Check the convergence of the series $\displaystyle{\sum_{n=1}^{+\infty}\frac{\left (n!\right )^2}{\left (2n+1\right )!}4^n}$ I want to check if the following series converge or not.

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*$\displaystyle{\sum_{n=1}^{+\infty}\frac{\left (n!\right )^2}{\left (2n+1\right )!}4^n}$
I suppose we have to find here an upper bound and apply then the comparison test. But I don’t really have an idea which bound we could take. Could you give me a hint?

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*$\displaystyle{\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}}$
We have a term that is a product of the form $\frac{2i-1}{2i}=1-\frac{1}{2i}$. To apply the comparison test we have to find an upper bound. Does it holds that $1-\frac{1}{2i}\leq \frac{1}{2}$ and so $$\prod_{i=1}^n\left (1-\frac{1}{2i}\right )\leq \prod_{i=1}^n \frac{1}{2}=\frac{1}{2^n}$$ Then taking the sum we get $$\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}\leq \sum_{n=1}^{+\infty} \frac{1}{2^n}=1$$ So from the comparison test the original sum must converge also.
Is everything correct?

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*$\displaystyle{\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n\cdot (2n+2)}}$
We have a term that is a product of the form $\frac{2i-1}{2i+2}$. Which upper bound could we use in this case?

Some hints:
For first we can use Raabe's test
$$n\left(\frac{a_n}{a_{n+1}}-1 \right) = \frac{n}{2(n+1)}$$
For second
$$\frac{1}{2\sqrt{n}} \leqslant \frac{1}{2} \frac{3}{4}  \cdots \frac{2n-1}{2n}  \leqslant \frac{1}{\sqrt{2n}}\quad (1)$$
Proof:
For $n=1$ we have $\frac{1}{2} \leqslant \frac{1}{2}  \leqslant \frac{1}{\sqrt{2}} $, so let's assume $n \geqslant 2$. We have
$$\frac{3}{4}>\frac{2}{3}, \frac{5}{6}>\frac{4}{5},\frac{7}{8}>\frac{6}{7}, \cdots, \frac{2n-1}{2n}>\frac{2n-2}{2n-1}$$
multiplication this inequalities gives
$$\frac{3}{4} \frac{5}{6}  \cdots \frac{2n-1}{2n} > \frac{2}{3} \frac{4}{5}  \cdots \frac{2n-2}{2n-1}$$
Now if we multiply left and right sides on left side, the we have
$$\left( \frac{3}{4} \frac{5}{6}  \cdots \frac{2n-1}{2n} \right)^2 > \frac{1}{n} $$
Which is left side of (1).
