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Let $\sum\limits_{n=0}^\infty a_n$ be absolutely convergent series in $\mathbb{R}^d$, for $d\ge 1$. Let $\mathbb{N}\cup \{0\} = A\cup B$, with $A\cap B=\emptyset$. Prove that $\sum\limits_{n=0}^\infty a_n = \sum\limits_{n\in A} a_n+\sum\limits_{n\in B} a_n$.

Proof:

$\sum\limits_{n\in A} \| a_n\| \le \sum\limits_{\min\limits_{n\in A}\{n\}}^\infty \|a_n\|\le \sum\limits_{n=0}^\infty \|a_n\|<\infty$, so this series converges. Similarly, $\sum\limits_{n\in B} \| a_n\|$ also converges.

Now, $\sum\limits_{n \in A} a_n+ \sum\limits_{n \in B}a_n = \sum\limits_{n \in A\cup B} a_n = \sum\limits_{n =0}^\infty a_n$.

I think my proof is not completely formalized. Can someone help me with that please? I'm new to this kind of work with series.

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    This is just the proposition that shuffling the terms of an absolutely convergent series doesn't change the sum. You have to completely make it clear where you are using absolute convergence. If you do, then I think this is fine. – 2017-02-16
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    @астонвіллаолофмэллбэрг Unfortunately, I'm not sure how to use absolute convergence here. – 2017-02-16

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The point is very simple. I'll illustrate it with an example.

We know that if $a_n = \frac{(-1)^n}{n}$, then $\displaystyle\sum_{n=1}^{\infty} a_n$ convergences (you can take $a_0=0$ if you like).

Now, suppose I take out the odd elements from this: these are forming the series $-1 - \frac 13 -\frac 15 - \frac 17 -\ldots$ which doesn't converge.

If I remove the even terms out, I'm getting $0 + \frac 12 + \frac 14 + \ldots$, which again, doesn't converge.

So the theorem does not apply for this situation, and the reason is very simple: I broke up the series into two subseries which do not converge. What if I took $A$ and $B$ similarly in the above question? I would be stuck, right?

Now think like this: what condition on $a_n$ prevents any subsequence of a convergent series from flying off?

The answer is: If the series is absolutely convergent. The reason is simple: If $A$ is a subset of $\mathbb N \cup \{ 0 \}$, then: $$\left|\sum_{n \in A} a_n\right| \leq \sum_{n \in A} |a_n| \leq \sum_{n = 0}^\infty |a_n| < \infty$$.

Hence, all you need to add to your answer is : by absolute convergence, $\sum_{n \in A} a_n$ and $\sum_{n \in B} a_n$ exist. Hence, since $A \cup B = \mathbb N \cup 0$, the conclusion follows.

This really is the crux of absolute convergence. I hope you have understood!

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    Yes, I have shown that both subseries converge absolutely. I have also done the other step you mentioned. That is, I said: $\sum\limits_{n \in A} a_n+ \sum\limits_{n \in B}a_n = \sum\limits_{n \in A\cup B} a_n = \sum\limits_{n =0}^\infty a_n$. It seems I did everything exactly as you said. But what is missing in my proof? – 2017-02-16
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    @sequence Yes, I'm just *completely* formalizing your proof, I saw the part, but one needs to mention the "term" absolute convergence. I'm telling you this because once I was reprimanded for not mentioning this in the exact same proof about a year ago. That's why. It may seem superfluous, though. At least, I tried to provide a counterexample to show why absolute convergence is necessary. Forgive me, I think I was a little doubtful when you said " I don't get where we used absolute convergence". – 2017-02-16
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    I see. There appears to be a peculiar point of justifying that $\sum_{n \in A} a_n + \sum_{n \in B} a_n = \sum_{n \in A\cup B} a_n$. – 2017-02-16
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    @sequence That's right. I just wanted to seal it with my answer. – 2017-02-16