${(1+1/n)}^n$ 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 twoconsecutive 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 twoconsecutive 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, .∎