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

  • $\begingroup$ What would $a_n^2$ mean if $a_n$ is a vector? $\endgroup$ Dec 17 '16 at 3:41
  • $\begingroup$ ah, good point! I think I delete the question ;) Big thanks @carmichael561 $\endgroup$
    – SAJW
    Dec 17 '16 at 3:41
  • $\begingroup$ Shouldn't matter what field. $\endgroup$
    – IAmNoOne
    Dec 17 '16 at 3:47
  • $\begingroup$ @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. $\endgroup$
    – SAJW
    Dec 17 '16 at 4:54

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.

  • $\begingroup$ 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. $\endgroup$
    – SAJW
    Dec 17 '16 at 4:15
  • $\begingroup$ How are you sure there really are finitely many terms with that lower bound? Doesn't this not work with any series? $\endgroup$
    – Hawk
    Dec 17 '16 at 4:20
  • 1
    $\begingroup$ Dear @Hawk: Note that $a_n \to 0$ as $n \to \infty$. $\endgroup$ Dec 17 '16 at 4:21
  • $\begingroup$ @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? $\endgroup$
    – Hawk
    Dec 17 '16 at 4:22
  • $\begingroup$ @ArpitKansal, there could be zero terms which are $|a_n| > 1$, so where is this $1$ coming from $\endgroup$
    – Hawk
    Dec 17 '16 at 4:23

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 $$ \sum a_n^2 < +\infty \quad \text{but} \quad \sum a_n = +\infty. $$


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