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!
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 $$\lim_{N\to\infty}S_N(a)=\sum_{n=0}^\infty\frac{(2n)!}{(n!)^2}\frac{1}{a^n}=\frac{2}{\pi}\int_0^\infty\frac{dt}{1+\Big(1-\frac{4}{a}\Big)t^2}=\frac{\sqrt a}{\sqrt{a-4}}$$
We have from first principles that
$$\bbox[5px,border:2px solid #00A000]{ S_n = \sum_{r=0}^n \frac{1}{4^r} {2r\choose r} = \; \underset{z}{\mathrm{res}} \; \frac{1}{z^{n+1}} \frac{1}{1-z} \frac{1}{\sqrt{1-z}}.}$$
Now put $1-\sqrt{1-z} = w$ so that $z=w(2-w)$ and $dz = 2(1-w)\; dw$ to get
$$\; \underset{w}{\mathrm{res}} \; \frac{1}{w^{n+1} (2-w)^{n+1}} \frac{1}{(1-w)^2} \frac{1}{1-w} 2(1-w) \\ = 2 (-1)^{n+1} \; \underset{w}{\mathrm{res}} \; \frac{1}{w^{n+1} (w-2)^{n+1}} \frac{1}{(w-1)^2}.$$
The residue at infinity is zero by inspection so we need the residues at $w=1$ and $w=2.$ For the former we get without the scalar in front
$$\left.\left( \frac{1}{w^{n+1}} \frac{1}{(w-2)^{n+1}} \right)'\right|_{w=1} \\ = \left.\left(- (n+1) \frac{1}{w^{n+2}} \frac{1}{(w-2)^{n+1}} - \frac{1}{w^{n+1}} (n+1) \frac{1}{(w-2)^{n+2}} \right)\right|_{w=1} \\= -(n+1) (-1)^{n+1} - (n+1) (-1)^{n+2} = 0.$$
With this our sum is minus the residue at $w=2.$ We write
$$2 (-1)^{n} \;\mathrm{Res}_{w=2} \frac{1}{((w-2)+2)^{n+1}} \frac{1}{(w-2)^{n+1}} \frac{1}{((w-2)+1)^2} \\ = \frac{(-1)^{n}}{2^n} \;\mathrm{Res}_{w=2} \frac{1}{(1+(w-2)/2)^{n+1}} \frac{1}{(w-2)^{n+1}} \frac{1}{(1+(w-2))^2}.$$
This will produce
$$\bbox[5px,border:2px solid #00A000]{ S_n = \frac{1}{2^n} \sum_{q=0}^n {n+q\choose n} \frac{1}{2^q} (n-q+1).}$$
Now we get two pieces here, where $S_n = A_n + B_n$, the first is
$$A_n = \frac{n+1}{2^n} \; \mathrm{Res}_{z=0} \; \frac{1}{z^{n+1}} \frac{1}{1-z} \frac{1}{(1-z/2)^{n+1}} \\ = (-1)^n 2 (n+1) \; \mathrm{Res}_{z=0} \; \frac{1}{z^{n+1}} \frac{1}{z-1} \frac{1}{(z-2)^{n+1}}.$$
We evaluate this using the residues at $z=1$ and $z=2.$ We get for the former the value $-2(n+1).$ We write for the latter
$$(-1)^n 2 (n+1) \; \mathrm{Res}_{z=2} \; \frac{1}{((z-2)+2)^{n+1}} \frac{1}{(z-2)+1} \frac{1}{(z-2)^{n+1}} \\ = (-1)^n \frac{n+1}{2^n} \; \mathrm{Res}_{z=2} \; \frac{1}{((z-2)/2+1)^{n+1}} \frac{1}{(z-2)+1} \frac{1}{(z-2)^{n+1}}$$
This yields
$$(-1)^n \frac{n+1}{2^n} \sum_{q=0}^n {n+q\choose q} (-1)^q \frac{1}{2^q} (-1)^{n-q}.$$
Simplify to obtain $A_n$. With residues adding to zero, we have established that for the first piece $A_n$ it evaluates to $A_n = n+1.$
For the second piece we find
$$B_n = - \frac{n+1}{2^n} \sum_{q=1}^n {n+q\choose n+1} \frac{1}{2^q} = -\frac{n+1}{2^{n+1}} \sum_{q=0}^{n-1} {n+1+q\choose n+1} \frac{1}{2^q} \\ = -\frac{n+1}{2^{n+1}} \; \mathrm{Res}_{z=0} \; \frac{1}{z^n} \frac{1}{1-z} \frac{1}{(1-z/2)^{n+2}} \\ = (-1)^{n} 2 (n+1) \; \mathrm{Res}_{z=0} \; \frac{1}{z^n} \frac{1}{z-1} \frac{1}{(z-2)^{n+2}}.$$
Again evaluate using residues at $z=1$ and $z=2.$ We get for the former the value $2(n+1).$ For the latter we write
$$(-1)^n \frac{n+1}{2^{n-1}} \; \mathrm{Res}_{z=2} \; \frac{1}{((z-2)/2+1)^{n}} \frac{1}{(z-2)+1} \frac{1}{(z-2)^{n+2}}$$
This yields
$$(-1)^n \frac{n+1}{2^{n-1}} \sum_{q=0}^{n+1} {n-1+q\choose q} (-1)^q \frac{1}{2^q} (-1)^{n+1-q} \\ = - \frac{n+1}{2^{n-1}} \sum_{q=0}^{n-1} {n-1+q\choose q} \frac{1}{2^q} - \frac{n+1}{2^{2n-1}} {2n-1\choose n} - \frac{n+1}{2^{2n}} {2n\choose n+1}.$$
The sum is
$$- \frac{n+1}{2^{n-1}} A_{n-1} \frac{2^{n-1}}{n} = - (n+1)$$
hence piece $B_n$ evaluates as
$$\frac{n+1}{2^{2n-1}} {2n-1\choose n} + \frac{n+1}{2^{2n}} {2n\choose n+1} - (n+1).$$
Adding the two pieces we have shown that
$$\bbox[5px,border:2px solid #00A000]{ S_n = \frac{2n+1}{2^{2n}} {2n\choose n} = {n+1/2\choose n}.}$$
as claimed. This may be seen from (evaluate LHS)
$$(2n+1) [z^n] \frac{1}{\sqrt{1-z}} = (2n+1) {-1/2\choose n} (-1)^n \\ = (2n+1) {n-1/2\choose n} = {n+1/2\choose n}.$$
