How to calculate the infinite sum of the following series, related to binomial expansion for rational number, $r$:


$$1-\frac{r}{1!}\cdot\frac{1}{3}+\frac{r(r-1)}{2!}\cdot\frac{1}{5}-\frac{r(r-1)(r-2)}{3!}\cdot\frac{1}{7}+\ \dots$$.


I know the limit: 


$$1-1/3+1/5-1/7+\ \dots = \pi/4$$ 


and I can calculate: 


$$1+\frac{r}{1!}+\frac{r(r-1)}{2!}+\frac{r(r-1)(r-2)}{3!}+\ \dots$$ 


but I'm ot sure if that helps finding the solution for a starting problem..? 


Thanx, Dragisa



Limit of an infinite series, related to (generalised Newton's) binomial expansion
You need 
$$\sum\_{k\geq 0} (-1)^{k}\binom{r}{k}\frac{1}{2k+1}=\int\_0^1 \sum\_{k\geq 0}(-1)^{k}\binom{r}{k} x^{2k}dx$$
$$=\int\_0^1 (1-x^2)^r dx$$
and this is a beta integral
$$\int\_0^1 (1-x^2)^r dx=\frac{1}{2}\int\_0^1 x^{-1/2}(1-x)^r dx=\frac{1}{2}B(\frac{1}{2},r+1)=\frac{\sqrt{\pi}\Gamma(r+1)}{2\Gamma(r+\frac{3}{2})}.$$


