Value of $\lim\limits_{n\rightarrow \infty}(a_{1}+a_{2}+\cdots +a_{n})$ 
If $\displaystyle a_{n}=\bigg(\frac{n!}{1\cdot 3 \cdot 5 \cdot 7\cdot...\cdot (2n+1)}\bigg)^2.$
Then $\displaystyle \lim_{n\rightarrow \infty}\bigg(a_{1}+a_{2}+...+a_{n}\bigg)$ is

Options:

$(a)$ Does not exists
$(b)$ Greater than $\displaystyle \frac{4}{27}$
$(c)$ Less than $\displaystyle \frac{4}{27}$
$(d)$ None of these

My Try: $$a_{n}=\bigg[\frac{n!\cdot 2\cdot 4 \cdot 6 \cdots (2n)}{1\cdot 2 \cdot 3\cdot 4\cdot \cdots (2n)\cdots (2n+1)}\bigg]^2$$
$$a_{n}=\bigg[\frac{n!\cdot 2 \cdot 4 \cdot 6 \cdots (2n)}{(2n+1)!}\bigg]^2$$
Could some help me to solve it , Thanks
 A: Why, why bounty when @gt6989b gave you an awesome hint :) ? 
$$a_n=\left(\left(\frac{1}{3}\right)\cdot\left(\frac{2}{5}\right)\cdot...\cdot\left(\frac{n}{2n+1}\right)\right)^2 < 
\frac{1}{2^{2n}}=\frac{1}{4^n}$$
and (calculating the first 3 terms and using infinite geometric progression) 
$$\sum\limits_{n=1}a_n =
\frac{1457}{11025}+\sum\limits_{n=4} a_n < 
\frac{1457}{11025}+\sum\limits_{n=4}\frac{1}{4^{n}}=\frac{1457}{11025}+\frac{1}{4^4}\cdot\frac{1}{1-\frac{1}{4}}=\\
\frac{1457}{11025}+\frac{1}{3\cdot 4^3}=\frac{96923}{705600}<\frac{4}{27}$$
A: Using the Wallis formula
$$\dfrac\pi2 = \dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot\dfrac{6\cdot6}{5\cdot7}\dots$$
in the form of
$$a_n=\left(\dfrac{1\cdot2\cdot3\dots n}{3\cdot 5\cdot7\dots(2n+1)}\right)^2 < \frac\pi{(2n+1)2^{2n+1}},$$
one can get
$$a_1+a_2+a_3+a_4+\dots < \frac19+\dfrac\pi{2^5}\left(\dfrac15+\dfrac1{7\cdot4}+\dfrac1{9\cdot4^2}+\dots\right)$$
$$ < \frac19+\dfrac\pi{5\cdot2^5}\cdot\dfrac1{1-\dfrac14}
 = \dfrac19+\dfrac\pi{120}\color{brown}{\mathbf{<\dfrac4{27}}}.$$
A: A standard trick is to look at the function 
$$f(x)=\sum \left(\frac{n!}{1.3...(2n+1)}\right)^2x^{2n+1}$$
Find a diiferential equation that $f(x)$ satisfies, solve it and find $f(1)$.
