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.
 A: 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 $|b_n-a|\lt \epsilon$. 
A: 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.
A: 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 $a$.
