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