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

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.

• Your proof is fine and quite professional too! Commented Feb 5, 2022 at 12:06
• ($A$ if $B$) means $B\implies A$. Commented Feb 5, 2022 at 12:14
• Personally, I don't really like the inequality $\sum_{n=n_0+1}^{\infty}a_n^2 < \sum_{n=n_0+1}^{\infty}|a_n|$ because that is equivalent to $+\infty < \sum_{n=n_0+1}^{\infty}|a_n|$. There is a small flaw with the strict inequality and it doesn't say which theorem you are using on sequences or series. Commented Feb 5, 2022 at 12:33
• How may I make it better then? What kind of theorem did I use? Commented Feb 5, 2022 at 12:39
• Here you have at least two possibilities: 1) $\forall n>n_0, 0\leq a_n^2 <|a_n|$, so from the comparison test of series with non-negative terms ... 2) $\forall N>n_0$, $\sum_{n=n_0+1}^{N}a_n^2 < \sum_{n=n_0+1}^{N}|a_n|$ and $\lim_{N\rightarrow +\infty}\sum_{n=n_0+1}^{N}a_n^2 =+\infty$, we deduce ... Commented Feb 5, 2022 at 12:49

To answer your first question, $$A$$ if $$B$$ means $$A\impliedby B$$, i.e. $$B\implies A$$.
$$\sum_{n=n_0+1}^N a_n^2 < \sum_{n=n_0+1}^N \left| a_n\right|.$$
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).