Proving this identity $\sum_k\frac{1}{k}\binom{2k-2}{k-1}\binom{2n-2k+1}{n-k}=\binom{2n}{n-1}$ using lattice paths 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?

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
$$\frac{1}{2}
\left(\frac{2n+1}{n}  + 1 - \frac{n+1}{n}\right)
{2n\choose n-1}
= {2n\choose n-1}.$$
