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.

  • $\begingroup$ It is unclear what is meant by "Is it possible to find explicitly symmetric form which would work for arbitrary integer $m$ and $n$?". Please edit the question to clarify. $\endgroup$ – parsiad Feb 1 '18 at 16:56

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).

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  • $\begingroup$ (+1). Verified. Nice to see Concrete Mathematics appear at MSE. $\endgroup$ – Marko Riedel Feb 1 '18 at 22:47
  • $\begingroup$ @G.Cab: Very nice derivation. (+1) $\endgroup$ – Markus Scheuer Feb 1 '18 at 22:47
  • $\begingroup$ @MarkoRiedel: oh yes, I found Concrete Mathematics to be an invaluable teacher ! $\endgroup$ – G Cab Feb 2 '18 at 0:09
  • $\begingroup$ I very like your derivation. (+1) It remains only to get rid of asymmetry between $m$ and $n$. $\endgroup$ – user Feb 2 '18 at 0:17
  • $\begingroup$ @MarkusScheuer: thanks Markus, actually I did not expect that the four binomials could slim down so much. The OP has done a good preliminary analysis. $\endgroup$ – G Cab Feb 2 '18 at 0:21

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.


  • 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.

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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.


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 $z_2>z_1$.

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  • $\begingroup$ Now I see what you were aiming to. Interesting, and much to dig around that, specially if transforming the above identities into matricial notation. $\endgroup$ – G Cab Feb 2 '18 at 15:30
  • $\begingroup$ Actually I did not know exactly what will come out but felt that this complicated equation with ugly $4^k$ and other terms is just a manifestation of something being much deeper and simpler. Thank you once more. $\endgroup$ – user Feb 2 '18 at 15:38

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