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?

The Stirling approximation implies $\binom{2n}{n}\sim\frac{4^n}{\sqrt{n\pi}}$, so $\frac{n!^24^n}{(2n+1)!}\sim\frac{\sqrt{\pi}/2}{\sqrt{n}}$, so the series diverges.
