Combinatorial/ Binomial Identity I am finding difficulty in showing the following identity:
$$\sum_{k=1}^n {n-1\choose k-1} {2n-1\choose k}^{-1} = \frac{2}{n+1}.$$
Mainly because of the inverse part.

We have by binomial coefficient algebra that
$$\sum_{k=1}^n {n-1\choose k-1} {2n-1\choose k}^{-1}
= \sum_{k=1}^n \frac{(n-1)!/(n-k)!/(k-1)!}
{(2n-1)!/(2n-1-k)!/k!}
\\ = {2n-1\choose n}^{-1}
\sum_{k=1}^n k \frac{(2n-1-k)!}{n! (n-k)!}
\\ = \frac{1}{n} {2n-1\choose n}^{-1}
\sum_{k=1}^n k {2n-1-k\choose n-k}
\\ = \frac{1}{n} {2n-1\choose n}^{-1}
[z^n] (1+z)^{2n-1} \sum_{k=1}^n k \frac{z^k}{(1+z)^k}.$$
Now the coefficient extractor enforces the range and we get
for the extractor only
$$[z^n] (1+z)^{2n-1} \sum_{k\ge 1} k \frac{z^k}{(1+z)^k}
= [z^n] (1+z)^{2n-1}
\frac{z/(1+z)}{(1-z/(1+z))^2}
\\ = [z^n] (1+z)^{2n-1}
\frac{z(1+z)}{(1+z-z)^2}
= [z^{n-1}] (1+z)^{2n}
= {2n\choose n-1}.$$
Restoring the scalar in front now yields
$$\frac{1}{n} {2n-1\choose n}^{-1} {2n\choose n-1}
= \frac{1}{n} 2n \frac{1}{n+1} = \frac{2}{n+1}.$$
This is the claim.
