Register to reply

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

by kishtik
Tags: 1 or nn, infinity, proof, prove
Share this thread:
kishtik
#1
Jan13-05, 08:47 AM
P: 115
Would you please proove that

(lim n -> infinity) (1 + 1/n)^n = e ?
Phys.Org News Partner Mathematics news on Phys.org
'Moral victories' might spare you from losing again
Fair cake cutting gets its own algorithm
Effort to model Facebook yields key to famous math problem (and a prize)
Galileo
#2
Jan13-05, 09:03 AM
Sci Advisor
HW Helper
Galileo's Avatar
P: 2,002
For me, that is the definition of the number e.

So if you want to prove it, you have to tell what your definition of e is to show it is equivalent to that limit.

If for example:

[tex]\mbox{ e is the number such that:} \quad \lim_{h \to 0} \frac{e^h-1}{h}=1[/tex]
is your definition. Then take the derivative of ln(x) at the point 1. (Expressed as a limit)
HallsofIvy
#3
Jan13-05, 09:09 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,357
Using what preliminaries? That is sometimes used as the definition of e!

Lacking that, the simplest way is to use L'Hopital's rule. Strictly speaking L'Hopital's rule only applies to limits of differentiable functions but the limit of (1+1/x)<sup>x</sup>, as x-> infinity, must be the same as the limit of the sequence.

At x= infinity this is of the form "1<sup>infinity</sup>" so let y= (1+ 1/x)<sup>x</sup> and take the logarithm: ln(y)= x ln(1+ 1/x) is now of the form "infinity*0". We can write that as \[\frac{ln(1+1/x)}{x}\] and differentiate numerator and denominator separately. The derivative of ln(1+1/x) is \[\frac{1}{1+1/x}(-\frac{1}{x^2})= -\frac{1}{x^2+x}\]. The derivative of 1/x is -1/x<sup>2</sup>. The ratio of those two is \[\frac{x^2}{x^2+x}= \frac{x}{x+1}= \frac{1}{1+\frac{1}{x}}\] and the limit of that, as x-> infinity, is 1.

Since the limit of ln y(x)= 1, and ln is continuous, the limit of y itself is e.

kishtik
#4
Jan13-05, 09:50 AM
P: 115
Prove (lim n -> infinity) (1 + 1/n)^n = e

Thanks but I couldn't understand the latex code.
vincentchan
#5
Jan13-05, 04:39 PM
P: 610
Okay, I'll use the definition in calculas to PROVED that.....

[tex] \frac{d}{dx} e^x = e^x [/tex] -------->>>difinition of e
substitude [tex] e = \lim (1+1/n)^n [/tex] into [tex] e^x[/tex]

we have [tex] e^x = \lim (1 + 1/n)^{nx} [/tex]..

let [tex]u=nx[/tex],

we got [tex]e^x = \lim (1 + 1/n)^{nx}=\lim (1 + x/u)^u=\lim (1 + x/n)^n[/tex]

[tex] \frac{d}{dx}\lim (1 + 1/n)^{nx} = \frac{d}{dx}\lim (1 + x/n)^n = \lim n(1+x/n)^{n-1}*(1/n) = \lim (1 + x/n)^n=\lim (1 + 1/n)^{nx}[/tex]

so..[tex] e=\lim (1 + 1/n)^n [/tex]
HallsofIvy
#6
Jan13-05, 07:55 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,357
Oh blast! I forgot which forum I was in!

At x= infinity this is of the form "1infinity" so let y= (1+ 1/x)x and take the logarithm: ln(y)= x ln(1+ 1/x) is now of the form "infinity*0". We can write that as [itex]\frac{ln(1+1/x)}{x}[/itex] and differentiate numerator and denominator separately. The derivative of ln(1+1/x) is [itex]\frac{1}{1+1/x}(-\frac{1}{x^2})= -\frac{1}{x^2+x}[/itex]. The derivative of 1/x is -1/x2. The ratio of those two is [itex]\frac{x^2}{x^2+x}= \frac{x}{x+1}= \frac{1}{1+\frac{1}{x}}[/itex] and the limit of that, as x-> infinity, is 1.

Since the limit of ln y(x)= 1, and ln is continuous, the limit is e.


Register to reply

Related Discussions
Proof: Compare two integral(Please look at my surgested proof) Calculus & Beyond Homework 11
Proof Calculus & Beyond Homework 4
Euler's formula proof help Calculus & Beyond Homework 1
Proof: One more irrationality proof Introductory Physics Homework 5
A proof is a proof-says Canadian Prime Minister General Math 0