Check: $\sum_{n=1}^{\infty}a_n$ converges, $\sum_{n=1}^{\infty}a^2_n$ diverges, then $\sum_{n=1}^{\infty}|a_n|$ diverges

Question

Please do check my proof, and in case, tell me how to make it more professional.

Additionally I'm not sure if the statement ($A$ if $B$) means $A\implies B$ or $B\implies A$.

Problem:

Let $\sum_{n=1}^{\infty}a_n$ be a convergent series. Show that $\sum_{n=1}^{\infty}|a_n|$ diverges if $\sum_{n=1}^{\infty}a^2_n$ diverges.

My Solution:

Since $\sum_{n=1}^{\infty}a_n$ converges, it must be true that $\lim_{n\to \infty}a_n=0.$ Then for some $n_0$: for all $n>n_0$, $|a_n|<1$.

Then it is true that $a^2_n<|a_n|$ for all $n>n_0$.

Hence $\sum_{n=n_0+1}^{\infty}a_n^2 < \sum_{n=n_0+1}^{\infty}|a_n| \implies \sum_{n=n_0+1}^{\infty}|a_n|$ diverges.

Answer 1

To answer your first question, $A$ if $B$ means $A\impliedby B$, i.e. $B\implies A$.

Secondly, your proof looks fine, and your idea is great. Here, however, it would probably also be a good idea to include what theorem you used for the comparison at the end, i.e. the comparison test for positive series. The main reason here is that your inequality is not completely justified in it's own right. So consider

$$\sum_{n=n_0+1}^N a_n^2 < \sum_{n=n_0+1}^N \left| a_n\right|.$$

This is all fine because you're dealing with finite sums. However if you want to take this to the limit as $N\to\infty$, this strict inequality becomes nonstrict, as the two could potentially be equal in the limit (and in fact here they are equal, as both are infinite).