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Let $l\ge 0$ be an integer.

In the process of solving A tough series related with a hypergeometric function with quarter integer parameters we have discovered the following identity: \begin{eqnarray} \frac{1}{4} \sum\limits_{j=0}^{l-1} \sum\limits_{\begin{array}{r} p=-l-1\\p\neq -1 \end{array}}^{l-1} \binom{-3/2}{j} \binom{-1/2}{l-1-j} \frac{\binom{2j+1}{p+j+1} - \binom{2j+1}{p+j+2}}{2^{2 j+1}}\cdot \frac{(-1)^{p+l}}{p+1}\cdot \left( 1+(-1)^p\right)=\frac{(1/2)^{(l)}}{l!}-1 \end{eqnarray} Is it possible to prove that identity in some other way, i.e. without resorting to the calculations in the aforementioned question?

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This question is actually easier than I thought and it can be solved using elementary methods. First of all note that : \begin{equation} \binom{2j+1}{p+j+1}- \binom{2j+1}{p+j+2}= \binom{2j+1}{p+j+1} \left(1-\frac{j-p}{p+j+2}\right)= \binom{2j+1}{p+j+1}\cdot \frac{2p+2}{p+j+2} \end{equation} Therefore the sum over $p$ gives: \begin{eqnarray} \sum\limits_{p=-l-1,p\neq-1}^{l-1} \frac{\binom{2j+1}{p+j+1}-\binom{2j+1}{p+j+2}}{2^{2j+1}}\cdot \frac{(-1)^{p+l}}{p+1} \cdot (1+(-1)^p)=\\ \sum\limits_{p=-l-1,p\neq-1}^{l-1} \frac{\binom{2j+1}{p+j+1}}{2^{2j+1}}\cdot \frac{2p+2}{p+j+2}\cdot \frac{(-1)^{p+l}}{p+1} \cdot (1+(-1)^p)=\\ \frac{1}{2^{2 j}}\cdot \frac{(-1)^l}{2j+2}\sum\limits_{p=-l-1}^{l-1} \binom{2j+2}{p+j+2} \cdot (1+(-1)^p)=\\ \frac{1}{2^{2 j}}\cdot \frac{(-1)^l}{2j+2}\sum\limits_{p=0}^{j+l+1} \binom{2j+2}{p} \cdot (1+(-1)^{p+j})= \frac{(-1)^l}{j+1} \cdot 2 \end{eqnarray} Therefore the sum in question reads: \begin{eqnarray} lhs&=& \frac{1}{4} \sum\limits_{j=0}^{l-1} \binom{-3/2}{j} \binom{-1/2}{l-1-j} \cdot \frac{(-1)^l}{j+1} \cdot 2\\ &=&-\sum\limits_{j=1}^l \binom{-1/2}{j} \binom{-1/2}{l-j} \cdot (-1)^l\\ &=&-1+\sqrt{\pi}(-1)^l \frac{1}{l! (-1/2+l)!}\\ &=&-1+\frac{(l-1/2)!}{l!(-1/2)!}\\ &=&-1+\frac{(1/2)^{(l)}}{l!} \end{eqnarray}

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