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.
 A: Let's consider, first of all, to rewrite the fraction putting aside ${m \choose k}$ and 
using the definition of the binomial in terms of the Rising and Falling Factorials
$$ \bbox[lightyellow] {  
\eqalign{
  & 4^{\,k} \left( \matrix{
  n \cr 
  k \cr}  \right)\;\mathop /\limits_{} \;\left( {\left( \matrix{
  2n \cr 
  2k \cr}  \right)\left( \matrix{
  2k \cr 
  k \cr}  \right)} \right) = 4^{\,k} {{n^{\,\underline {\,k\,} } \left( {2k} \right)!k!k!} \over {k!\left( {2n} \right)^{\,\underline {\,2k\,} } \left( {2k} \right)!}} = 
  4^{\,k} {{n^{\,\underline {\,k\,} } } \over {\left( {2n} \right)^{\,\underline {\,2k\,} } }}k! =   \cr 
  &  = 4^{\,k} {{\prod\limits_{0\, \le \,j\, \le \,k - 1} {\left( {n - j} \right)} } \over {\prod\limits_{0\, \le \,j\, \le \,2k - 1} {\left( {2n - j} \right)} }}k! =
   {{4^{\,k} k!\prod\limits_{0\, \le \,j\, \le \,k - 1} {\left( {n - j} \right)} } \over {\prod\limits_{0\, \le \,2j\, \le \,2k - 1} {\left( {2n - 2j} \right)} 
  \prod\limits_{0\, \le \,2j + 1\, \le \,2k - 1} {\left( {2n - 2j - 1} \right)} }} =   \cr 
  &  = {{4^{\,k} k!\prod\limits_{0\, \le \,j\, \le \,k - 1} {\left( {n - j} \right)} } \over {2^{\,k} \prod\limits_{0\, \le \,j\, \le \,k - 1} {\left( {n - j} \right)} \;
   \quad 2^{\,k} \prod\limits_{0\, \le \,j\, \le \,k - 1} {\left( {n - 1/2 - j} \right)} }} = {{k!} \over {\prod\limits_{0\, \le \,j\, \le \,k - 1} {\left( {n - 1/2 - j} \right)} }} =   \cr 
  &  = {{k!} \over {\left( {n - 1/2} \right)^{\,\underline {\,k\,} } }} = 1^{\,\overline {\,k\,} } \left( {n + 1/2} \right)^{\,\overline {\, - \,k\,} }  = 1\;\mathop 
  /\limits_{} \;\left( \matrix{n - 1/2 \cr  k \cr}  \right) \cr} 
 } \tag{1}$$
Then we employ the law of exponents addition for the Rising factorial, to get
$$ \bbox[lightyellow] {  
\eqalign{
  & \left( {n - m + 1/2} \right)^{\,\overline {\,m - \,k\,} }  = \left( {n - m + 1/2} \right)^{\,\overline {\,m\,} } \left( {n + 1/2} \right)^{\,\overline {\, - \,k\,} } \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {n + 1/2} \right)^{\,\overline {\, - \,k\,} }  = {{\left( {n - m + 1/2} \right)^{\,\overline {\,m - \,k\,} } } \over {\left( {n - m + 1/2} \right)^{\,\overline {\,m\,} } }} \cr} 
 } \tag{2}$$
so that we can take advantage of the fact that the generalized binomial expansion also applies to the Factorials, and conclude
$$ \bbox[lightyellow] {  
\eqalign{
  & \sum\limits_{0\, \le \,k\, \le \,m} {4^{\,k} {{\left( \matrix{
  n \cr 
  k \cr}  \right)\left( \matrix{
  m \cr 
  k \cr}  \right)} \over {\left( \matrix{
  2n \cr 
  2k \cr}  \right)\left( \matrix{
  2k \cr 
  k \cr}  \right)}}}  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {{{\left( \matrix{
  m \cr 
  k \cr}  \right)} \over {\left( \matrix{
  n - 1/2 \cr 
  k \cr}  \right)}}}  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{
  m \cr 
  k \cr}  \right)1^{\,\overline {\,k\,} } \left( {n + 1/2} \right)^{\,\overline {\, - \,k\,} } }  = \cr 
  &  = {1 \over {\left( {n - m + 1/2} \right)^{\,\overline {\,m\,} } }}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{
  m \cr 
  k \cr}  \right)1^{\,\overline {\,k\,} } \left( {n - m + 1/2} \right)^{\,\overline {\,m - \,k\,} } }  =   \cr 
  &  = {{\left( {n - m + 3/2} \right)^{\,\overline {\,m\,} } } \over {\left( {n - m + 1/2} \right)^{\,\overline {\,m\,} } }} = {{n + 1/2} \over {n - m + 1/2}} = {{2n + 1} \over {2(n - m) + 1}} \cr} 
 } \tag{3}$$
Now, identity (3) is valid for any non-negative integer $m$, and for $n$ that can take even a real or complex value,
except for $n=m-1/2$.
