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

If $$m < n$$ such that $$m$$ and $$n$$ are natural numbers then prove $$\left(1+\frac{1}{m}\right)^m < \left(1+\frac{1}{n}\right)^n$$

## closed as off-topic by Cameron Williams, José Carlos Santos, Robert Wolfe, RRL, Theo BenditFeb 9 at 7:20

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• You could just look at the function $f(x)=(1+\frac{1}{x})^x$ and its derivative, and see that the derivative is positive. – Matti P. Feb 8 at 12:08

$$\frac{d}{dx} \exp \left( \ln\left(1 + \frac{1}{x}\right) x \right) = \exp \left( \ln\left(1 + \frac{1}{x}\right) x \right) \left( \ln\left(1 + \frac{1}{x}\right) - \frac{1}{x + 1} \right) > 0$$ 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 $$\frac{1}{x + 1} = \frac{1}{x} - \frac{1}{x^2} + \frac{1}{x^3} \mp \cdots.$$

• I think you missed $\ln x$ in derivative of $1+1/x$. – didgogns Feb 9 at 5:26
• Thanks, yes, I made a mistake. It should now be fine. – AlgebraicsAnonymous Feb 9 at 8:54

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

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., $$(1+\frac{1}{m})^m>(1+\frac{1}{n})^n$$

If $$m, then for all $$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}