$\sum_{n=1}^\infty \frac{(2n+1)!}{2^{2n}(n!)^2}$ converges? Does the series $$\sum_{n=1}^\infty \frac{(2n+1)!}{2^{2n}(n!)^2}$$ converges? I tried D' Alembert's quotient criterion but the $\displaystyle\lim_n \frac{a_{n+1}}{a_n}$ equals to $1$ and then by Cauchy's n-th root criterion $\displaystyle\lim_n \sqrt[n]{a_n}$ it gives an expession that is $\frac{\infty}{\infty}$ so I use the inequality $\sqrt[n]{n!}>2,\ \forall n\geq 4$ as a result $\displaystyle\lim_n \sqrt[n]{a_n}=+\infty$, and consequently the series diverges. However, I don't know if it is correct (if yes, then somebody may confirm my thoughts). Also, does anybody see another way to decide if the previous series converges or not? Thanks
 A: Set $a_n := \frac{(2n+1)!}{2^{2n}(n!)^2}$. I'm sure that you have done the following computation:
$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(2n+3)! \cdot 2^{2n} \cdot (n!)^2}{(2n+1)! \cdot 2^{2n+2} \cdot ((n+1)!)^2}.$$
Now,
$$\frac{(2n+3)!}{(2n+1)!} = 2(n+1)(2n+3),$$
$$\frac{2^{2n}}{2^{2n+2}} = \frac14,$$
$$\frac{(n!)^2}{((n+1)!)^2} = \frac1{(n+1)^2},$$
so that the above expression simplifies to
$$\left|\frac{a_{n+1}}{a_n}\right| = \frac{2n+3}{2n+2} > 1.$$
This shows that $(a_n)$ is (strictly) increasing. Since $a_1 = \frac32$, this means that $\lim a_n \neq 0$. Thus, the series diverges.
A: Note that for $n\ge 1$ we have
$$\begin{align}
\frac{(2n+1)!}{(n!)^2}&=\frac{(2n+1)(2n)(2n-1)(2n-2)\cdots (n+1)n(n-1)(n-2)\cdots4\cdot 3\cdot 2}{(n!)^2}\\\\
&>\frac{(2n)^2 (2n-2)^2 \cdot (2)^2}{(n!)^2}\\\\
&=4^n
\end{align}$$
Hence, the general term of the series, $\displaystyle a_n=\frac{(2n+1)!}{4^n(n!)^2}$, is bounded below by $1$ and fails to approach $0$ as $n\to \infty$.  We conclude that the series diverges.

NOTE:  Using Stirling's Formula shows that the terms of the series are $O\left(\sqrt{n}\right)$ as $n\to \infty$.  So, not only do the general terms fail to approach $0$, they are unbounded.
A: Hint:
The term $T_5$ is $$\frac{2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11}{(2\cdot4\cdot6\cdot8\cdot10)^2}=\frac{3\cdot5\cdot7\cdot9\cdot11}{2\cdot4\cdot6\cdot8\cdot10}.$$ You see that every new factor is larger than $1$ and the general term does not decrease. (In fact, even the general term itself diverges.)
A: Let $$f(x) := (1-4x^2)^{-1/2} = \sum_{n=0}^{\infty}\binom{2n}{n}x^{2n}.$$
The radius of convergence is $\tfrac{1}{2}$, so for $x<\tfrac{1}{2}$ we have:
$$\frac{d}{dx}(xf(x)) = \sum_{n=0}^{\infty}(2n+1)\binom{2n}{n}x^{2n}$$
which we can find analytically:
$$(xf(x))' = \frac{1+4x^2}{(1-4x^2)^2}$$
So as $x \to 1/2$ from below, the power series can be made as large as we like.
A: I would use equivalents * with Stirling's' formula:
\begin{align}
\frac{(2n+1)!}{2^{2n}(n!)^2}&=\frac{\sqrt{2\pi(2n+1)}\Bigl(\cfrac{2n+1}{\mathrm e}\Big)^{\!2n+1}}{2^{2n}2\pi\mkern1.5mu n\Bigl(\cfrac{n}{\mathrm e}\Bigr)^{\!2n}} \\
&=\frac{(2n+1)^{\tfrac 32}}{\sqrt{2\pi}\,n}\,\biggl(\frac{2n+1}{2n}\biggr)^{\mkern-5mu2n}\sim_\infty 2\mkern1.5mu  \mathrm e\sqrt{\cfrac n\pi },
\end{align}
which is tends to $\infty$, so the series diverges trivially.
