${(1+1/n)}^n$ is an increasing sequence

Theorem 1.

The sequence ${\mathrm{(}\mathrm{1}\mathrm{+}\mathrm{1}\mathrm{/}n\mathrm{)}}^n$ is increasing.

Proof.

To see this, rewrite $1+(1/n)=(1+n)/n$ and divide two consecutive terms of the sequence:

${\displaystyle \frac{{\left(1+\frac{1}{n}\right)}^n}{{\left(1+\frac{1}{n-1}\right)}^{n-1}}}$ $=$ ${\displaystyle \frac{{\left(\frac{n+1}{n}\right)}^n}{{\left(\frac{n}{n-1}\right)}^{n-1}}}$

$=$ ${\left({\displaystyle \frac{(n-1)(n+1)}{n^2}}\right)}^{n-1}{\displaystyle \frac{n+1}{n}}$

$=$ ${\left(1-{\displaystyle \frac{1}{n^2}}\right)}^{n-1}\left(1+{\displaystyle \frac{1}{n}}\right)$

Since ${(1-x)}^n\ge 1-nx$, we have

${\displaystyle \frac{{\left(1+\frac{1}{n}\right)}^n}{{\left(1+\frac{1}{n-1}\right)}^{n-1}}}$ $\ge$ $\left(1-{\displaystyle \frac{n-1}{n^2}}\right)\left(1+{\displaystyle \frac{1}{n}}\right)$

$=$ $1+{\displaystyle \frac{1}{n^3}}$

$>$ $1,$

hence the sequence is increasing. ∎

Theorem 2.

The sequence ${\mathrm{(}\mathrm{1}\mathrm{+}\mathrm{1}\mathrm{/}n\mathrm{)}}^{n\mathrm{+}\mathrm{1}}$ is decreasing.

Proof.

As before, rewrite $1+(1/n)=(1+n)/n$ and divide two consecutive terms of the sequence:

${\displaystyle \frac{{\left(1+\frac{1}{n}\right)}^{n+1}}{{\left(1+\frac{1}{n-1}\right)}^n}}$ $=$ ${\displaystyle \frac{{\left(\frac{n+1}{n}\right)}^{n+1}}{{\left(\frac{n}{n-1}\right)}^n}}$

$=$ ${\left({\displaystyle \frac{(n-1)(n+1)}{n^2}}\right)}^n{\displaystyle \frac{n+1}{n}}$

$=$ ${\left(1-{\displaystyle \frac{1}{n^2}}\right)}^n\left(1+{\displaystyle \frac{1}{n}}\right)$

Writing $1+1/n$ as $1+n/n^2$ and applying the inequality $1+n/n^2\le {(1+1/n^2)}^n$, we obtain

${\displaystyle \frac{{\left(1+\frac{1}{n}\right)}^{n+1}}{{\left(1+\frac{1}{n-1}\right)}^n}}$ $\le$ ${\left(1-{\displaystyle \frac{1}{n^2}}\right)}^n{\left(1+{\displaystyle \frac{1}{n^2}}\right)}^n$

$=$ ${\left(1-{\displaystyle \frac{1}{n^4}}\right)}^n$

$1,$

hence the sequence is decreasing.

∎

Theorem 3.

For all positive integers $m$ and $n$, we have .

Proof.

We consider three cases.

Suppose that $m=n$. Since $n>0$, we have $1/n>0$ and . Hence, .

Suppose that . By the previous case, . By theorem 1, . Combining, .

Suppose that $m>n$. By the first case, By theorem 2, . Combining, . ∎

Title	${(1+1/n)}^n$ is an increasing sequence
Canonical name	11nnIsAnIncreasingSequence
Date of creation	2013-03-22 15:48:39
Last modified on	2013-03-22 15:48:39
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	14
Author	rspuzio (6075)
Entry type	Theorem
Classification	msc 33B99