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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?

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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 $$(1+x)^{\alpha}<1+\frac{\alpha x}{1-(\alpha-1)x}\text{.}$$

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

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