Prove $\sum_{n=0}^{\infty}\frac{(2n)!}{4^n(n+1)!(n)!} = 2$ Hi my homework question is to prove
$$
\sum_{n = 0}^{\infty}\frac{\left(2n\right)!}{4^{n}\left(n+1\right)!\left(n\right)!} = 2.
$$

*

*I know the sequence of partial sums $S_n$ converges from trying the ratio test.

*I tried to prove that $S_n\rightarrow 2$ as $n\rightarrow \infty$ by expressing writing $\frac{1}{1-x}$ with taylor expansion so that I can represent 2 with an infinite summation and subtract that. That didn't work.

*Next I tried to just consider each term in the summation and write $4^n=2^n2^n$ and then write $2^n(n)!=(2n)(2n-2)(2n-4)...(4)(2)$ and try to cancel terms and then write the leftover $2^n(n+1)!=(2n+2)(2n)...(4)(2))$. I was left with alternating products on the nominator and denominator, I belive it looked something like this $\frac{(2n-1)(2n-3)...(3)(1)}{(2n+2)(2n)(2n-2)...(4)(2)}$.

I don't know where to go from here and don't have other ideas to try. Please help.
 A: The binomial theorem gives
$$\frac 1{\sqrt{1-x}}=(1-x)^{-1/2} = \sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}(-1)^n x^n =\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}x^n$$
and hence
$$\int_{0}^{z} \frac 1{\sqrt{1-x}}\;\text{d}x = \int_{0}^{z} \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}x^n\;\text{d}x = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\int_{0}^{z} x^n\;\text{d}x$$
which gives
$$2-2\sqrt{1-z} = \sum_{n\geq 0}\frac{\binom{2n}{n}}{(n+1)4^n}\,z^{n+1}$$
Putting $z=1$, we have what you want.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
& \color{#44f}{\sum_{n = 0}^{\infty}{\pars{2n}! \over
4^{n}\pars{n + 1}!\pars{n}!}} =
\sum_{n = 0}^{\infty}{1 \over
4^{n}}{\pars{2n}! \over n!\ n!}{1 \over n + 1} =
\sum_{n = 0}^{\infty}{1 \over 4^{n}}
\overbrace{2n \choose n}^{\ds{{-1/2 \choose n}\pars{-4}^{n}}}\overbrace{\int_{0}^{1}t^{n}\,\dd t}
^{\ds{\quad1/\pars{n + 1}}}
\\[5mm] = & \
\int_{0}^{1}\sum_{n = 0}^{\infty}{-1/2 \choose n}\pars{-t}^{n}\,\dd t =
\int_{0}^{1}\pars{1 - t}^{-1/2}\,\,\dd t = \bbx{\color{#44f}{\large 2}}
\\ &
\end{align}
