Using tha integral representation of the reciprocal of the binomial co-efficient: $${n \choose j}^{-1}=(n+1)\int_{0}^{1} x^j (1-x)^{n-j}~ dx,~~~~(1)$$ here, we state and prove an interesting sum that $$\sum_{k=0}^{\infty} \frac{2^k}{{2k \choose k}}=2+\frac{\pi}{2}~~~~~~(2)$$ Proof: $$S=\sum_{k=0}^{\infty} 2^{k}{2k \choose k}^{-1}= \int_{0}^{1} \sum_{k=0}^{\infty} (2k+1)~ [2x(1-x)]^{k}$$ Using $\sum_{k=0}^{\infty} (2k+1)r^k=\frac{1+r}{(1-r)^2},$ we can write $$S=\int_{0}^{1} \frac{1+2x(1-x)}{(2x^2-2x+1)^2} dx =\int_{0}^{1} \left (\frac{2}{(2x^2-2x+1)^2} -\frac{1}{2x^2-2x+1} \right)dx $$ The first integral in above has been obtained by assuming $$\int \frac{2}{(2x^2-2x+1)^2}dx=\frac{Ax+B}{2x^2-2x+1},$$ then by differentiation, we find $A=2, B=-1$. $$\implies S=\left(\frac{2x-1}{2x^2-2x+1}-\tan^{-1}(1-2x)\right)_{0}^{1}=2+\frac{\pi}{2}.$$
The question is: how to prove this using other approaches?
