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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.

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marked as duplicate by Sil, Saucy O'Path, RRL, Lord Shark the Unknown, Pedro Feb 24 at 11:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The hint: $$0<a_n=\prod_{k=1}^n\left(1-\frac{1}{\sqrt{k+1}}\right)<\prod_{k=1}^n\left(1-\frac{1}{k+1}\right)=\frac{1}{n+1}\rightarrow0$$

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  • $\begingroup$ Michael.In 2 lines! :) $\endgroup$ – Peter Szilas Feb 24 at 9:28
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As we multiply positive factors below $1$, the sequence is positive and strictly decreasing, hence convergent (ruling out B). As already $a_1<1$, we also rule out A.

If we multiply $a_n$ by $b_n:=\left(1+\frac1{\sqrt 2}\right)\cdots \left(1+\frac1{\sqrt {n+1}}\right)$, note that the product of corresponding factors is $\left(1-\frac1{\sqrt {k}}\right)\left(1+\frac1{\sqrt {k}}\right)=1-\frac1k<1$, hence $a_nb_n<1$. On the other hand, by expanding the product and dropping lots of positive terms $$b_n\ge 1+\frac1{\sqrt 2}+\frac1{\sqrt 3}+\ldots+\frac1{\sqrt {n+1}} $$ so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_n\to 0$.

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  • $\begingroup$ Hagen.Very nice. $\endgroup$ – Peter Szilas Feb 24 at 9:27
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Here is a simple approach to rule out the wrong options and therefore find the correct one. I stress that this is not a proof that the limit is what it is, but a quick way of reasoning your way through a multiple choice question.

The numbers $1-1/\sqrt{k}$ are all in $(0,1)$, so your sequence is positive and strictly decreasing. Therefore it has a limit. You have already computed enough terms to rule out the limits $1$ and $1/\sqrt{\pi}$. Remember that it's decreasing. The only remaining option is that the limit is zero.

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