Convergence of series $(\frac{1}{3})^{2}+(\frac{1.2}{3.5})^{2}+(\frac{1.2.3}{3.5.7})^{2}+...$

Question

$$\left(\frac{1}{3}\right)^{2}+\left(\frac{1\cdot2}{3\cdot5}\right)^{2}+\left(\frac{1\cdot2\cdot3}{3\cdot5\cdot7}\right)^{2}+...$$

I am not able to find a general equation and that's creating problem for me as I can't proceed further without it.

Answer 1

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?

Answer 2

Yes it converges of course! Use $$\frac{n}{2n+1}<\frac{1}{2}$$

We obtain: $$\left(\frac{1}{3}\right)^{2}+\left(\frac{1.2}{3.5}\right)^{2}+\left(\frac{1.2.3}{3.5.7}\right)^{2}+...<\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}.$$

Answer 3

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: $$ \sum_{n\geq 1}\left(\frac{n!}{(2n+1)!!}\right)^2= -1+\int_{0}^{\pi/6}\frac{4\theta\,d\theta}{\sqrt{1-4\sin^2\theta}}.\tag{4}$$