Convergence of $sumlimits_{n=1}^inftyleft{frac{1cdot 3dots 2n-1 }{2cdot 4dots 2n}cdotfrac{4n+3}{2n+2}right}^2$

Show if the inf series
$sumlimits_{n=1}^inftyleft{frac{1cdot3dots2n-1 }{2cdot 4dots2n}cdotfrac{4n+3}{2n+2}right}^2$
converges.

My thought:

When $2n=2^k$, $frac{1cdot 3dots2n-1 }{2cdot 4dots2n}cdot frac{4n+3}{2n+2}
approx (1-1/4)^1cdot (1-1/8)^2dots(1-1/(2^k))^{2^{k-2}} approx (1-1/4)^{k-1},$
and so the $n$th item =
$((1-1/4)^{k-1})^2=(9/16)^{k-1}$.

When $2n neq 2^k$, we have $2^{k-1}<2n<2^k$,
begin{align*}
frac{1cdot 3dots2n-1 }{2cdot 4dots2n}cdot frac{4n+3}{2n+2}
approx (1-1/4)^1cdot (1-1/8)^2dots(1-1/(2^{k-1}))^{2^{k-3}}cdot(1-1/(2^k))^{n-(2^{k-2})}\
approx (1-1/4)^{k-2}cdot (1-1/(2^k))^{n-(2^{k-2})},
end{align*}
where $0<n-2^{k-2}<2^{k-2}$.

Then sum $(2^{k-2}+1)$ th to $n’$ th terms, for $2^{k-1}<2n'<=2^k$,
begin{align*}
sumlimits_{n=2^{k-2}+1}^{n’}left{frac{1cdot3dots2n-1 }{2. 4dots2n}cdot frac{4n+3}{2n+2}right}^2
approx sumlimits_{n=2^{k-2}+1}^{n’} (1-1/4)^{2(k-2)}.(1-1/(2^k))^{2(n-(2^{k-2}))}\
=(1-1/4)^{2(k-2)}cdot frac{(1-1/(2^k))^2}{1-(1-1/(2^k))^2}cdot(1-(1-1/(2^k))^2)^{n’-(2^{k-2})})\
=(9/16)^{(k-2}cdot (2^{k-1}-1)cdot (1-frac{n’-(2^{k-2})}{2^{k-1}}),$
end{align*}
which, when $2n’=2^k$, approximates
$(9/16)^{k-2}cdot (2^{k-1}-1)cdot(1-1/2)approx (9/16)^{k-2}cdot (2^{k-2}).$

(All the terms approximates $(9/16)^{k-1}$, and so the sum is roughly $(9/16)^{k-1}cdot2^{k-2}$. This gives a little larger estimation.)

Therefore the infinite series equals
$$sumlimits_{k=2}^{infty}sumlimits_{n=2^{k-2}+1}^{n’} approx sumlimits_{k=2}^{infty} (9/16)^{k-2}cdot (2^{k-2})=sumlimits_{k=2}^{infty} (9/8)^{k-2},$$
which diverges, and so the series possibly diverges.$

Given the final approximation being $sum a^n$ where a is near 1, a little difference in the approximations above could change the convergence. Were the series convergent and had the approximations made it not, which one causes that? Besides, are there other methods?