I saw a question about proving the divergence of a function $f(n)$, defined as $\sum_{r=0}^{n}{\frac{1}{4^r}\binom{2r}{r}}$. I'm very curious if there is a closed form for $f(n)$, but I didn't manage to get anywhere after an hour.
I did not find any similar questions on the internet. Wolfram Alpha suggests that $f(n) = \binom{n+0.5}{n}$, which I verified and is true, but I have no idea how can I prove this result, and I would greatly appreciate some hints.
Thank you!
Best Answer
Let's denote $\displaystyle S_N(a)=\sum_{n=0}^N\frac{(2n)!}{(n!)^2}\frac{1}{a^n}$
Next, we use $$\frac{1}{2\pi}\int_0^{2\pi}\sin^{2n}(x)dx=\frac{1}{4^n}\binom{2n}{n}=\frac{1}{4^n}\frac{(2n)!}{(n!)^2}$$ It can be proved, for example, via direct integration: $$\frac{1}{2\pi}\int_0^{2\pi}\sin^{2n}(x)dx=\frac{(-1)^n}{2\pi\,4^n}\int_0^{2\pi}\big(e^{it}-e^{-it}\big)^{2n}dt$$ and noting, that $\int_0^{2\pi}e^{ikt}dt=2\pi$, if $\,k=0$; otherwise is zero.
Therefore,
$\displaystyle S_N(a)=\frac{1}{2\pi}\sum_{n=0}^N\Big(\frac{4}{a}\Big)^n\int_0^{2\pi}\sin^{2n}(x)dx\tag*{}$
Let's denote for a while $\alpha=\frac{4}{a}$, then $\displaystyle S_N(\alpha)=\frac{1}{2\pi}\int_0^{2\pi}\frac{1-\alpha^{N+1}\sin^{{2N+2}}(x)}{1-\alpha\sin^2(x)}dx\tag*{}$ $\displaystyle \stackrel{t=\tan x}{=}\,\frac{2}{\pi}\int_0^{\infty}\frac{1-\alpha^{N+1}\Big(\frac{t^2}{1+t^2}\Big)^{N+1}}{1+(1-\alpha)t^2}dt\tag*{}$ For $\alpha=1$ $\displaystyle S_N(\alpha=1)=\frac{2}{\pi}\int_0^{\infty}\Big(1-\Big(\frac{t^2}{1+t^2}\Big)^{N+1}\Big)dt\tag*{}$ Making the substitution $x=\frac{1}{1+t^2}$ $\displaystyle S_N(\alpha=1)=\frac{1}{\pi}\int_0^1\Big(x^{-\frac{3}{2}}(1-x)^{-\frac{1}{2}}-x^{-\frac{3}{2}}(1-x)^{N+\frac{1}{2}}\Big)dx\tag*{}$
Using the Beta-function of negative argument (for example, here )
$\displaystyle S_N(\alpha=1)=\frac{1}{\pi}\Big(B\big(-\frac{1}{2};\frac{1}{2}\big)-B\big(-\frac{1}{2};N+\frac{3}{2}\big)\Big)\tag*{}$ Using $\displaystyle B\Big(-\frac{1}{2};\frac{1}{2}\Big)=\frac{\Gamma\big(-\frac{1}{2}\big)\Gamma\big(\frac{1}{2}\big)}{\Gamma\big(\frac{1}{2}-\frac{1}{2}\big)}=0\tag*{}$
and $\Gamma\big(-\frac{1}{2}\big)=-2\Gamma\big(\frac{1}{2}\big)$, we finally get $\displaystyle S_N(a=4)=\sum_{n=0}^N\frac{(2n)!}{(n!)^24^n}=\frac{2}{\sqrt\pi}\frac{\Gamma\Big(N+\frac{3}{2}\Big)}{\Gamma(N+1)}=\frac{2}{\sqrt\pi}\frac{\Gamma\Big(N+\frac{3}{2}\Big)}{N!}\tag*{}$
As a bonus, for $a>4$ we can also find $$\lim_{N\to\infty}S_N(a)=\sum_{n=0}^\infty\frac{(2n)!}{(n!)^2}\frac{1}{a^n}=\frac{2}{\pi}\int_0^\infty\frac{dt}{1+\Big(1-\frac{4}{a}\Big)t^2}=\frac{\sqrt a}{\sqrt{a-4}}$$