<s>[INST] Series $\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\cdots$ Find the sum of the series to infinity$$\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\cdots$$
Attempt-
I wrote the general term as $$\frac{\binom{2n}{n}}{2^{2n}\cdot (n+1)}$$
I don't know what to do next
 [/INST] It is not trivial, but your series is a telescopic one. Let $a_n=\frac{1}{4^n}\binom{2n}{n}$. We have:
$$ a_{n}-a_{n+1} = \frac{1}{4^n}\binom{2n}{n}-\frac{1}{4^{n+1}}\binom{2n+2}{n+1} =\frac{a_n}{2n+2}$$
so if we set $b_n = \frac{1}{(n+1)4^n}\binom{2n}{n}$ it follows that:
$$\sum_{n\geq 1}b_n = 2\sum_{n\geq 1}\frac{a_n}{2n+2} = 2\sum_{n\geq 1}\left(a_n-a_{n+1}\right) = 2a_1 = \color{red}{1}$$
since $\lim_{n\to +\infty}a_n = 0$.
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