From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: When was e first defined?
Date: 4 Oct 1999 06:04:22 -0400
Newsgroups: sci.math
Keywords: Bernoulli inequality proves  e  exists

In article <37F877BF.A251AB7B@nospam.interchange.ubc.ca>,
Shawn  <shawnc@NOSPAM.interchange.ubc.ca> wrote:
:I have a question related to e: can anyone give a proof that for the
:sequence An =
:(1+(1/n))^n , 2=<An<3 for all n being elements of natural numbers, and
:that An < A(n+1) ?

With pleasure. The main tool is Bernoulli inequality:

If r > -1 and r is not 0, and if n >= 2 then

 (1 + r)^n > 1 + r*n

Proof goes by induction.

Now for A(n) = (1+1/n)^n, manipulate  A(n+1) / A(n) (planning to prove
that it is >1): (Fill in the gaps for exercise)

A(n+1) / A(n) = (n*(n+2)/(n+1)^2)^(n+1) * (1 + 1/n)

 = (1 - 1/(n+1)^2)^(n+1) * (1+1/n)

and by Bernoulli inequality, this is

 > (1 - 1/(n+1)) ( 1 + 1/n) = (n/(n+1)) * (n+1)/n = 1

proved.

Since A(1)=2, we have A(n)>2 for n>1.

Trick: Define B(n) = (1+1/n)^(n+1) and prove that B decreases.

This is done by proving that B(n)/B(n+1) > 1. Again,

B(n)/B(n+1) = ... = (1+1/((n*(n+2)))^(n+2) / (1 + 1/n)

> (1 + 1/n) / (1 + 1/n) = 1

Finally, A(n) < B(n) evidently, and we can prove A(n) < B(m) for all m, n.
How? Suppose n<m, then A(n)<A(m)<B(m), and similarly if n>=m (using B
terms twice).

Here B(1)=4, B(2)=27/8 < 4. Your claim is proved.
(And both A(n), B(n) converge to e, from different sides.)

Exercise: Is the sequence C(n) = (1 + 1/(n-1/2))^n increasing or
decreasing? (It also converges to e, and faster than A(n) or B(n).)

Cheers, ZVK(Slavek).