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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

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up vote 6 down vote accepted

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 $$\int_0^1 (1-x^2)^r dx=\frac{1}{2}\int_0^1 x^{-1/2}(1-x)^r dx=\frac{1}{2}B(\frac{1}{2},r+1)=\frac{\sqrt{\pi}\Gamma(r+1)}{2\Gamma(r+\frac{3}{2})}.$$

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Thanks Gjergji, seems that this is what I was searching for. I stumbled upon this problem when trying to define the measure of superellipticity of shapes. I'll run experiments, hopefully it'll work. –  Dragisa Zunic Nov 30 '11 at 22:26
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