Sum of Little o(1/n) Terms

Question

I have a question about the little-oh notation: If every $h_i=o(1/n)$, then consider $ \sum_i^n h_i$. Is the sum $o(1)$?

Currently, I am trying to derive it by definition, but I am quite unsure about exchanging limit and sum:

$\lim_{n\rightarrow \infty} \frac{|\sum_i^n h_i|}{1/n}\leq \lim_{n\rightarrow \infty} \frac{\sum_i^n |h_i|}{1/n}=\sum_i^n \lim_{n\rightarrow \infty} \frac{| h_i|}{1/n}=\sum_i^n0=0$

Is the above derivation right? How can I correct it?

Thank you!

======================= Thank you very much for the counter-example.

Answer 1

Without knowing anything about how $h_i$ depends on $i$, we can't say anything. For example, if $h_i = \frac{i^2}{n^2}$, we'd say that $h_i = o(\frac1n)$, however, $$ \sum_{i=1}^n h_i = \frac1{n^2} \sum_{i=1}^n i^2 = \frac{(n+1)(2n+1)}{6n} = \Theta(n). $$

Answer 2

You can't interchange the limit and sum, because the summation goes up to $n$ and that is not defined if the summation is outside the limit.

Consider now $h_i=i^2/n^2$. Then $h_i=o(1/n)$, but $\sum_{i=n}^n h_i\to \infty$.