Mean Value Theorem Problem

sr3056

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) = e^{60*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 > e^x but am not sure where to go from there.

Thanks!

chiro

Hey sr3056 and welcome to the forums.

Try subtracting 1/59 from both sides (let x = 1)

mfb

Do you know e^x >= 1+x (with "=" just for x=0)? You can use it to show e^x/(1-x) > 1/(1-x)

1/1-x > e^x... maybe the derivatives can give you some way to prove this.

@chiro: x=1 is not in the range where the equation is well-defined.

haruspex

Try inverting each term in that line (adjusting the relationships as appropriate), then plugging in a value for x.

sr3056

Thank you for your responses.

Inverting, I get e^(x/x-1) < 1 - x < e^-x

Letting x = 1/60 I then get e^-1/59 < 59/60 < e^-1/60

This is starting to look a bit more like it, but I'm not sure how my f(60) and f(infinity) values are going to come in...

mfb

"f(60)/f(infinity)" = e*(1-1/60)^60
Take the log, and your inequality reads:

0 > 1+60 log(1-1/60) > -1/59
You get the left ">" with "-log(1-x) > x" and the right ">" with "x/(1-x) > -log(1-x)", both with the obvious choice for x.

sr3056

Thank you very much