1 votes 1 votes 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 $0$ Calculus engineering-mathematics calculus userisi2015 usermod sequence-series limits + – ankitgupta.1729 asked Feb 21, 2019 ankitgupta.1729 1.3k views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Satbir commented Feb 21, 2019 i edited by Satbir Feb 21, 2019 reply Follow Share there is + or * in between brackets ? I am getting 1. 0 votes 0 votes ankitgupta.1729 commented Feb 21, 2019 reply Follow Share @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. 0 votes 0 votes Satbir commented Feb 21, 2019 reply Follow Share 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 0 votes 0 votes ankitgupta.1729 commented Feb 21, 2019 reply Follow Share @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. 2 votes 2 votes Satbir commented Feb 21, 2019 reply Follow Share yes correct. 0 votes 0 votes ankitgupta.1729 commented Feb 24, 2019 reply Follow Share https://math.stackexchange.com/questions/3124615/limiting-value-of-a-sequence-when-n-tends-to-infinity 3 votes 3 votes Shiva Sagar Rao commented May 7, 2021 reply Follow Share https://gateoverflow.in/321855/isi2015-mma-22 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes 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. venkatesh pagadala answered Feb 21, 2019 venkatesh pagadala comment Share Follow See 1 comment See all 1 1 comment reply ankitgupta.1729 commented Feb 21, 2019 reply Follow Share @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 $a_{n}$? 1 votes 1 votes Please log in or register to add a comment.