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Is it true that every real sequence that converges to zero has the property that every rearrangement of it also converges to zero?

I have a proof in mind, but I'm not 100% sure it's correct (although I'm pretty sure), so I just want a yes/no answer.

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3 Answers 3

up vote 6 down vote accepted

Yes. A sequence converges to $0$ if and only if all but finitely many terms in it are less than $\epsilon$ in absolute value for any $\epsilon > 0$, and this condition is manifestly invariant under rearrangement.

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Well thats more than a "yes", but thats pretty much the proof I also had in my mind :) –  user38523 Aug 23 '12 at 9:08

I should think so. Given a real sequence $\{a_{n}\} \to 0$ and an arbitrary $\epsilon > 0$, we have that there exists some $N$ such that for all $n > N$, $|a_{n}| < \epsilon$. That being said, let $\sigma: \mathbb{N} \to \mathbb{N}$ be a permutation, and consider the rearranged sequence $\{a_{\sigma(n)}\}$. Since the convergence of the initial sequence implies that all but finitely many elements in the sequence can be more then $\epsilon$ away from $0$, after a finite number $N'$ of terms in this new, rearranged sequence, we will have exhausted all those cases, and for $n > N'$ we will have $|a_{\sigma(n)}| < \epsilon$. Note that this works with $0$ replaced by an arbitrary finite limit $L$.

Note that this fails for series, as a result of the Riemann rearrangement theorem.

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To extend the answer by Qiaochu.

Well, the limit does not matter provided it is finite - let me show it for the case $\lim x_n = 0$. You know that for any $\delta>0$ there is $N(\delta)$ such that $|x_n|<\delta$ for all $n>N(\delta).$

Let $\hat x_n$ be some rearrangement of $x_n$ and suppose that it does not converge to zero, i.e. there exists $\delta$ such that for any $m$ there is $n(m)>m$ and such that $|\hat x_{n(m)}|>\delta$. But that means that there are infinitely many such numbers $n(m)$, hence the convergence of $x_n$ does not hold.

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