TITLE: Series $\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\cdots$
QUESTION [8 upvotes]: 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

REPLY [3 votes]: $$S = 2\left[\frac{1}{2\cdot 4}+\frac{1\cdot 3}{2\cdot 4\cdot 6}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}+\cdots\right]$$
Here $$T_{n}\bf{(n^{th}) term } = 2\cdot \left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6.........(2n)(2n+2)}\right]$$
$$ = 2\left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6...(2n)(2n+2)}((2n+2)-(2n+1))\right]$$
So $$T_{n} = 2\left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6.........(2n)}-\frac{1\cdot 3 \cdot 5......(2n+1)}{2\cdot 4 \cdot 6.........(2n)}\right]$$
So $$S_{\infty} = 2\sum^{\infty}_{r=1}\left[\frac{1\cdot 3 \cdot 5......(2r-1)}{2\cdot 4 \cdot 6.........(2r)}-\frac{1\cdot 3 \cdot 5......(2r+1)}{2\cdot 4 \cdot 6.........(2r)}\right] = 2\cdot \frac{1}{2} = 1$$