ISI MMA-2015

Question

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 $0$

Comments

there is + or * in between brackets ?

I am getting 1.

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@Satbir It is * between brackets.

It is like $\left ( 1-\frac{1}{\sqrt{1+1}} \right ) * \left ( 1-\frac{1}{\sqrt{2+1}} \right )*\left ( 1-\frac{1}{\sqrt{3+1}} \right )*.........*\left ( 1-\frac{1}{\sqrt{n+1}} \right )$

Answer is not given. Could you please explain how you are getting 1.

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I don't know whether this approach is correct or not.

write the series in reverse order

$\left ( 1- \frac{1}{\sqrt(n+1)} \right )$ * $\left ( 1- \frac{1}{\sqrt(n)} \right )$ * $\left ( 1- \frac{1}{\sqrt(n-1)} \right )$ $\left ( 1- \frac{1}{\sqrt(n-2)} \right )$ .... $\left ( 1- \frac{1}{\sqrt(n-(n-2))} \right )$

applying limit

$\left ( 1- \frac{1}{\sqrt(\infty +1)} \right )$ * $\left ( 1- \frac{1}{\sqrt(\infty )} \right )$ * $\left ( 1- \frac{1}{\sqrt(\infty -1)} \right )$ $\left ( 1- \frac{1}{\sqrt(\infty -2)} \right )$ .... $\left ( 1- \frac{1}{\sqrt(\infty -(\infty -2))} \right )$

$\Rightarrow$  $\left ( 1- \frac{1}{(\infty)} \right )$ * $\left ( 1- \frac{1}{(\infty )} \right )$ * $\left ( 1- \frac{1}{(\infty)} \right )$ $\left ( 1- \frac{1}{(\infty)} \right )$ .... $\left ( 1- \frac{1}{(\infty -(\infty ))} \right )$

$\Rightarrow$  $( 1- 0 )$ * $( 1- 0 )$ * $( 1- 0 )$ * $( 1- 0 )$ * .... $( 1- 0 )$

$\Rightarrow$ 1

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@Satbir In last term , you have put $\frac{1}{\infty -\infty }$ = $0$ but $\infty-\infty$ is an indeterminate form means we can't determine its value. It is not infinity.

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yes correct.

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https://math.stackexchange.com/questions/3124615/limiting-value-of-a-sequence-when-n-tends-to-infinity

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https://gateoverflow.in/321855/isi2015-mma-22

Answer 1

Limit tends to zero.

Just take a calculator and verify for some range of values, you can observe that as we increase the value of n we will get a result close to zero.

Comments

@venkatesh pagadala  here, n tends to $\infty$ , So, to observe the nature of the function $a_{n}$ , we should have to take  very large values of n... but for large value of n , finding $a_{n}$ is difficult.

Do you have any procedure to simplify and calculate limiting value of $a_{n}$?