Double binomial series simplification Could someone please check that this double binomial series (with a delta function) simplification makes sense: 
$$\sum_{m_J = -N/2}^{N/2} \sum_{m_K = -N/2}^{N/2} \dbinom{N}{N/2 +m_J}^{1/2}\dbinom{N}{N/2 + m_K}^{1/2} \delta_{m_K, m_J +1} = \bigg[\sum_{m_J = - N/2}^{N/2}\dbinom{N}{N/2 +m_J}^{1/2}\bigg]\bigg[ \sum_{m_J = - N/2-1}^{N/2} \dbinom{N}{N/2 + m_J + 1}^{1/2} \bigg] = \bigg[\sum_{m_J = - N/2}^{N/2}\dbinom{N}{N/2 +m_J}^{1/2}\bigg]^{2}$$
Please advise on the correct simplification if there are any inconsistencies. 
Thanks for the assistance.
 A: We consider the case $N=2M$

We obtain
  \begin{align*}
\sum_{m_J=-M}^M&\sum_{m_K=-M}^M\binom{2M}{M+m_J}^\frac{1}{2}\binom{2M}{M+m_K}^\frac{1}{2}\delta_{m_K,m_J+1}\\
&=\sum_{m_J=-M}^M\sum_{{-M\leq m_K\leq M}\atop{m_K=m_J+1}}\binom{2M}{M+m_J}^\frac{1}{2}\binom{2M}{M+m_K}^\frac{1}{2}\\
&=\sum_{m_J=-M}^{M-1}\binom{2M}{M+m_J}^\frac{1}{2}\binom{2M}{M+m_J+1}^\frac{1}{2}
\end{align*}
The inner sum reduces to one summand with $m_K=m_J+1$. Since $\binom{2M}{2M+1}=0$ the upper limit of the outer sum can be set to $M-1$.

The calculation $N=2M+1$ follows similarly.

[Add-On 2017-04-10]: We consider according to OPs comment
  \begin{align*}
\frac{1}{2^N}\sum_{m_J=-N/2}^{N/2}&\sum_{m_K=-N/2}^{N/2}\binom{N}{N/2+m_J}^\frac{1}{2}\binom{N}{N/2+m_K}^\frac{1}{2}\delta_{m_K,m_J}\\
&=\frac{1}{2^N}\sum_{m_J=-N/2}^{N/2}\binom{N}{N/2+m_J}^\frac{1}{2}\sum_{{-N/2\leq m_K\leq N/2}\atop{m_K=m_J}}\binom{N}{N/2+m_K}^\frac{1}{2}\\
&=\frac{1}{2^N}\sum_{m_J=-N/2}^{N/2}\binom{N}{N/2+m_J}^\frac{1}{2}\binom{N}{N/2+m_J}^\frac{1}{2}\\
&=\frac{1}{2^N}\sum_{m_J=-N/2}^{N/2}\binom{N}{N/2+m_J}\\
&=\frac{1}{2^N}\sum_{m_J=0}^{N}\binom{N}{m_J}\\
&=1
\end{align*}

