# Showing that $\left(1+\frac{\varepsilon^2}{17n}\right)^n-1\leq \frac{\varepsilon^2}{16}$ when $0\leq \varepsilon\leq 1$

Question Show that $$\left(1+\frac{\varepsilon^2}{17n}\right)^n-1\leq \frac{\varepsilon^2}{16}$$ when $$0\leq \varepsilon\leq 1$$.

This inequality appeared in the middle of an argument I was reading and was unsure of the justification I came up with. By the binomial theorem we can write \begin{align} \left(1+\frac{\varepsilon^2}{17n}\right)^n-1&= \sum_{k=1}^n\left(\frac{\varepsilon^2}{17}\right)^k\frac{(n-1)\dotsb(n-k+1)}{n^{k-1}k!}\leq\sum_{k=1}^\infty\left(\frac{\varepsilon^2}{17}\right)^k=\frac{\varepsilon^2}{17-\varepsilon^2}\leq\frac{\varepsilon^2}{16} \end{align} since $$0\leq \varepsilon\leq 1$$. Is this argument correct?

Yes. An important scholium here is a form of Bernoulli's inequality: if $$x> -1$$, and if either $$\alpha> 1$$ and $$x< \tfrac{1}{\alpha-1}$$ or $$\alpha< 0$$, then $$(1+x)^{\alpha}<1+\frac{\alpha x}{1-(\alpha-1)x}\text{.}$$

• (+1) for the nice term "scholium". – Markus Scheuer Nov 12 '18 at 14:13