Find the smallest $a>0$ such that $(1+\frac{1}{x})^{x+a}>e$ for all $x\geq 1$

Question

Which value for $a>0$ ist the smallest one such that $\displaystyle (1+\frac{1}{x})^{x+a}>e$ for all $x\geq 1$ ?

It seems to be $\displaystyle a=\frac{1}{2}$, but that’s not obviously.

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Note: a proof with $a:=\frac{1}{2}$

It’s $\displaystyle \frac{e^{2x}-1}{x}<e^{2x}+1$ and therefore $x\coth x>1$ for $x\in\mathbb{R}\setminus\{0\}$.

With $\displaystyle \tilde{x} :=\ln(1+\frac{1}{x})$ one gets $\,\displaystyle e<\exp(\frac{\tilde{x}}{2} \coth \frac{\tilde{x} }{2})=(1+\frac{1}{x})^{x+\frac{1}{2}}\,$.

Answer 1

First note that $$ \lim_{x\rightarrow \infty}(1+1/x)^{x+a}=e $$ So, provided we can insure that $(1+1/x)^{x+a}$ is decreasing for our $a$, we will have the desired result. Let's find the derivative $$ f(x)=(1+1/x)^{x+a}\Rightarrow f'(x)=(1+1/x)^{x+a}\left(\log(1+1/x)-\frac{x+a}{x^2+x}\right) $$ Which we require to be less than zero. So let's see how this restricts $a$ $$ 0>f'(x)=(1+1/x)^{x+a}\left(\log(1+1/x)-\frac{x+a}{x^2+x}\right)\Rightarrow \log(1+1/x)<\frac{x+a}{x^2+x}\\ \Rightarrow a>(x^2+x)\log(1+1/x)-x $$ Then when is $g(x)=(x^2+x)\log(1+1/x)-x$ maximized? By another derivative computation (which I will spare) you can check that the derivative of this function is positive $[1,\infty)$, so our only hope is for a horizontal asymptote, but we are in luck as

$$ \lim_{x\rightarrow \infty}[(x^2+x)\log(1+1/x)-x]= \lim_{x\rightarrow \infty}[x^2\log(1+1/x)+x\log(1+1/x)-x] $$ For which we use Taylor's to find $$ \lim_{x\rightarrow \infty}[x^2\log(1+1/x)+x\log(1+1/x)-x]=x^2\left(1/x-\frac{1}{2x^2}\right)+x(1/x)-x=1-\frac{1}{2}=\frac{1}{2} $$ So picking $a=\frac{1}{2}$ will suffice.