Convergence of series $(\frac{1}{3})^{2}+(\frac{1.2}{3.5})^{2}+(\frac{1.2.3}{3.5.7})^{2}+...$ $$\left(\frac{1}{3}\right)^{2}+\left(\frac{1\cdot2}{3\cdot5}\right)^{2}+\left(\frac{1\cdot2\cdot3}{3\cdot5\cdot7}\right)^{2}+...$$
I am not able to find a general equation and that's creating problem for me as I can't proceed further without it. 
 A: The generic term of the series is
$$a_n=\prod_{k=1}^n \left(\frac{k}{2k+1}\right)^2.$$
Hence, as $n\to+\infty$,
$$\frac{a_{n+1}}{a_n}= \left(\frac{(n+1)}{2(n+1)+1}\right)^2\to \frac{1}{4}.$$
What may we conclude?
A: We have to deal with:
$$ \sum_{n\geq 1}\left(\frac{n!}{(2n+1)!!}\right)^2=\sum_{n\geq 1}\left(\frac{2^n n!^2}{(2n+1)!}\right)^2 = \sum_{n\geq 1}\frac{4^n}{(2n+1)^2\binom{2n}{n}^2} $$
and since $\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$, the above series is clearly convergent. We may recall that
$$ \frac{\arcsin z}{\sqrt{1-z^2}}=\sum_{n\geq 0}\frac{z^{2n+1}4^n}{(2n+1)\binom{2n}{n}} \tag{1}$$
$$ \int_{0}^{\pi/2}\left(\sin\theta\right)^{2n+1}\,d\theta = \frac{4^n}{(2n+1)\binom{2n}{n}}\tag{2} $$
hence we have:
$$ \sum_{n\geq 1}\left(\frac{n!}{(2n+1)!!}\right)^2=\tfrac{1}{9}\cdot\phantom{}_3 F_2\left(1,2,2;\tfrac{5}{2},\tfrac{5}{2};\tfrac{1}{4}\right)=-1+\int_{0}^{\pi/2}\frac{4\arcsin\frac{\sin\theta}{2}}{\sqrt{4-\sin^2\theta}}\,d\theta.\tag{3} $$
or, in a more compact form:
$$ \sum_{n\geq 1}\left(\frac{n!}{(2n+1)!!}\right)^2= -1+\int_{0}^{\pi/6}\frac{4\theta\,d\theta}{\sqrt{1-4\sin^2\theta}}.\tag{4}$$
A: Yes it converges of course! Use $$\frac{n}{2n+1}<\frac{1}{2}$$
We obtain:
$$\left(\frac{1}{3}\right)^{2}+\left(\frac{1.2}{3.5}\right)^{2}+\left(\frac{1.2.3}{3.5.7}\right)^{2}+...<\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}.$$
