Proving $\left(1+\frac{1}{m}\right)^m < \left(1+\frac{1}{n}\right)^n$

Question

Let $m,n\in \mathbb{N}$. If $m > n$ show that

$$\left(1+\frac{1}{m}\right)^m > \left(1+\frac{1}{n}\right)^n$$

My works: I tried to show if $g(x)=\left(1+\frac{1}{x}\right)^x$ then $g'(x) > 0$.

\begin{align} g'(x) &= \frac{d e^ { x \ln(1+\frac{1}{x}) }} {dx} \\ &= e^{x\ln(1+1/x)} \left(\ln(1+\frac{1}{x}) - \frac{1}{1+x} \right) >0 \end{align}

Answer 1

Normally the proof goes something like, let $a_n=(1+\frac{1}{n})^n$ for all $n\in\mathbb Z^+$. Then, $$\frac{a_{n+1}}{a_n}=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^n}=\frac{(\frac{n+2}{n+1})^{n+1}}{(\frac{n+1}{n})^n}=\frac{(n+2)^{n+1}n^n}{(n+1)^n(n+1)^{n+1}}=\frac{(n^2+2n)^n}{(n+1)^{2n}}\frac{n+2}{n+1}=\frac{((n+1)^2-1)^n}{((n+1)^2)^n}\frac{n+2}{n+1}=\bigg(1-\frac{1}{(n+1)^2}\bigg)^n\frac{n+2}{n+1}\geq\bigg(1-\frac{n}{(n+1)^2}\bigg)\frac{n+2}{n+1}=\frac{n^2+n+1}{(n+1)^2}\frac{n+2}{n+1}=\frac{(n^2+n+1)(n+2)}{(n+1)^3}=\frac{n^3+3n^2+3n+2}{n^3+3n^2+3n+1}> 1$$

(the first inequality that appears above comes from Bernoulli's inequality which can be proven by simple induction as seen in the link)

So, $\frac{a_{n+1}}{a_n}>1\implies a_{n+1}>a_n$. This means if $m$ is any integer greater than $n$, $$a_m>a_{m-1}>a_{m-2}>...>a_{n+1}>a_n$$
giving us $a_m>a_n$, i.e., $(1+\frac{1}{m})^m>(1+\frac{1}{n})^n$

Answer 2

$$ \frac{d}{dx} \exp \left( \ln\left(1 + \frac{1}{x}\right) x \right) = \exp \left( \ln\left(1 + \frac{1}{x}\right) x \right) \left( \ln\left(1 + \frac{1}{x}\right) - \frac{1}{x + 1} \right) > 0 $$ for $x > 1$, so that the function is strictly increasing on $(1, \infty)$. Indeed, $$ \ln\left(1 + \frac{1}{x}\right) = \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3} \mp \cdots, $$ whereas $$ \frac{1}{x + 1} = \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x^3} \mp \cdots. $$

Answer 3

If $m<n$, then for all $j=0...m$ we have $1-\frac jm \le 1-\frac jn$ and so $\frac{m-j}{m} \le \frac{n-j}{n}$ (for $j=0$ we have equality) and therefore:

$$ \begin{align} \left(1+\frac1m\right)^m & = \sum_{k=0}^m \binom{m}{k}\frac{1}{m^k} = \sum_{k=0}^m \frac {1}{k!}\prod_{j=0}^{k-1}\frac{m-j}{m}\\ & \le \sum_{k=0}^m \frac {1}{k!}\prod_{j=0}^{k-1}\frac{n-j}{n} = \sum_{k=0}^m \binom{n}{k}\frac{1}{n^k}\\ & < \sum_{k=0}^n \binom{n}{k}\frac{1}{n^k} = \left(1+\frac1n\right)^n\\ \end{align} $$

Answer 4

By Bernoulli’s inequality, for $n>m>0$, $$\left(1+\frac1n\right)^{\frac{n}m} >1+\frac1m$$