a question about order estimation

Question

Does $$o(\frac{1}{n^{2}})＋\cdots＋o(\frac{n}{n^{2}})＝o(1)$$hold？

I know that when there is only finite terms the addition is valid. But in this situation, I don't know whether I can use the addition of limit. Hope for your hint.

Answer 1

To address the OP's original problem, if $f$ is differentiable at $0$, there is some $\epsilon:x\mapsto \epsilon(x)$ such that $f(x) = f(0)+xf'(0)+x\epsilon(x)$, $\epsilon(0)=0$ and $\epsilon$ is continuous at $0$.

Assuming $f(0)=0$, $\sum_{k=1}^n f(\frac{k}{n^2})=f'(0)\frac 1n\sum_{k=1}^n \frac{k}{n} + \sum_{k=1}^n \frac{k}{n^2}\epsilon(\frac{k}{n^2})$

Consider $\gamma >0$. There exists some $\delta >0$ such that $|x|\leq \delta \implies |\epsilon(x)|\leq \gamma $. For $n$ larger than $\frac{1}{\delta} +1$, $$\left| \sum_{k=1}^n \frac{k}{n^2}\epsilon(\frac{k}{n^2}) \right|\leq \gamma \frac 1n n = \gamma$$

Thus $\lim_n \sum_{k=1}^n \frac{k}{n^2}\epsilon(\frac{k}{n^2}) = 0$

Since $\frac 1n\sum_{k=1}^n \frac{k}{n}$ is a Riemann sum that converges to $\frac 12$, $\sum_{k=1}^n f(\frac{k}{n^2})$ converges to $\frac{f'(0)}{2}$.

Answer 2

No it does't hold in general since

$$f(n)=o\left(\frac{1}{n^{2}}\right)＋\cdots＋o\left(\frac{n}{n^{2}}\right)\iff f(n)=\omega_1 \left(\frac{1}{n}\right) \frac{1}{n^{2}} ＋ \cdots ＋\omega_n \left(\frac{1}{n}\right) \frac{1}{n}$$

and we don't have information about $\omega_i \left(\frac{i}{n}\right)\to 0$.