tight bounds for $(1-x)^{1/x}$ for $x\in[0,1]$

Question

It is well known that $0\le (1-x)^{1/x}\le exp(-1)$ for $x\in[0,1]$. However, are there tighter upper and lower bounds that also capture the rate by which the value drops w.r.t $x$?

Answer 1

As @FengShao points out, $$ \frac{\log(1-x)}{x}=-1-\frac{x}2-\frac{x^2}3-\frac{x^3}4-\cdots $$ which means that (taking $\exp$ of both sides) $$ (1-x)^{1/x} = \exp\left(-1-\frac{x}2-\frac{x^2}3-\frac{x^3}4-\cdots \right) $$ Now because $\exp$ is an increasing function, and all the terms involving $x$ are negative, we can further write $$ (1-x)^{1/x} = \exp\left(-1-\frac{x}2-\frac{x^2}3-\frac{x^3}4-\cdots \right) \le \exp(-1), $$ which is your known bound. But we can do better. Because each term involving $x$ is nonnegative, we can also say that \begin{align} (1-x)^{1/x} &\le\exp\left(-1-\frac{x}2\right)\\ (1-x)^{1/x} &\le\exp\left(-1-\frac{x}2-\frac{x^2}3\right)\\ (1-x)^{1/x} &\le\exp\left(-1-\frac{x}2-\frac{x^2}3-\frac{x^3}4\right)\\ \end{align} a sequence of upper bounds that converges, in the limit, to my trivial upper bound. Away from $x = 0$, each of these bounds can also be replaced by a strict inequality (i.e., on the interval $0 < x < 1$).