Sum of little o and big o Consider the following expression:
$$o\left(\frac{1}{nh_n^p}\right) + O\left(\frac{1}{n}\right) \ \text{as} \ n \rightarrow \infty$$
where $p$ is a positive integer and $h_n$ is a function of $n$. We assume that $nh_n \rightarrow \infty$ as $n \rightarrow \infty$ and $h_n \rightarrow 0$ as $n \rightarrow \infty$. I am told that the expression is equal to $$o\left(\frac{1}{nh_n^p}\right)$$
Why is this the case? 
What I did was the following. We know that $O\left(\frac{1}{n}\right) = o(1)$ and so $$o\left(\frac{1}{nh_n^p}\right) + O\left(\frac{1}{n}\right) = o\left(\frac{1}{nh_n^p}\right) + o\left(1\right) = o\left(\frac{1+nh_n^p}{nh_n^p} \right)$$.
 A: This has no definite answer without knowing more about $h_n^p$. Consider the case $h_n^p = \frac{1}{\sqrt n}$, which still allows for $nh_n^p\rightarrow\infty$ for $n\rightarrow\infty$. Then:
$$o\left(\frac{1}{nh_n^p}\right)=o\left(\frac{1}{\sqrt n}\right)$$
So this function grows strictly slower than $n^{-\frac{1}{2}}$, but it may grow like $n^{-\frac{2}{3}}$, for example, which can be faster than a function of $O\left(n^{-1}\right)$ (which may be $\frac{1}{n}$, for example).
On the other hand, if $h_n^p=n$, we have:
$$o\left(\frac{1}{nh_n^p}\right)=o\left(\frac{1}{n^2}\right)$$
This may be a function that grows like $n^{-\frac{5}{2}}$, for example, and thus slower than $O(\frac{1}{n})$, which would then be the dominant term for $n\rightarrow\infty$.
A: Under the assumption that $h_n\to0$ and $p\geq 1$, we have
$$
h_n^p = o(1) 
$$
so that $nh_n^p = o(n)$ and therefore $1/nh_n^p = \omega(1/n)$, i.e.,
$$
\frac{1}{n} = o\!\left(\frac{1}{nh_n^p}\right)\,.
$$
It follows that
$$o\!\left(\frac{1}{nh_n^p}\right)+O\!\left(\frac{1}{n}\right) = o\!\left(\frac{1}{nh_n^p}\right)+o\!\left(\frac{1}{nh_n^p}\right) = o\!\left(\frac{1}{nh_n^p}\right).$$
