Applying squeeze theorem to conditionally convergent series Suppose that the series ∑_n≥1(a_n) converges conditionally. Then by the Riemann Series theorem, for any real number L there exists a rearrangement of a_n(let's call it b_n) that converges to L. For a finite sum, ∑(a_n)=∑(b_n), for any rearrangement b_n. But if I apply the squeeze theorem, since I have that ∑(a_n)≤∑(b_n)≤∑(a_n), and ∑(a_n) converges to a number A, shouldn't I have ∑(b_n) converging to A? Doesn't this mean that the infinite sum of any rearrangent b_n converges to A. What did I do wrong?
 A: How did you know that for a finite sum $\sum a_n=\sum b_n$? To see why this is wrong in general, consider $a_n=\frac{1}{n^2}$ from $n=1$ to $n=10$. Now, we have:
$$\sum_1^5 a_n=\frac{5269}{3600}$$
Consider $b_n$ where the first 5 terms and the others right after them are swapped, so we get
$$\sum_1^5 b_n=\sum_6^{10} a_n=\frac{547129}{6350400}$$
Thus, your initial step was wrong. Done.
Edit: 
After seeing your comment, I shall reflect on what you said.
Technically, yes, in the case where a rearrangement is as you said, then you are correct. It will converge to $A$. The proof is way simpler by just saying that $|\sum_1^N a_n - A|=|\sum_1^N b_n - A|$ since $\sum_1^Na_n=\sum_1^N b_n$ since all you did was rearrange the terms before $N$. 
However, I must note that this isn't exactly what was done in the proof of the Riemann series theorem. As you can see in this proof, the rearrangement is done while choosing terms arbitrarily, since the first $N$ terms can be negative while we are allowed to pick the positive terms which might be starting from $N+1$ for example, all of course for some $N$.
You might have gotten the definition of a rearrangement wrong as it isn't what you described in your comment. All it means is that there exists a bijective mapping between $a_n$ and $b_n$ terms. That is, there is a bijection $\sigma: \mathbb{N}\to\mathbb{N}$ such that $b_n=a_{\sigma (n)}$ 
I hope this helps.
