I was having trouble with the sum
$$\sum^{\infty}_{n=1} \frac{1}{n {2n\choose n} }$$
My Attempt $$S=\sum^{\infty}_{n=1} \frac{1}{n {2n\choose n} }=\sum^{\infty}_{n=1} \frac{n!\; n!}{n\;(2n)! }=\sum^{\infty}_{n=1} \frac{n(n-1)!\; n(n-1)!}{2n^2\;(2n-1)! } = \frac 1 2 \sum^{\infty}_{n=1}\frac{\Gamma(n)^2}{\Gamma(2n)}$$
$$S=\sum^{\infty}_{n=1}\frac{\Big(\int_0^{\infty}x^{n-1}e^{-x}dx\Big)^2}{\int_0^{\infty}x^{2n-1}e^{-x}dx}$$
What am I doing wrong? How do I proceed? How do I solve this integral?
Use $${n\choose k}^{-1}=(n+1) \int_{0}^{1} x^k(1-x)^{n-k} dx.$$ Then $$S=\sum_{n=1}^{\infty} \frac{1}{n{2n \choose n}}= \int_{0}^{1} dx ~\sum_{n=1}^{\infty}\frac{2n+1}{n} (x(1-x))^n=\int_{0}^{1} \frac{2x(1-x)}{1-x+x^2}dx-\int_{0}^{1}\log(1-x+x^2) dx =$$ $$\implies S=-2+2 \int_{0}^{1} \frac{dx}{1-x+x^2} -x \log(1-x+x^2)|_{0}^{1}+\int_{0}^{1} \frac{x(2x-1)dx}{1-x+x^2}.$$ Next, these integrals can be done easily to get the result.