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

My attempt-
write down the Taylor series for $(1-x)^{\frac12}$
$$(1-x)^{\frac12}=1-\frac12x-\frac12\cdot\frac12\cdot\frac1{2!}x^2-\frac12\frac12\frac32\frac1{3!}x^3...$$
After putting x=1 and simplifying, I am getting value as 1.
