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

###### Theorem 1.

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

###### Proof.

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

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

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

 $\displaystyle{\left(1+{1\over n}\right)^{n}\over\left(1+{1\over n-1}\right)^{n% -1}}$ $\displaystyle\geq$ $\displaystyle\left(1-{n-1\over n^{2}}\right)\left(1+{1\over n}\right)$ $\displaystyle=$ $\displaystyle 1+{1\over n^{3}}$ $\displaystyle>$ $\displaystyle 1,$

hence the sequence is increasing. ∎

###### Theorem 2.

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

###### Proof.

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

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

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

 $\displaystyle{\left(1+{1\over n}\right)^{n+1}\over\left(1+{1\over n-1}\right)^% {n}}$ $\displaystyle\leq$ $\displaystyle\left(1-{1\over n^{2}}\right)^{n}\left(1+{1\over n^{2}}\right)^{n}$ $\displaystyle=$ $\displaystyle\left(1-{1\over n^{4}}\right)^{n}$ $\displaystyle<$ $\displaystyle 1,$

hence the sequence is decreasing.

###### Theorem 3.

For all positive integers $m$ and $n$, we have $(1+1/m)^{m}<(1+1/n)^{n+1}$.

###### Proof.

We consider three cases.

Suppose that $m=n$. Since $n>0$, we have $1/n>0$ and $1<1+1/n$. Hence, $(1+1/n)^{n}<(1+1/n)^{n+1}$.

Suppose that $m. By the previous case, $(1+1/n)^{n}<(1+1/n)^{n+1}$. By theorem 1, $(1+1/m)^{m}<(1+1/n)^{n}$. Combining, $(1+1/m)^{m}<(1+1/n)^{n+1}$.

Suppose that $m>n$. By the first case, $(1+1/m)^{m}<(1+1/m)^{m+1}$ By theorem 2, $(1+1/m)^{m+1}<(1+1/n)^{n+1}$. Combining, $(1+1/m)^{m}<(1+1/n)^{n+1}$. ∎

Title $(1+1/n)^{n}$ is an increasing sequence 11nnIsAnIncreasingSequence 2013-03-22 15:48:39 2013-03-22 15:48:39 rspuzio (6075) rspuzio (6075) 14 rspuzio (6075) Theorem msc 33B99