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

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

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. 
