Prove an equation about binomial coefficients

Question

Could we prove:

$ \sum_{k} \binom{2k}{k}\binom{n+k}{m+2k} \frac{(-1)^k}{k+1} = \binom{n-1}{m-1}$ when $m,n \in N$

Answer 1

When $n=1$ it's easy to evaluate for all $m\ge1$. Then it's easy to prove the $n+1$ by induction, given the $n$ case (for all $m$): just write $$ \binom{n+k}{m+2k} = \binom{n-1+k}{m+2k} + \binom{n-1+k}{m-1+2k}. $$

Answer 2

The generating function for the Central Binomial Coefficients is $$ \sum_{k=0}^\infty\binom{2k}{k}x^k=(1-4x)^{-1/2}\tag{1} $$ Integrating $(1)$ and dividing by $x$ yields $$ \sum_{k=0}^\infty\binom{2k}{k}\frac{x^k}{k+1}=\frac{1-\sqrt{1-4x}}{2x}\tag{2} $$ Sum the formula against $x^m$: $$ \begin{align} &\sum_{m=-\infty}^n\left(\sum_{k=0}^{n-m}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}\right)x^m\tag{3}\\ &=\sum_{k=0}^\infty\sum_{m=-2k}^{n-k}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}x^m\tag{4}\\ &=\sum_{k=0}^\infty x^{-2k}\sum_{m=0}^{n+k}\binom{2k}{k}\binom{n+k}{m}\frac{(-1)^k}{k+1}x^m\tag{5}\\ &=\sum_{k=0}^\infty x^{-2k}\binom{2k}{k}\frac{(-1)^k}{k+1}(x+1)^{n+k}\tag{6}\\ &=(x+1)^n\sum_{k=0}^\infty\binom{2k}{k}\frac{(-1)^k}{k+1}\left(\frac{x+1}{x^2}\right)^{k}\tag{7}\\ &=(x+1)^n\frac{-1+\sqrt{1+4\frac{x+1}{x^2}}}{2\frac{x+1}{x^2}}\tag{8}\\ &=x(x+1)^{n-1}\frac{-x+(x+2)}{2}\tag{9}\\[9pt] &=x(x+1)^{n-1}\tag{10}\\[6pt] &=\sum_{m=1}^n\binom{n-1}{m-1}x^m\tag{11} \end{align} $$ Justification:
$\:\ (3)$: sum against $x^m$
$\:\ (4)$: change order of summation
$\:\ (5)$: substitute $m\mapsto m-2k$
$\:\ (6)$: apply the Binomial Theorem
$\:\ (7)$: reorganize
$\:\ (8)$: apply $(2)$
$\:\ (9)$: simplify
$(10)$: simplify
$(11)$: apply the Binomial Theorem

Equating coefficients of $x^m$ yields $$ \sum_{k=0}^{n-m}\binom{2k}{k}\binom{n+k}{m+2k}\frac{(-1)^k}{k+1}=\binom{n-1}{m-1}\tag{12} $$

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Note that the sum has non-zero terms for $m\lt1$, but these cancel; e.g. $n=3,m=-1$: $$ \begin{align} &\binom{0}{0}\binom{3}{-1}\frac11-\binom{2}{1}\binom{4}{1}\frac12+\binom{4}{2}\binom{5}{3}\frac13-\binom{6}{3}\binom{6}{5}\frac14+\binom{8}{4}\binom{7}{7}\frac15\\[6pt] &=0-4+20-30+14\\[9pt] &=0 \end{align} $$

Answer 3

Here is a similar approach using basic complex variables. Suppose we seek to evaluate $$\sum_{k\ge 0} {2k\choose k} {n+k\choose m+2k} \frac{(-1)^k}{k+1}$$ where $n\ge m.$

Introduce the integral representation $${n+k\choose m+2k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+2k+1}} (1+z)^{n+k} \; dz.$$

This yields the following integral for the sum: $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \sum_{k\ge 0} {2k\choose k} \frac{(-1)^k}{k+1} \frac{1}{z^{m+2k+1}} (1+z)^{n+k} \; dz \\= \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m+1}} \sum_{k\ge 0} {2k\choose k} \frac{(-1)^k}{k+1} \frac{(1+z)^k}{z^{2k}} \; dz.$$

Now the OGF of the Catalan numbers is $$\sum_{k\ge 0} {2k\choose k} \frac{1}{k+1} w^k = \frac{1-\sqrt{1-4w}}{2w}.$$ Substituting this closed form that we computed into the integral yields

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m+1}} \frac{1-\sqrt{1+4(1+z)/z^2}}{(-1)\times 2(1+z)/z^2} \; dz \\= - \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m-1}} \frac{1-\sqrt{1+4(1+z)/z^2}}{2(1+z)} \; dz \\= - \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^m} \frac{z-\sqrt{z^2+4(1+z)}}{2(1+z)} \; dz.$$

Observe that $z^2+4(1+z) = (z+2)^2$ so the integral becomes $$- \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^m} \frac{z-(z+2)}{2(1+z)} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n-1}}{z^m} \; dz.$$

This last integral can be evaluated by inspection and yields $$[z^{m-1}] (1+z)^{n-1} = {n-1\choose m-1}.$$

Note that for the convergence of the Catalan GF series we require $|(1+z)/z^2|\lt 1/4$ (distance to the nearest singularity). Now $|(1+z)/z^2|\le (1+\epsilon)/\epsilon^2$ so we may take $\epsilon = N \ge 5.$

Remark: the same functions appear here as in @robjohn's answer, who was first.

A trace as to when this method appeared on MSE and by whom starts at this MSE link.