Introduce the asymptotic notations $O$ and $o$. Let $a_n=(nh)^{-1}$ and $b_n=h^2$ be sequences and assume $h:=h_n\to 0, nh\to\infty$ as $n\to\infty$. In my problem, I was left with a reminder $O(a_n)+o(b_n)$, i.e., $O((nh)^{-1})+o(h^2)$. My hypothesis is that $O(a_n)+o(b_n)=o(b_n)$.
My argument
$$\frac{h^2}{1/(nh)}=\frac{n}{h}\to\infty\implies \frac{a_n}{b_n}\to 0.$$ In other words, $a_n=o(b_n)$ and $a_n$ goes faster than $b_n$ to zero (thus, is dominated by $b_n$). So, a sequence of order $O(a_n)$ is certainly $o(b_n)$ (see, $O(a_n)=O(o(b_n))=o(b_n)$), but there could be a sequence in the set $o(b_n)$ which is not $O((nh)^{-1})$ (i.e., that goes to zero slower than any sequence in $O((nh)^{-1})$). Thinking $o$ and $O$ as sets of sequences, I conclude that $O((nh^{-1}))\subseteq o(h^2)$. It means that $O(a_n)+o(b_n)=o(b_n)+o(b_n)=o(b_n).$
Can you give me feedbacks?
Thanks you in advance.
We have that
$$O\left(\frac1{nh}\right)+o(h^2)=h^2\cdot \left(O\left(\frac1{nh^3}\right)+o(1)\right)\neq o(h^2)$$
we can only infer that
$$O\left(\frac1{nh}\right)+o(h^2)= o(1)=O\left(\frac1{nh}\right) $$