n}{u_{n+1}}-1}\right)=\frac{3}{2}\ge 1$$ Therfore this series $\sum_{n=0}^{\infty} \frac{(2n+1)!!}{(2n+2)!!}\frac{1}{n+1}$ converge

We have : $$\sum_{n=0}^{\infty} \frac{(2n+1)!!}{(2n+2)!!}\int^1_0 x^{n} dx=\int^1_0\sum_{n=0}^{\infty} \frac{(2n+1)!!}{(2n+2)!!}x^n$$ and we have : $$\frac{1}{\sqrt{1-x}} =1+\sum_{n=0}^{\infty} \frac{(2n+1)!!}{(2n+2)!!}x^{n +1}$$ therfore: $$\sum_{n=0}^{\infty} \frac{(2n+1)!!}{(2n+2)!!}\frac{1}{n+1}=\int^1_0 \frac{1}{x\sqrt{1-x}}-\frac{1}{x} dx=2\int^1_0 \frac{1}{x+1} dx=2\ln(2)$$

I started thinking about this problem for more than three hours. I started writing the question on this site, and when I submitted my attempt, I found the idea for the solution (it's the second time). Thank you. If anyone has any other ideas, they are welcome

Answer 1

$$ \sum_{n=0}^\infty \frac{(2n+1)!!}{(2n+2)!!(n+1)} = \sum_{n=0}^\infty \frac{(2n+1)!}{(2n+2)!!(2n)!!(n+1)} = \sum_{n=0}^\infty \frac{(2n+1)!}{(n+1)!2^{n+1}(2n)!!(n+1)} $$ $$ = \sum_{n=0}^\infty \frac{(2n+1)!}{(n+1)!2^{n+1}2^n n!(n+1)} = \sum_{n=0}^\infty \frac{\Gamma(2n+2)}{\Gamma(n+2) 2^{2n+1} n!(n+1)} $$ $$ = \sum_{n=0}^\infty \frac{(2)_{2n}}{(2)_n(n+1)}\frac{1}{2^{2n+1}n!} = \sum_{n=0}^\infty \frac{(1)_n(3/2)_n2^{2n}}{(2)_n (n+1)}\frac{1}{2^{2n+1}n!} $$ $$ = \sum_{n=0}^\infty \frac{(1)_n(3/2)_n2^{2n}(1)_n}{(2)_n (2)_n}\frac{1}{2^{2n+1}n!} = \frac12 {}_3F_2(1,1,3/2 ; 2,2 ; 1) $$ Then use the integral representation for $_3F_2(a,b,b+1/2 ; a+1,2b,z]$ with $a=b=1$, $z=1$ by Gottschalk and Maslen https://dx.doi.org/10.1088/0305-4470/21/9/015 $$ = \frac{1}{2}\times 4\int_0^1 (1+y)^{-1} dy = 2\ln 2. $$