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.


1 Answer 1


Either you are missing some hypotheses and/or context, or the article is blatantly wrong.

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.


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