# Divergent and convergent series of positive terms

I recently read an article which contains the following facts :---

Let $$\{a_n\}$$ be a sequence of positive number such that $$\sum_{n=1}^{\infty} a_n$$ diverges, so we must have $$a_n\sim \frac{1}{n^p}$$ where $$p\leq 1$$. Here $$\sim$$ means nearby, for instance $$\frac{k}{n^p}$$ for some constant $$k$$.

Another fact is that , let $$\sum_{n=1}^{\infty} b_n$$ be a series of positive terms and $$\sum_{n=1}^{\infty} b_n$$ is convergent, that means $$a_n=O(\frac{1}{n^p})$$ for some $$p>1$$, where $$O$$ means order.

I cannot understand how do I prove these facts rigoursly, even I don't know whether they are true or not.

Take for instance $$a_n=2^{-n}$$ if $$n\neq 2^{p^2}$$ for any integer $$p$$ and $$a_{2^{p^2}}=\frac{1}{p}$$. The sum of $$a_n$$ is divergent, but there is no real number $$a>0$$ such that $$a_n=O(n^{-a})$$ or $$n^{-a} = O(a_n)$$ (however $$a_n$$ does go to $$0$$).
Take now the same sequence, except that $$a_{2^{p^2}}=p^{-2}$$. Then there is no real number $$a>0$$ such that $$a_n=O(n^{-a})$$ but the sum of the $$a_n$$ converges.