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

Say $f(x) = \frac{1}{x^{1+\frac{\epsilon}{2}}}$

Think about another approach to show that f is integrable