convergence of $\sum \limits_{n=1}^{\infty }\bigl\{ \frac {1\cdot3 \cdots (2n-1)} {2\cdot 4\cdots (2n)}\cdot \frac {4n+3} {2n+2}\bigr\} ^{2}$

I am investigating the convergence of
$$\begin{split}\sum _{n=1}^{\infty }\left\{ \dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)}\cdot \dfrac {4n+3} {2n+2}\right\} ^{2} &= \sum _{n=1}^{\infty }\left\{ \dfrac {\prod _{t=1}^n (2t-1)} {\prod _{t=1}^n (2t)}\cdot \dfrac {4n+3} {2n+2}\right\} ^{2} \\
&=\sum _{n=1}^{\infty }\left\{ \prod _{t=1}^n\left( 1-\dfrac {1} {2t}\right) \dfrac {4n+3} {2n+2}\right\} ^{2}
\end{split}$$
which after some manipulations I have reduced to
$$\sum _{n=1}^{\infty }e^ \left\{ 2\ln \left(2 -\dfrac {1} {2n+2}\right) +2\cdot \sum _{t=1}^{n}\ln \left( 1-\dfrac {1}{2t}\right) \right\} $$
and from an alternative approach I was able to reduce it to
$$\sum _{n=1}^{\infty } \dfrac{\left( 4n+3\right) ^{2}}{4\left(n+1\right)^{2}} \prod _{t=1}^n\left( 2+\dfrac{1}{2t^{2}}-\dfrac{2}{t}\right)$$
I am unsure how to proceed from here in either of the two cases. Any help would be much appreciated.

Solutions Collecting From Web of "convergence of $\sum \limits_{n=1}^{\infty }\bigl\{ \frac {1\cdot3 \cdots (2n-1)} {2\cdot 4\cdots (2n)}\cdot \frac {4n+3} {2n+2}\bigr\} ^{2}$"

We can prove by induction that

$$\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)} \ge \frac{1}{\sqrt{4n}}$$

and so your series diverges.

You can also notice that

$$\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n)} = \dfrac{\binom{2n}{n}}{4^n}$$

and try using the approximation

$$ \dfrac{\binom{2n}{n}}{4^n} = \frac{1}{\sqrt{\pi n}} \left(1 + \mathcal{O}\left(\frac{1}{n}\right)\right)$$

Denote by $a_n$ the general term, which is positive. We can rewrite it as $\left(\frac{(2n)!}{4^nn!n!}\right)^2\left(\frac{4n+3}{2n+2}\right)^2$, which is equivalent to $b_n:=4\left(\frac{(2n)!}{4^nn!n!}\right)^2$. Now we use Stirling’s formula, which states that $n!\overset{+\infty}{\sim}\left(\frac ne\right)^n\sqrt{2\pi n}$. We get
\begin{align*}
b_n&\overset{+\infty}{\sim} 4\left(\frac{\left(\frac{2n}e\right)^{2n}\sqrt{4n\pi}}{4^n\left(\frac ne\right)^{2n}2\pi n}\right)^2\\
&=\frac 4{n\pi},
\end{align*}
and using the fact that the harmonic series diverges, we get that the series $\sum_n a_n$ is divergent.