Question

prove to me sum of (n!)/(n)^(n) converge or diverge

Answer 1

it'll converge...use ratio test

Answer 2

lim n---> infintiy a(n+1) /a(n) = 1/e, which is less than 1 so the given series will converge

Answer 3

a(n)= (n!)/(n)^(n)

Answer 4

it will converge only when n(!)/(n)^(n)is greater than one

Answer 5

.... the result of the limit is very baffling to me, how do you get 1/e?

Answer 6

lim n---> infintiy a(n+1) /a(n)
=lim n---->infinity [(n+1)!)/(n+1)^(n+1) ] / [(n!)/(n)^(n)]
=lim n---->infinity 1/ ( 1+ 1/n)^n

Answer 7

fine?

Answer 8

oh yeh then it will diverge

Answer 9

[(n+1)!)/(n+1)^(n+1) ] / [(n!)/(n)^(n)]

= {n/(n+1) } ^n

Answer 10

ha bhai! yeh to bata ki diverge hoga ya converge

Answer 11

and use lim n---> infinity (1 + 1/n) ^n = e

Answer 12

abe bas maths dekh li teri

Answer 13

:)

Answer 14

good job, I don't remember that limit for e. Anyways kudos. Are you a physics major?

Answer 15

yes

Answer 16

lol, it was bad enough. Akash is exceptional in solving that problem.

Answer 17

alright, let's see. I've got plenty of hard questions