<s>[INST] 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
 [/INST] 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) 
$$\sum\limits_{n=1}a_n =
\frac{1457}{11025}+\sum\limits_{n=4} a_n < 
\frac{1457}{11025}+\sum\limits_{n=4}\frac{1}{4^{n}}=\frac{1457}{11025}+\frac{1}{4^4}\cdot\frac{1}{1-\frac{1}{4}}=\\
\frac{1457}{11025}+\frac{1}{3\cdot 4^3}=\frac{96923}{705600}<\frac{4}{27}$$
</s>