1

I am pretty sure that

Absolute convergence of $\sum_{n=0}^{\infty}a_n\implies\sum_{n=0}^{\infty}a_n^2$ is absolute convergent too

is a true statement, but before I proof it I ask wether this exercise is probably meant for series in $\mathbb{R}$ respectivly $\mathbb{C}$ or is it meant for ANY series. Like series of vectors in an unknown vectorspace.

Could someone clarify me?

I don't know what I mumble so here is what I stumbled on, while searching for definitions of absolute convergence: https://proofwiki.org/wiki/Definition:Absolute_Convergence/General

SAJW
  • 1,327
  • What would $a_n^2$ mean if $a_n$ is a vector? – carmichael561 Dec 17 '16 at 03:41
  • ah, good point! I think I delete the question ;) Big thanks @carmichael561 – SAJW Dec 17 '16 at 03:41
  • Shouldn't matter what field. – IAmNoOne Dec 17 '16 at 03:47
  • @Gaffney I indeed looked both topics before I made mine. My question tho is about the validity area of the statement, not the proof itself. My understanding problem was that $|a|$ is always a real number, no matter which field $a$ belongs to. Note that this only became clear for me after some discussion. – SAJW Dec 17 '16 at 04:54

2 Answers2

15

Since $\sum_n a_n$ converges absolutely, only finitely many $|a_n| > 1$. Hence, except for these finitely many terms, $|a_n^2| = |a_n|^2 \le |a_n|$, and thus $\sum_n a_n^2$ converges absolutely. Note this solution works over $\mathbb{R}$, $\mathbb{C}$, and I guess any normed algebra.

mathworker21
  • 34,399
  • Now, bear with me. What would |apple| mean where apple is an element of a field of fruits? |z| with z in C means some real number. But I don't understand why we can argue with inequalities in any field, and therefore not ordered fields. – SAJW Dec 17 '16 at 04:15
  • How are you sure there really are finitely many terms with that lower bound? Doesn't this not work with any series? – Lemon Dec 17 '16 at 04:20
  • 1
    Dear @Hawk: Note that $a_n \to 0$ as $n \to \infty$. – Arpit Kansal Dec 17 '16 at 04:21
  • @ArpitKansal, yes so for $n \geq N$, he has $|a_n| < \epsilon$, but how does he know for $n < N$, the lower bound of $1$ works? – Lemon Dec 17 '16 at 04:22
  • @ArpitKansal, there could be zero terms which are $|a_n| > 1$, so where is this $1$ coming from – Lemon Dec 17 '16 at 04:23
  • @Null : a norm has codomain $\mathbb{R}$ – mathworker21 Dec 17 '16 at 04:23
  • @Hawk: Suppose conclusion does not hold,then $a_n$ can't go to zero as $n \to \infty$ – Arpit Kansal Dec 17 '16 at 04:24
  • @Hawk 1 is arbitrary. As you said, $|a_n| < \epsilon$ past some $N$, so there are only finitely many possible terms which have absolute value $> 1$. – mathworker21 Dec 17 '16 at 04:24
  • Dear @mathworker21: Your solution is neat,note that 'similar' result hold in more generality.For instance you can find it here and here – Arpit Kansal Dec 17 '16 at 04:27
  • @mathworker21 But there could be zero terms that satisfy $|a_n| > 1$? That is included in the word "finite" right? – Lemon Dec 17 '16 at 04:29
  • @mathworker21, I mean if we pick $a_n \in [0,1/2]$ for all $n$, that condition can't hold right? That's what i mean – Lemon Dec 17 '16 at 04:31
  • @Hawk, correct. that is included in finite. – mathworker21 Dec 17 '16 at 04:34
  • @mathworker21, thanks sorry if my comment is red-herring. – Lemon Dec 17 '16 at 04:39
0

Consider a series on a normed vector space $(V,\|\cdot\|)$ where for every $a\in V$ is well defined $a^2$ with $a^2\in V$ and $\|a^2\| = {\|a\|}^2$. The series $\sum a_n = {\left\{ \sum_{n=1}^k a_n \right\}}_{k\ge 1}$ is absolute convergent if $$ \sum \|a_n\| = \lim_{k\to \infty} \sum_{n=1}^k \|a_n\| < +\infty. $$ Let us now turn to the statement made. Since the series $\sum a_n$ converge absolute, then only a finite number of terms $a_n$ have the property $ \|a_n\|>1$, otherwise there would be infinite terms $a_n$ with $\|a_n\|>1$, then $\sum_{\|a_n\|>1} 1 = +\infty$ and follows $$ \sum \|a_n\| = \sum_{n\colon \|a_n\|>1} \|a_n\| + \sum_{n\colon \|a_n\|\le 1} \|a_n\| > \sum_{n\colon \|a_n\|>1} 1 + \sum_{n\colon \|a_n\|\le 1} 0 \ge +\infty $$ which is a contradiction. Then, it follows \begin{align*} \sum \|a_n^2\| = \sum {\|a_n\|}^2 &= \sum_{n\colon \|a_n\|>1} {\|a_n\|}^2 + \sum_{n\colon \|a_n\|\le 1} {\|a_n\|}^2\\ &\le \sum_{n\colon \|a_n\|>1} {\|a_n\|}^2 + \sum_{n\colon \|a_n\|\le 1} \|a_n\| \\ &\le \sum_{n\colon \|a_n\|>1} {\|a_n\|}^2 + \sum \|a_n\| < +\infty, \end{align*} since $\sum_{n\colon \|a_n\|>1} {\|a_n\|}^2$ is a finite sum, ${\|a_n\|}^2 \le \|a_n\|$ si $\|a_n\|\le 1$ and $\sum_{n\colon \|a_n\|\le 1} \|a_n\| \le \sum \|a_n\|$. This it, $\sum a_n^2$ is absolute convergent.\ Posdata: I use notation $$ n\colon \|a_n\|\le 1 = \{n\in \mathbb{N}\colon \|a_n\|\le 1\}. $$ Also, review this example, if $a_n=\frac{1}{n},~n\ge 1$, then $$ \sum a_n^2 < +\infty \quad \text{but} \quad \sum a_n = +\infty. $$

Daniel Camarena Perez
  • 730