# Ngô Quốc Anh

## March 31, 2021

### Monotonicity of (1+1/x)^(x+α)

Filed under: Uncategorized — Ngô Quốc Anh @ 1:11

This post concerns the monotonicity of $\displaystyle f: x \mapsto \Big(1+\frac 1x\Big)^{x+a }$

on $(0,+\infty)$ with $a \in \mathbb R$. The two cases $a =0$ and $a =1$ are of special because these are always mentioned in many textbooks as $\displaystyle \lim_{n \to \infty} \Big(1+\frac 1n\Big)^n= \lim_{n \to \infty} \Big(1+\frac 1n\Big)^{n+1} =e.$

Clearly, $f$ is monotone increasing with respect to $a$. Hence we are left with the monotonicity of $f$ with respect to $x$. From the above discussion, the function $f$ is monotone increasing with respect to $x$ when $a = 0$. When $a<0$, as the function $(1+1/x)^{-|a|}$ is monotone increasing with respect to $x$, we deduce that the function $f$ is monotone increasing with respect to $x$. Hence, we are left with the case $a > 0$. To study this problem, we examine $f'$ with respect to $x$.

Derivative of $f$. It is easy to get $\displaystyle f'(x) = \Big(1+\frac 1x\Big)^{x+a } \Big[ \log \Big(1+\frac 1x\Big)- \frac{x+a }{x(x+1)}\Big].$

Hence the sign of $f'$ is determined by the sign of $\displaystyle g(x) = \log \Big(1+\frac 1x\Big)- \frac{x+a }{x(x+1)} .$

on $(0, \infty)$.

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