Knowing that $m\to \infty$, $n\to \infty$, $m^2<n^{2/3}$, I came across the following expression: $$\frac{2x^2}{\frac{x^2m^2}{n^{2/3}}+m^4n^{2/3}}$$ Playing a bit with said expression, I saw it increases until $x=mn^{2/3}$ and then decreases once again, all the while being $o(1)$ in one form or another.
I was wondering if summing it from $x=1$ to $x=n$ will be $o(1)$ (and if not, how far can I sum it so it will still be $o(1)$). I tried to see what WolframAlpha was suggesting, and it led to this:
An expression involving the Digamma function, which I never came across. I was wondering if anyone has a suggestion as to the original question, or how to deal with what WolframAlpha suggested?
I do not know if this is better.
You can use generalized harmonic numbers and get $$\sum_{i=1}^n \frac{2 x}{m^4 n^{2/3}+\frac{m^2 x^2}{n^{2/3}}}=\frac{n^{2/3}} {m^2}\left(H_{n-i m n^{2/3}}+H_{n+i n^{2/3} m}-H_{-i m n^{2/3}}-H_{i m n^{2/3}}\right)$$