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