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 $$o\left(\frac{1}{nh_n^p}\right) + O\left(\frac{1}{n}\right) = o\left(\frac{1}{nh_n^p}\right) + o\left(1\right) = o\left(\frac{1+nh_n^p}{nh_n^p} \right)$$.

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
$$o\!\left(\frac{1}{nh_n^p}\right)+O\!\left(\frac{1}{n}\right) = o\!\left(\frac{1}{nh_n^p}\right)+o\!\left(\frac{1}{nh_n^p}\right) = o\!\left(\frac{1}{nh_n^p}\right).$$
