- #1
sr3056
- 10
- 0
Given f(n) = (1 - (1/n))n
I calculate that the limit as n -> infinity is 1/e.
Also given that x/(1-x) > -log(1-x) > x with 0<x<1 (I proved this in an earlier part of the question) I want to show that:
1 > (f(60)/f(infinity)) > e-1/59 > 58/59
I have tried using my value for f (infinity) and f(60) = e60*log(59/60) in the original inequality but cannot seem to rearrange it to the last one.
I get e(x/1-x) > 1/1-x > ex but am not sure where to go from there.
Thanks!
I calculate that the limit as n -> infinity is 1/e.
Also given that x/(1-x) > -log(1-x) > x with 0<x<1 (I proved this in an earlier part of the question) I want to show that:
1 > (f(60)/f(infinity)) > e-1/59 > 58/59
I have tried using my value for f (infinity) and f(60) = e60*log(59/60) in the original inequality but cannot seem to rearrange it to the last one.
I get e(x/1-x) > 1/1-x > ex but am not sure where to go from there.
Thanks!