I am in the process of showing that a sequence of functions is Cauchy. After a long trail of reasoning I find myself needing to prove that $\left(\frac{m}{n}\right)^{\left(\frac{1}{m-n}\right)}$ is in $[1,2]$ for $m,n \geq 2$.
But I am having a difficult time. Is there a simple way?
Question:
Show that $\left(\frac{m}{n}\right)^{\left(\frac{1}{m-n}\right)}$ is in $[1,2]$ for $m>n \geq 2$.
Attempt:
One way is easy. $$\left(\frac{m}{n}\right)^{\left(\frac{1}{m-n}\right)}\geq\left(\frac{n}{n}\right)^{\left(\frac{1}{m-n}\right)}=1$$ How can I show $\left(\frac{m}{n}\right)^{\left(\frac{1}{m-n}\right)} \leq 2$?
The inequality is equivalent to $\frac{m}{2^m}\leq\frac{n}{2^n}$ with $m>n$. Thus it suffices to prove that $\frac{x}{2^x}$ is a decreasing function when $x\geq2$. Differentiating in terms of $x$ gives $(1-x\log2)2^{-x}$, which is indeed negative for $x\geq2$, as desired.