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Q) Let, $a_{n} \;=\; \left ( 1-\frac{1}{\sqrt{2}} \right ) ... \left ( 1- \frac{1}{\sqrt{n+1}} \right )$ , $n \geq 1$. Then $\lim_{n\rightarrow \infty } a_{n}$

(A) equals $1$

(B) does not exist

(C) equals $\frac{1}{\sqrt{\pi }}$

(D) equals $0$

My Approach :- I am not getting a particular direction or any procedure to simplify $a_{n}$ and find its value when n tends to infinity.
So, I tried like this simple way to substitute values and trying to find the limiting value :-
$\left ( 1-\frac{1}{\sqrt{1+1}} \right ) * \left ( 1-\frac{1}{\sqrt{2+1}} \right )*\left ( 1-\frac{1}{\sqrt{3+1}} \right )*\left ( 1-\frac{1}{\sqrt{4+1}} \right )*\left ( 1-\frac{1}{\sqrt{5+1}} \right )*\left ( 1-\frac{1}{\sqrt{6+1}} \right )*\left ( 1-\frac{1}{\sqrt{7+1}} \right )*\left ( 1-\frac{1}{\sqrt{8+1}} \right )*.........*\left ( 1-\frac{1}{\sqrt{n+1}} \right )$

=$(0.293)*(0.423)*(0.5)*(0.553)*(0.622)*(0.647)*(0.667)* ....$ =0.009*...

So, here value is tending to zero. I think option $(D)$ is correct.
I have tried like this
$\left ( \frac{\sqrt{2}-1}{\sqrt{2}} \right )*\left ( \frac{\sqrt{3}-1}{\sqrt{3}} \right )*\left ( \frac{\sqrt{4}-1}{\sqrt{4}} \right )*.......\left ( \frac{\sqrt{(n+1)}-1}{\sqrt{n+1}} \right )$
= $\left ( \frac{(\sqrt{2}-1)*(\sqrt{3}-1)*(\sqrt{4}-1)*.......*(\sqrt{n+1}-1)}{{\sqrt{(n+1)!}}} \right )$
Now, again I stuck how to simplify further and find the value for which $a_{n}$ converges when $n$ tends to infinity . Please help if there is any procedure to solve this question.