<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] 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
$$a_1+a_2+a_3+a_4+\dots < \frac19+\dfrac\pi{2^5}\left(\dfrac15+\dfrac1{7\cdot4}+\dfrac1{9\cdot4^2}+\dots\right)$$
$$ < \frac19+\dfrac\pi{5\cdot2^5}\cdot\dfrac1{1-\dfrac14}
 = \dfrac19+\dfrac\pi{120}\color{brown}{\mathbf{<\dfrac4{27}}}.$$
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