Jtdylktuy

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

(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.

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

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
$$b_nge 1+frac1{sqrt 2}+frac1{sqrt 3}+ldots+frac1{sqrt {n+1}} $$
so that $b_n$ gets arbitrarily large. We conclude that $a_n$ gets arbitrarily small positive. In other words $a_nto 0$.

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.