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I know $4^{n}$ diverges, and $frac{(n!)^2}{(2n)!}$ converges. I also think that their product diverges since the term of this series is increasing but I don't know how to prove it. I tried ratio test which is inconclusive, and some other tests.

Let $a_n:=4^ncdotfrac{(n!)^2}{(2n)!}$, and note that $2^ncdot n!=2cdot4cdot6cdots 2n$. Hence

$$a_n=4^n frac{(n!)^2}{(2n)!}=prod_{k=1}^{n}frac{2k}{2k-1}=expleft(sum_{k=1}^nlogleft(frac{2k}{2k-1}right)right)ge expleft(nlogleft(frac{2n}{2n-1}right)right)tag1$$

and

$$lim_{xtoinfty}xlogleft(frac{2x}{2x-1}right)=lim_{xtoinfty}frac{frac{2(2x-1)-4x}{(2x-1)^2}}{-left(frac{2x}{2x-1}right)frac1{x^2}}=lim_{xtoinfty}frac{2x^2}{(2x-1)2x}=frac12tag2$$

Thus $lim a_nge sqrt e$, so $sum a_n=infty$.

A more simple approach is expanding and chopping the product on $(1)$, that is

$$prod_{k=1}^nfrac{2k}{2k-1}=prod_{k=1}^nleft(1+frac1{2k-1}right)=1+sum_{k=1}^nfrac1{2k-1}+ldotstag3$$

Then we have the lower bound

$$a_nge 1+sum_{k=1}^nfrac1{2k-1}=1+frac12sum_{k=1}^nfrac1{k-1/2}ge 1+frac12sum_{k=1}^nfrac1ktag4$$

Hence $lim a_n=infty$ (because $sumfrac1k=infty$).

Stirling's approximation? Naah, let us go for a greater overkill. Since the series
$$ sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3 $$
is convergent to $frac{pi}{Gammaleft(frac{3}{4}right)^4}$ due to the relation with the squared complete elliptic integral of the first kind (identity $(7)$ at $k=frac{1}{sqrt{2}}$), its main term is convergent to zero and your sequence is divergent.

Seriously, an elementary approach. Since $2costheta=e^{itheta}+e^{-itheta}$ and $int_{0}^{2pi}e^{nitheta}e^{-mitheta},dtheta = 2pidelta(m,n)$,
$$ frac{1}{4^n}binom{2n}{n}=frac{2}{pi}int_{0}^{pi/2}cos^{2n}theta,dtheta. $$
By the dominated/monotone convergence theorem, the limit of both sides as $nto +infty$ is zero, hence your sequence is divergent. We also have that $left{frac{1}{4^n}binom{2n}{n}right}_{ngeq 1}$ is log-convex due to the Cauchy-Schwarz inequality and the previous integral representation.

Yet another elementary approach. You may prove in a combinatorial fashion that
$$ sum_{k=0}^{n}binom{2k}{k}binom{2n-2k}{n-k} = 4^n tag{Convolution}$$
hence it follows that
$$ left[sum_{k=0}^{n}frac{1}{4^k}binom{2k}{k}right]^2 leq sum_{k=0}^{2n} 1 = 2n $$
and
$$ sum_{k=0}^{n}frac{1}{4^k}binom{2k}{k} leq sqrt{2n}. $$
On the other hand we also have

$$ left[sum_{k=0}^{n}frac{1}{4^k}binom{2k}{k}right]^2 geq sum_{k=0}^{n} 1 = n $$
hence
$$ sqrt{n}leq sum_{k=0}^{n}frac{1}{4^k}binom{2k}{k} leq sqrt{2n}.tag{SumInequality} $$
Since the sequence $left{frac{1}{4^n}binom{2n}{n}right}_{ngeq 1}$ is decreasing, the previous inequality implies that $frac{1}{4^n}binom{2n}{n}llfrac{1}{sqrt{n}}$ as $nto +infty$. The correct asymptotic behaviour is given by Wallis' product and it is
$$ frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}.tag{GoodToKnow}$$

By Stirling's approximation

$$n! sim sqrt{2 pi n}left(frac{n}{e}right)^n$$

we have

$$4^n frac{(n!)^2}{(2n)!}sim 4^n frac{2 pi nleft(frac{n}{e}right)^{2n}}{sqrt{4 pi n}left(frac{2n}{e}right)^{2n}}=sqrt{pi n}to infty$$

then the series diverges.