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I am trying to prove the following for a very, very long time:

$$\sum_{k=0}^j \frac{ (-1)^k}{k! (j-k)!} \frac{1}{2k+1} = \frac{1}{2} \frac{\sqrt{\pi}}{\Gamma(3/2 + j)}$$

or, equivalently

$$\sum_{k=0}^j \frac{ (-1)^k}{k! (j-k)!} \frac{1}{2k+1} = \frac{(j+1)! 4^{j+1}}{2\cdot (2(j+1))!}$$

I would be extremely happy if somebody could help me with this!

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2 Answers 2

$$ \begin{eqnarray*}\sum_{k=0}^j (-1)^k \binom{j}{k} \frac{1}{2k+1} & = & \sum_{k=0}^j (-1)^k \binom{j}{k} \int^1_0 t^{2k} dt \\ & = & \int^1_0 \sum_{k=0}^j \binom{j}{k} (-t^2)^k dt \\ & = & \int^1_0 (1-t^2)^j dt \\ & = & \frac{1}{2} \int^1_0 u^{-1/2} (1-u)^j du \\ & = & \frac{1}{2} \beta(1/2, j+1) \\ & = & \frac{1}{2} \frac{ \Gamma(1/2) \Gamma(j+1) }{\Gamma( 3/2 + j) } \\ & = & \frac{1}{2} \frac{\sqrt{\pi} j!}{\Gamma(3/2 + j) } \end{eqnarray*}$$

where $\beta$ was the Beta function. Dividing both sides by $j!$ gives the result.

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Thank you Ragib Zaman, from now on you shall be my personal hero! –  Gaußmaster Apr 12 '12 at 13:20

I don't know if this helps, but it looks like the obvious way to start:

Let $$f(x)=\sum_{k=0}^j{(-1)^kj!\over k!(j-k)!}{x^{2k+1}\over2k+1}$$ Then $f(0)=0$, and $f(1)=j!\times{\rm\ your\ sum}$. Differentiate: $$f'(x)=\sum_{k=0}^j{(-1)^kj!\over k!(j-k)!}x^{2k}=\sum_{k=0}^j{j\choose k}(-x^2)^k=(1-x^2)^j$$ so $$f(1)=\int_0^1(1-x^2)^j\,dx$$

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