TITLE: Prove that $\frac{1}{n + 1}{2n\choose n}$ is a positive integer for $n \ge 0$.
QUESTION [5 upvotes]: I attempted to use Pascal's triangle identity to help out, but I do not know how to deal with $\frac{1}{n+1}$.

REPLY [3 votes]: Approach 1:
Note that
$$
\frac{2n+1}{n+1}\overbrace{\ \ \binom{2n}{n}\ \ }^{\large\frac{(2n)!}{n!\,n!}}=\overbrace{\binom{2n+1}{n+1}}^{\large\frac{(2n+1)!}{(n+1)!\,n!}}
$$
Then, because $\frac1{n+1}=2-\frac{2n+1}{n+1}$, we have
$$
\begin{align}
\frac1{n+1}\binom{2n}{n}
&=2\binom{2n}{n}-\frac{2n+1}{n+1}\binom{2n}{n}\\
&=2\binom{2n}{n}-\binom{2n+1}{n+1}
\end{align}
$$

Approach 2:
Note that
$$
\frac{n}{n+1}\overbrace{\ \ \binom{2n}{n}\ \ }^{\large\frac{(2n)!}{n!\,n!}}=\overbrace{\binom{2n}{n+1}}^{\large\frac{(2n)!}{(n+1)!\,(n-1)!}}
$$
Then, because $\frac1{n+1}=1-\frac{n}{n+1}$, we have
$$
\begin{align}
\frac1{n+1}\binom{2n}{n}
&=\binom{2n}{n}-\frac{n}{n+1}\binom{2n}{n}\\
&=\binom{2n}{n}-\binom{2n}{n+1}
\end{align}
$$