Can rearranging a SEQUENCE (not a series) change the limit?

I have this question on a homework assignment. I sat down with two other people for a long time and we derived the alternating harmonic series example, but I don't think that's valid because the question explicitly asks about sequences and not series.

Note that it's for an analysis class and so far we've covered open and closed sets and balls, preimages, and cluster points.

Note that if you view a sequence as a set (which means forgetting the ordering altogether), the limit of the sequence becomes the unique cluster point of the set (assuming the sequence converges). Since the set does not depend on the order of the sequence, the cluster point and hence the limit does not either.

• Technically, a cluster point of a sequence $(a_n)$ need not be a cluster point of the set $\{a_n\}$ if the point itself occurs infinitely many times in the sequence. (Consider e.g. the case $a_n = 0$ for all $n$.) But it's easy enough to fix that gap, either by treating it as a special case, or just by considering cluster points of the sequence itself rather than of the corresponding set, which are still independent of the order. Commented Sep 14, 2013 at 11:22
• @IlmariKaronen It depends on whether by "cluster point" one means "limit point" or "accumulation point". I wasn't sure, since I hadn't heard the term before. Commented Sep 14, 2013 at 18:59
• I think this is what my professor was looking for. It doesn't involve $\epsilon$-$\delta$ and it directly applies what we've done in class. Thanks! Commented Sep 14, 2013 at 20:04

No, the behavior of the sequence does not change when rearranging its terms. As an example, let's prove that if the sequence $(a_n)_n$ has limit $a$ then every rearranging of it converges and to the same limit $a$.

Consider a rearranging of $(a_n)_n$, that is, a sequence $\left(a_{\phi(n)}\right)_n$ where $\phi:\mathbb N \rightarrow \mathbb N$ is a bijection.

Fix an arbitrary $\epsilon > 0$. As $(a_n)_n$ is convergent to $a$, there exists $n_0 \in \mathbb N$ such that for every $n \geq n_0$ we have $|a_n - a | < \epsilon$. Now, you know that the only terms in $(a_n)_n$ that possibly violate the condition $|a_n - a | < \epsilon$ are in the set $\{ a_1, a_2,\ldots , a_{n_0} \}$ so taking $n_1 = \max \{ n : 1\leq \phi(n) \leq n_0 \} + 1$ will be enough to have $|a_{\phi(n)} - a | < \epsilon$ for every $n \geq n_1$. As $\epsilon$ was arbitrary this proves that $\left(a_{\phi(n)}\right)_n$ also converges to $a$.

• Well, I sarted writing before André's answer and finished editing now... so I guess that mine will be an unnecesary one Commented Sep 14, 2013 at 3:13
• Although the ideas are the same, the procedure you used was substantially more formal, so the two versions may be informative to readers. Commented Sep 14, 2013 at 3:21

Let us make a standard $\epsilon$-$N$ argument.

If $(a_n)$ is the original sequence, then given $\epsilon\gt 0$, there is an $N$ such that if $N\lt n$ then $|a_n-a|\lt \epsilon$.

Now consider the permuted sequence $(b_n)$, where $b_n=a_{\pi(n)}$, for some permutation $\pi$ of $\mathbb{N}$. Use the same $\epsilon$.

Let $N^\ast$ be the maximum of the $\pi(i)$, as $i$ ranges from $1$ to $N$. If $n\gt N^\ast$, then $|b_n-a|\lt \epsilon$.