Samyak Jain
Last Activity: 5 Years ago
Ans is e^(1/2).
Let the limit be L. Now take natural logarithm on both sides.
lnL = lim (1/n).ln{(1 + 1/n)(1 + 2/n)...(1 + n/n)}
n
lnL = lim (1/n).[ln(1 + 1/n) + ln(1 + 2/n) +...+ ln(1 + n/n)]
n
Now use expansion of ln(1 + x) which is x – x2/2 + x3/3 – … and neglect higher powers of x here as n is tending to infinity.
lnL = lim (1/n).[1/n – 1/2n2 + 2/n – 2/2n2 +...+ n/n – 2/n.n2)]
n
lnL = lim (1/n).[(1/n) n(n + 1)/2 – (1/2n2).n(n + 1)/2]
n
= lim n(n + 1)/2n
2 – lim n (n + 1)/4n
3 = 1/2 – 0 = 1/2
L = e^(1/2)
