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I am trying to prove the following equality (which has arised when I was trying to prove that $E|S_n| \rightarrow \sqrt{\frac{2n}{\pi}}$):

$$ \sum_{x=1}^k \frac{x}{(k+x)!(k-x)! } = \frac{1}{2 \Gamma(k+1)\Gamma(k)}.$$

I tried to expanding the left side different ways but the closest I got is this:

$$\frac{1}{2 \Gamma(k+1)\Gamma(k)} = \frac{k}{2 (k!)^2}, \quad \sum_{x=1}^k \frac{x}{(k+x)!(k-x)! } = \frac{k}{(k!)^2}\left( 1\cdot\frac{1}{k+1} + 2\cdot\frac{(k-1)}{(k+1)(k+2)} +...+k \cdot \frac{(k-1)!}{(k+1)(k+2)...(2k)}\right).$$

So, now I am stuck proving that latter sum is $\frac{1}{2}.$

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Sure, just a typo, thanks! – grozhd Dec 16 '12 at 17:54
up vote 4 down vote accepted

The sum $S$ on the LHS is $$ S=\sum_{x=0}^k\frac{k+x-k}{(k+x)!(k-x)!}=T-kR, $$ where $$ T=\sum_{x=0}^k\frac1{(k+x-1)!(k-x)!},\qquad R=\sum_{x=0}^k\frac1{(k+x)!(k-x)!}, $$ hence $$ T=\frac1{(k-1)!k!}+\sum_{i+j=2k-1}\frac{\mathbf 1_{i\gt j}}{i!j!},\qquad R=\sum_{i+j=2k}\frac{\mathbf 1_{i\geqslant j}}{i!j!}. $$ If $i+j=2k-1$, then $i\ne j$ hence, by symmetry, $$ 2T-\frac2{(k-1)!k!}=\sum_{i+j=2k-1}\frac1{i!j!}=[t^{2k-1}](\mathrm e^t\cdot\mathrm e^t)=\frac{2^{2k-1}}{(2k-1)!}, $$ where $[t^n](\varphi(t))$ is the coefficient of $t^n$ in the power expansion of $\varphi(t)$. Likewise, doubling $R$ counts twice the term $i=j=k$ hence $$ 2R-\frac1{(k!)^2}=\sum_{i+j=2k}\frac1{i!j!}=[t^{2k}](\mathrm e^t\cdot\mathrm e^t)=\frac{2^{2k}}{(2k)!}. $$ One sees that $$ 2kR-\frac{k}{(k!)^2}=k\frac{2^{2k}}{(2k)!}=\frac{2^{2k-1}}{(2k-1)!}=2T-\frac{2k}{(k!)^2}, $$ hence $$ 2S=2T-2kR=\frac{k}{(k!)^2}=\frac1{k!(k-1)!}. $$

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