find the minimun value of $\alpha$ such that $(1+\frac{1}{x})^{x+\alpha}>e$ for every $x>0$ Find the minimun $\alpha$ such that $(1+\frac{1}{x})^{x+\alpha}>e$ for every $x>0$
I tried a lot but nothing was helpful :/

If you want $$f=(1+\frac{1}{x})^{x+\alpha}-e>0$$ for any value of $x>0$, it must be true for very large values of $x$. For large values of $x$, the Taylor expansion of $f$ is given by $$\frac{e a-\frac{e}{2}}{x}+\frac{\frac{e a^2}{2}-e a+\frac{11
   e}{24}}{x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ Then $a \geq \frac {1}{2}$ fullfills the condition.
A plot of the function $f$ for $a =\frac {1}{2}$ shows that it is always positive. If $a=0.49$ is used $f$ is negative for any $x \gt 7.84$; if $a=0.48$ is used $f$ is negative for any $x \gt 3.68$;  if $a=0.47$ is used $f$ is negative for any $x \gt 2.30$.
