Does e^1/n converge?

Question

I'm not sure how to go about solving this. Can someone please explain? Thanks in advance :)

Answer 1

As 1/n=>0 as n=> to infinity e^1/n=>1 so S e^1/n (S=sum)
does not converge because it a necessary condition for convergence of S a_n that lim a_n =0

Answer 2

Yes, it converges to 1. Since e^(1/n) is continuous, one
can consider the limit of ln(e^(1/n)) = 1/n instead. As
n approaches infinity, 1/n approaches 0. Hence, e^(1/n)
approaches the inverse image of ln at 0, which is 1.

Answer 3

It should converge. You can use the ratio test:


abs( e^(1/n+1) / e^(1/n)) = abs (e^(1/n+1 - 1/n))
= abs (e^(n/n*(n+1) - (1+n)/n*(n+1))
= abs(e^(-1/n*(n+1)))< 1 as n goes to infinity.

Answer 4

e^(1/n) as n goes to infinity will go to e^0, which equals 1.

Answer 5

as n tends to infinity f(n) tends to 0