Show the convergence of $\sum_{n=0}^{\infty}\frac{1}{2}\cdot \frac{(2n)!}{n!(n+1)!}\cdot \left(\frac{1}{4}\right)^n$

In fact, if $|x|\le\tfrac14$ then $\sum_{n\ge0}\frac{(2n)!}{n!(n+1)!}x^n=\frac{1-\sqrt{1-4x}}{2x}$, so the given sum is $1$.

One possible solution using Raabe's Test $$n\cdot\left(\frac{a_n}{a_{n+1}} -1\right) = n\cdot\frac{6n +6}{(2n+1)(2n+2)}\to \frac{3}{2}>1$$