This under the acception that the two binomials in $n$, when null get simplified. That is that
the two binomials be rewritten as above in terms of Falling Factorials (or Gamma function) with $n$ real, 
and simplify the fraction (take the limit).
A: Following very profound and enlightening answer of G Cab the following form of the identity seems to be most general:
$$
\sum_{k=0}^m\frac{\binom{m}{k}}{\binom{z-1}{k}}
=\frac{z}{z-m},\tag{1}
$$
where $m$ is a non-negative integer and $z$ is arbitrary complex number excluding integer values from $1$ to $m$. The original identity which have inspired the question corresponds to $z=n+1/2$.
Among other possible means the equality can be proved by induction over $m$. Let $\mathbb{C}_m=\mathbb{C}\setminus\left\{1,\dots,m\right\}$. Obviously if $m=0$ the equality (1) holds for arbitrary $z$. Suppose the equality is true for $m-1$ and arbitrary $z\in\mathbb{C}_{m-1}$. Then it is true for $m$ and arbitrary $z\in\mathbb{C}_m$ as well:
$$
S^m_z\stackrel{\text{def}}{=}\sum_{k=0}^m\frac{\binom{m}{k}}{\binom{z-1}{k}}=
1+\sum_{k=1}^m\frac{\frac{m}{k}\binom{m-1}{k-1}}{\frac{z-1}{k}\binom{z-2}{k-1}}\\=1+\frac{m}{z-1}\sum_{k=0}^{m-1}\frac{\binom{m-1}{k}}{\binom{z-2}{k}}=1+\frac{m}{z-1}S^{m-1}_{z-1}\\
\stackrel{I.H.}{=}1+\frac{m}{z-1}\cdot\frac{z-1}{(z-1)-(m-1)}=\frac{z}{z-m}.
$$
Setting in (1) $z_1=-m-\alpha$, $z_2=-n-\alpha$ one obtains the "symmetric identity" suggested in the question:
$$
\sum_{k=0}^\infty\left[\frac{\binom{n}{k}}{\binom{-m-1-\alpha}{k}}+\frac{\binom{m}{k}}{\binom{-n-1-\alpha}{k}}\right]=1+\frac{\alpha}{m+n+\alpha}.\tag{2}
$$
For $\alpha=0$ it degenerates to:
$$
\sum_{k=0}^\infty\left[\frac{\binom{n}{k}}{\binom{-m-1}{k}}+\frac{\binom{m}{k}}{\binom{-n-1}{k}}\right]=1+\delta_{m0}\delta_{n0}.\tag{3}
$$
This can be seen as the "decomposition of 1" being asked in the question.
UPDATE:
As shown elsewhere the most general form of the identity is
$$
\sum_{k=0}^\infty\frac{\binom{z_1}{k}}{\binom{z_2-1}{k}}=\frac{z_2}{z_2-z_1},\tag{*}
$$
where $z_1$ and $z_2$ are complex numbers, $z_2$ is not a positive integer, and $\Re(z_2-z_1)<0$. The last restriction does not apply if $z_1$ is a non-negative integer. In this case $z_2$ can also take on positive integer values $z_2>z_1$. 
A: 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.
