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Let $n\ge 2$ postive integer, show that $$I=\sum_{k=1}^{n-1}\dfrac{1}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\binom{2(n-1)}{n-1}$$

I have done this works $$I=\sum_{k=1}^{n-1}\dfrac{1}{k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}+\sum_{k=1}^{n-1}\dfrac{1}{n-k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\sum_{k=1}^{n-1}\dfrac{2}{k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}$$

darij grinberg
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math110
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  • Hint: $\dfrac{1}{u}\dbinom{2\left(u-1\right)}{u-1} = 2 \dbinom{2\left(u-1\right)}{u-1} - \dbinom{2u-1}{u-1}$ for each positive integer $u$. Applying this twice allows you to get rid of the denominator in $I$, at the cost of introducing two differences. Expand the product of the differences, and apply the Chu-Vandermonde identity four times. Then simplify. – darij grinberg Jan 04 '17 at 16:08
  • @darijgrinberg Thanks,I have use your hint,I only use two times chu-vandermonde identity,$=4\binom{2n-4}{n-2}-2\binom{2n-3}{n-2}$,wihci wrong? can you post it – math110 Jan 04 '17 at 16:23
  • Ah, I meant to apply my argument to the original form of $I$, not to your modified version. The answer doesn't look too bad; you probably made a little typo somewhere. – darij grinberg Jan 04 '17 at 16:28
  • @functionsug: There seems to be a typo at the RHS which should be multiplied with $\frac{1}{n}$. – Markus Scheuer Jan 04 '17 at 17:55

2 Answers2

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Note the Catalan numbers are defined as \begin{align*} C_{n}&=\frac{1}{n+1}\binom{2n}{n}\qquad\quad n\geq 0\\ \end{align*} with generating function \begin{align*} C(z)=\sum_{n=0}^\infty C_nz^n=\frac{1}{2z}\left(1-\sqrt{1-4z}\right) \end{align*}

The LHS of OPs binomial identity is a Cauchy-product of Catalan numbers

\begin{align*} \sum_{k=1}^{n - 1} C_{k-1}C_{n -k-1}\tag{1} \end{align*}

Since Cauchy products occur when multiplying series we could work with generating functions:

\begin{align*} C^2(z)=\sum_{n=0}^\infty\left(\sum_{k=0}^{n}C_kC_{n-k}\right)z^n\tag{2} \end{align*}

Let $[z^n]$ denote the coefficient operator.

We observe with the help of (1) and (2) for $n\geq 2$

\begin{align*} [z^{n-2}]C^2(z)&=[z^{n-2}]\sum_{n=0}^\infty\left(\sum_{k=0}^{n}C_kC_{n-k}\right)z^n\\ &=\sum_{k=0}^{n-2}C_kC_{n-2-k}\\ &=\sum_{k=1}^{n-1}C_{k-1}C_{n-1-k}\\ \end{align*} on the other hand we obtain \begin{align*} [z^{n-2}]C^2(z)&=[z^{n-2}]\left(\frac{1}{2z}\left(1-\sqrt{1-4z}\right)\right)^2\\ &=[z^{n-2}]\frac{1}{2z^2}\left(1-\sqrt{1-4z}\right)-1\\ &=[z^{n-1}]\frac{1}{2z}\left(1-\sqrt{1-4z}\right)\\ &=[z^{n-1}]C(z)\\ &=C_{n-1}\\ &=\frac{1}{n}\binom{2n-2}{n-1}\\ \end{align*} and the claim follows.

Markus Scheuer
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2

It is a convolution identity. You may consider that: $$ \sum_{k\geq 1}\frac{x^k}{k}\binom{2k-2}{k-1} = \frac{1-\sqrt{1-4x}}{2}\tag{1}$$ holds as a consequence of the generating function of Catalan numbers.
If you square both sides of $(1)$ and consider the coefficient of $x^n$, you get: $$ \sum_{k=1}^{n-1}\frac{1}{k(n-k)}\binom{2k-2}{k-1}\binom{2(n-k)-2}{(n-k)-1} = [x^n]\left(\frac{1-\sqrt{1-4x}}{2}-x\right) \tag{2}$$ and for $n\geq 2$, the RHS of $(2)$ is just $\color{red}{\frac{1}{n}\binom{2n-2}{n-1}}=\color{blue}{\frac{1}{2n-1}\binom{2n-1}{n}}$, by $(1)$.

Jack D'Aurizio
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