Another summation identity with binomial coefficients Recently I have encountered on this page a rather nice identity:
$$
\sum_{k=0}^m4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}=\frac{2n+1}{2n-2m+1},
$$
which however is valid only for $n\ge m$. This motivated me to try finding out a more symmetric expression with the aim to get $\frac{1}{2n+2m+1}$ on the RHS. Simple negation of $m$ did not help until the expression $\binom{2n}{2k}$ in the denominator was replaced with that one involving $m$:
$$
\sum_{k=0}^{n}(-4)^k\frac{\binom{n}{k}\binom{m+k}{k}}{\binom{2m+2k+1}{2k}\binom{2k}{k}}=\frac{2m+1}{2n+2m+1}.
$$
After swapping $n$ and $m$ the aim was achieved:
$$
\sum_{k=0}^{\infty}\frac{(-4)^k}{\binom{2k}{k}}\left[\frac{\binom{n}{k}\binom{m+k}{k}}{\binom{2m+2k+1}{2k}}+\frac{\binom{n+k}{k}\binom{m}{k}}{\binom{2n+2k+1}{2k}}\right]=1+\frac{1}{2n+2m+1}.
$$
But I am still not quite satisfied. Is there a suitable decomposition of 1 which could help to simplify the expression? Or maybe some different approach can better clarify the origin of the identity? I would appreciate any hint.

We prove a generalisation of  OPs two binomial identities. The first being
\begin{align*}
\sum_{k=0}^m4^k\frac{\binom{\color{blue}{n}}{k}\binom{m}{k}}{\binom{\color{blue}{2n}}{2k}\binom{2k}{k}}=\frac{2n+1}{2n-2m+1}\tag{1}
\end{align*}
In the second identity we exchange the variables $m$ and $n$ to get the identity more close to (1) and we also use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$. We obtain
\begin{align*}
\sum_{k=0}^{m}&(-4)^k\frac{\binom{n+k}{k}\binom{m}{k}}{\binom{2n+2k+1}{2k}\binom{2k}{k}}
=\sum_{k=0}^m4^k\frac{\binom{\color{blue}{-n-1}}{k}\binom{m}{k}}{\binom{\color{blue}{-2n-2}}{2k}\binom{2k}{k}}=\frac{2n+1}{2n+2m+1}\tag{2}
\end{align*}
Comparison of (1) and (2) strongly indicates a generalisation where let's say $\alpha=n$ in the first case and $\alpha=-n-1$ in the second case.

The following is valid for non-negative integer values $m$ and $\alpha\in\mathbb{C}\setminus\left\{m-\frac{1}{2}\right\}$:
  \begin{align*}
\color{blue}{\sum_{k=0}^m4^k\frac{\binom{\alpha}{k}\binom{m}{k}}{\binom{2\alpha}{2k}\binom{2k}{k}}
=\frac{2\alpha+1}{2\alpha-2m+1}}
\end{align*}
We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^m}\color{blue}{4^k\frac{\binom{\alpha}{k}\binom{m}{k}}{\binom{2\alpha}{2k}\binom{2k}{k}}}
&=\sum_{k=0}^m\binom{m}{k}4^k\frac{\alpha^{\underline{k}}}{k!}\cdot\frac{(2k)!}{(2\alpha)^{\underline{2k}}}
\cdot\frac{k!k!}{(2k)!}\tag{1}\\
&=\sum_{k=0}^m\binom{m}{k}4^k\frac{\alpha^{\underline{k}}k!}{2^{2k}\alpha^{\underline{k}}\left(\alpha-\frac{1}{2}\right)^{\underline{k}}}\tag{2}\\
&=\sum_{k=0}^m\binom{m}{k}\binom{\alpha-\frac{1}{2}}{k}^{-1}\tag{3}\\
&=\left(\alpha+\frac{1}{2}\right)\int_0^1\sum_{k=0}^m\binom{m}{k}z^k(1-z)^{\alpha-\frac{1}{2}-k}\,dz\tag{4}\\
&=\left(\alpha+\frac{1}{2}\right)\int_0^1(1-z)^{\alpha-\frac{1}{2}}\sum_{k=0}^m\binom{m}{k}\left(\frac{z}{1-z}\right)^k\,dz\tag{5}\\
&=\left(\alpha+\frac{1}{2}\right)\int_0^1(1-z)^{\alpha-\frac{1}{2}}\left(1+\frac{z}{1-z}\right)^m\,dz\\
&=\left(\alpha+\frac{1}{2}\right)\int_0^1(1-z)^{\alpha-m-\frac{1}{2}}\,dz\\
&=\frac{\alpha+\frac{1}{2}}{\alpha-m+\frac{1}{2}}\left[-(1-z)^{\alpha-m-\frac{1}{2}}\right]_0^1\\
&=\color{blue}{\frac{2\alpha+1}{2\alpha-2m+1}}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we use the falling factorial
$\alpha^{\underline{k}}=\alpha(\alpha-1)\cdots(\alpha-k-1)$ notation and the identity
\begin{align*}
\binom{n}{k}=\frac{n^{\underline{k}}}{k!}
\end{align*}

*In (2) we do some simplifications and use the identity
\begin{align*}
(2\alpha)^{\underline{2k}}=2^{2k}\alpha^{\underline{k}}\left(\alpha-\frac{1}{2}\right)^{\underline{k}}
\end{align*}

*In (3) we cancel terms and use again the representation with binomial coefficients.

*In (4) we write the reciprocal of a binomial coefficient using the beta function
\begin{align*}
\binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz
\end{align*}

*In (5) we do some rearrangements to prepare for the usage of the binomial theorem in the next step.
