Supposing we start from
$$\sum_{q=0}^n {n\choose q} \frac{q}{q+1} {2n\choose q+1}^{-1}
= \frac{1}{n+1}.$$
The LHS is
$$n \sum_{q=1}^n {n-1\choose q-1}
\frac{1}{q+1} {2n\choose q+1}^{-1}.$$
Recall from MSE
4316307 the
following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$
In the present case we get
$$n\sum_{q=1}^n {n-1\choose q-1}
[z^{2n}] \log\frac{1}{1-z} (z-1)^{2n-q-1}
\\ = n [z^{2n}] \log\frac{1}{1-z} (z-1)^{2n-2}
\sum_{q=1}^n {n-1\choose q-1} \frac{1}{(z-1)^{q-1}}
\\ = n [z^{2n}] \log\frac{1}{1-z} (z-1)^{2n-2}
\left[ 1 + \frac{1}{z-1} \right]^{n-1}
\\ = n [z^{2n}] \log\frac{1}{1-z} (z-1)^{n-1}
z^{n-1}
= n [z^{n+1}] \log\frac{1}{1-z} (z-1)^{n-1}.$$
Apply the identity again (with $k=2$) to get
$$n \frac{1}{2} {n+1\choose 2}^{-1}
= \frac{1}{2} n \frac{2}{(n+1)n} = \frac{1}{n+1}.$$
This is the claim.