Determine the smallest $k$ such that $\log (1 +e^x) < k + x ?$ Determine the  smallest  $k$ such that  $\log (1  +e^x)  < k  + x  $  where $k$ is constant and $x \in (0,\infty)$
My attempt :$$\log(1+e^x)= e^x -\frac{e^{x}}{2}  +\frac{e^{2x}}{3}+.....$$
$$e^x  = 1 +x+x^2/2^2+...$$
$$\log(1+e^x)= (1+x+x^2/2^2+....)-\frac{1}{2}(1+x+x^2/2^2+...)\\+\frac{1}{3}(1+2x+(2x)^2/2^2+...)-\frac{1}{4}(1+3x+(3x)^2/2^2+...$$
$$=(1-\frac{1}{2}  +\frac{1}{3} -\frac{1}{4}+...)  + x(1-1/2+2/3-3/4+....)  +  \text{higher degree} $$
$$\implies\log(1+e^x)   < \log 2  + x \implies  k=\log 2$$

$f(x)=e^x (e^k -1)>1$ is strictly increasing.
$\\ $ $f(x\to +0)>1 \implies k>\ln 2 $ or $\inf k =\ln 2 $.
