Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I prove the identity $\sum_k\frac{1}{k}\binom{2k-2}{k-1}\binom{2n-2k+1}{n-k}=\binom{2n}{n-1}$?

I have to prove it using lattice paths, it should be related to Catalan numbers

The $n$th Catalan number $C_n$ counts the number of monotonic paths along the edges of a grid with $n\times n$ square cells, which do not pass above the diagonal.

See for example this link

For example $\frac{1}{k}\binom{2k-2}{k-1}$ is exactly $C_{k-1}$, and the other terms can also be expressed in terms of the Catalan numbers.

The second part of the exercise ask to prove the recurrence formula $C_n=\sum_{k=1}^n C_{k-1}C_{n-k}$ using similar reasoning (i.e. lattice paths). So we can't use this formula to prove the first.

Could you help me please?

share|cite|improve this question
The right hand side counts the number of paths from $(0,0)$ to $(n+1,n-1)$, so I suspect the lhs does too, but in a different way. – TMM Sep 20 '11 at 1:46
$\dfrac{1}{k}\dbinom{2k-2}{k-1}$ is actually $C_{k-1}$. – Brian M. Scott Sep 20 '11 at 1:53
@Brian: yeah, you're right, I will fix the question – Alex M Sep 20 '11 at 2:34

2 Answers 2

up vote 7 down vote accepted

The right-hand side counts the number of monotonic paths from $(0,0)$ to $(n-1,n+1)$. Since $(n-1,n+1)$ is above the diagonal, every one of these paths must cross the diagonal at some point. Suppose that the first ‘bad’ step is from $(k-1,k-1)$ to $(k-1,k)$.

  • How many ways are there to get from $(0,0)$ to $(k-1,k-1)$ without going above the diagonal?

  • How many ways are there to get from $(k-1,k)$ to $(n-1,n+1)$?

share|cite|improve this answer

This one can also be done using complex variables. Suppose we seek to evaluate $$\sum_{k=1}^n \frac{1}{k} {2k-2\choose k-1} {2n-2k+1\choose n-k}.$$

Introduce the integral representation $${2n-2k+1\choose n-k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n-2k+1}}{z^{n-k+1}} \; dz.$$ This has the property that it is zero when $k\gt n.$

We obtain for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{n+1}} \sum_{k\ge 1} \frac{1}{k} {2k-2\choose k-1} \frac{z^k}{(1+z)^{2k}} \; dz.$$

Recall the generating function for the Catalan numbers $$\sum_{q\ge 0} \frac{1}{q+1} {2q\choose q} w^q = \frac{1-\sqrt{1-4w}}{2w}$$ This is equal to $$\sum_{q\ge 1} \frac{1}{q} {2q-2\choose q-1} w^{q-1}$$ so that $$\sum_{q\ge 1} \frac{1}{q} {2q-2\choose q-1} w^q = \frac{1-\sqrt{1-4w}}{2}.$$

Substitute this into the integral to obtain $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{n+1}} \frac{1-\sqrt{1-4z/(1+z)^2}}{2} \; dz.$$

This has two components, the first is $$\frac{1}{2} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{n+1}} \; dz = \frac{1}{2} {2n+1\choose n}.$$

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

This evaluates to $$\frac{1}{2} {2n\choose n-1} - \frac{1}{2} {2n\choose n}.$$

Factoring the sum of the two contributions to reveal the target term we obtain $$\frac{1}{2} \left(\frac{2n+1}{n} + 1 - \frac{n+1}{n}\right) {2n\choose n-1} = {2n\choose n-1}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.