# ${(1+1/n)}^{n}$ 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:

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

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

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

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

$\frac{{\left(1+\frac{1}{n}\right)}^{n}}{{\left(1+\frac{1}{n-1}\right)}^{n-1}}$ | $\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:

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

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

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

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

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