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.
Best Answer
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.