@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.
@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.
https://math.stackexchange.com/questions/3124615/limiting-value-of-a-sequence-when-n-tends-to-infinity
@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 $a_{n}$?