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
Best Answer
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) $$\sum\limits_{n=1}a_n = \frac{1457}{11025}+\sum\limits_{n=4} a_n < \frac{1457}{11025}+\sum\limits_{n=4}\frac{1}{4^{n}}=\frac{1457}{11025}+\frac{1}{4^4}\cdot\frac{1}{1-\frac{1}{4}}=\\ \frac{1457}{11025}+\frac{1}{3\cdot 4^3}=\frac{96923}{705600}<\frac{4}{27}$$