Proof writing: $\sum_{n=1}^{\infty}| a_n|<\infty $ implies $\sum_{n=1}^{\infty} a_n^2<\infty $.

Question

Let $\sum_{n=1}^{\infty} a_n $ be an absolutely converging series. By definition, this means $\sum_{n=1}^{\infty} \lvert a_n\rvert $ converges. We want to show that $\sum_{n=1}^{\infty} a^2_n $ converges absolutely as well.

Here is what I tried:

Let's say $\sum_{n=1}^{\infty} \lvert a_n^2\rvert $ converges. Let $\epsilon > 0$ be given and there exists $N \in \mathbb{N} $ such that

$\lvert a^2_{m+1}\rvert+ \lvert a^2_{m+2}\rvert + \dots+ \lvert a^2_{n}\rvert< \epsilon$ for all $ n > m \geq N $. By the triangle inequality,

$$\lvert a^2_{m+1} + a^2_{m+2} + \dots + a^2_{n}\rvert \leq \lvert a^2_{m+1}\rvert+ \lvert a^2_{m+2}\rvert + \dots + \lvert a^2_{n}\rvert $$

so the Cauchy Criterion guarantees that $\sum_{n=1}^{\infty} a^2_n $ also converges.

Here is my question:

Is this correct, or do I need to prove anything else?

"Let's say $\sum |a_n^2|$ converges" this is what you want to prove, you can't start with this. — ThePortakal, Jun 07 '16 at 23:25
Possible duplicate of If $\sum_{n=1}^{\infty} a_n$ is absolutely convergent, then $\sum_{n=1}^{\infty} (a_n)^2$ is convergent — Clement C., Jun 07 '16 at 23:33
@Clement: No, it’s not a duplicate, because it asks specifically whether the argument given is correct. — Brian M. Scott, Jun 07 '16 at 23:35
@BrianM.Scott My bad. Clicked before actually thinking. (I retracted the closing vote.) — Clement C., Jun 07 '16 at 23:36
Apparently someone would prefer that the OP’s confusion not be remedied, since the question and all three answers have been downvoted. — Brian M. Scott, Jun 07 '16 at 23:36
Yes, downvoting seems to be a very contagious disease these days... — Clement C., Jun 07 '16 at 23:36
I've edited the question, could someone take a look at it now? — skancharla, Jun 07 '16 at 23:58
I'm still a little confused as to why $|a^2_n| \leq |a_n|$. It seems intuitive to me that it would be the opposite. Could someone explain this part? — skancharla, Jun 07 '16 at 23:59
(1): the last inequality (the first) is wrong, it should be $0 \leq a_n^2 < \lvert a_n\rvert$ (for $n\geq N$). (the reason is: when $x\in[0,1]$, we have $0\leq x^2 = x\cdot x \leq 1\cdot x = x$). (2) I am not entirely sure about the policy, but completely changing your question after it got several answers seems slightly off to me. Instead, you could for instance have added the new content after the previous one, prefixing it with an "Edit" notice? — Clement C., Jun 08 '16 at 00:00
Regarding $|a_n^2| \leq |a_n|$. There are two kinds of $|a_n|$: those that are less then $1$ and the others. How can you show there are only finitely many of "the others"? — Eric Towers, Jun 08 '16 at 04:32

score 3 · Answer 1 · answered Jun 07 '16 at 23:27

No, this is not correct: you’ve proved that if $\sum_{n\ge 1}a_n^2$ converges absolutely, then it converges, which is not at all what you’re supposed to be doing. You have assumed precisely the result that you are supposed to be proving, and you’ve ignored the hypothesis.

You should be assuming that $\sum_{n\ge 1}a_n$ converges absolutely and using that assumption to prove that $\sum_{n\ge 1}a_n^2$ converges absolutely.

HINT: If $\sum_{n\ge 1}a_n$ converges absolutely, then there is an $m\in\Bbb N$ such that $|a_n|<1$ for each $n\ge m$ (why?), and hence $0\le|a_n^2|<|a_n|$ for each $n\ge m$.

Actually addressing the question... (+1) – Clement C. Jun 07 '16 at 23:37 — Clement C., Jun 07 '16 at 23:37

score 2 · Answer 2 · edited Oct 10 '17 at 13:06

2

Hint: Since $(a_n)$ converges absolutely, $\lim_n\mid a_n\mid=0$. This implies there exists $N$ such that $n>N$ implies that $\mid a_n\mid <1$. This implies that for $n>N$, $\mid a^2_n\mid<\mid a_n\mid$.

edited Oct 10 '17 at 13:06

answered Jun 07 '16 at 23:24

Tsemo Aristide

87,475

The hint is valid, but the question is "Is this correct, or do I need to prove anything else?" not "give me a way to do it." (emphasis mine) – Clement C. Jun 07 '16 at 23:39
hello can you explain how did you get $|a_n|<1$? did you use the geometric criterion? thnk you. – Elina Jun 24 '16 at 21:21
@Elina Take $\epsilon=1/2$. Since $|a_n| \to 0$, there exists an $m\in \mathbb{N}$ such that beyond this $m$ all $a_n$ lie in the $\epsilon=1/2$ interval of $0$ and hence $||a_n|-0|<1/2\rightarrow|a_n|<1$ – ZSMJ Nov 11 '18 at 07:08

score 0 · Answer 3 · answered Jun 08 '16 at 00:50

Brian has already pointed out what is wrong with your reasoning. Here is another approach.

Note that if $\sum_{n=1}^\infty|a_n|<\infty$, then $$\lim_{n \to \infty} \frac{a^2_n}{|a_n|} = \lim_{n \to \infty} |a_n| = 0.$$

Then take a look at the limit comparison test.

[Might not be useful for now though] In real analysis, this is nothing but a simple case of a more general theorem (see for instance Folland's Real Analysis, Chapter 6) $$ \ell^p(\mathbb{N})\subset\ell^q(\mathbb{N}) $$ for $0<p<q\leq\infty$.

Proof writing: $\sum_{n=1}^{\infty}| a_n|<\infty $ implies $\sum_{n=1}^{\infty} a_n^2<\infty $.

3 Answers3