Q:

Given $f$ is $o(1/x)$ as $x\to \infty$, is it true that $f$ is $o(1/x^{1+\epsilon})$ as well? ( for some small enough $\epsilon>0$)

I'm basically trying to show that $f$ is summable $1$ to $\infty$ and that $f$ is integrable $1$ to $\infty$. Also, it necessary that $f$ be monotonic or continuous? 

A:

No.   Let $f(x)=\frac 1{x \log x}$.  This is $o(\frac 1x)$ but not $o(\frac 1{x^{1+\epsilon}})$

