# Rearrangement of terms Can someone explain the last sentence to me?

It says we get a contradiction if only one of the sums have an unbounded sum. I don't see an issue.

$$\sum a_n = \sum (p_n + q_n)$$

$$\sum |a_n| = \sum (p_n - q_n)$$

If both of the sums $\sum p_n$ and $\sum q_n$ diverge, this still makes the whole sum $\sum a_n$ diverge, yielding another contradiction.

• If the series doesn't converge absolutely, you can't split it into the sum of positive and the sum of negative terms, so you can't write $\sum a_n = \sum p_n + \sum q_n$ then (that would be an indeterminate form $\infty - \infty$). – Daniel Fischer Aug 18 '13 at 21:52
• Suppose I have the edited sum. I still can't see the problem if both have unbounded partial sums. – Hawk Aug 18 '13 at 21:55
• Your edit doesn't solve the problem -- you are rearranging terms, while at the same time trying to prove that rearrangements can give any real number. Therefore, you still have $\sum a_n\neq \sum(p_n+q_n)$. – M Turgeon Aug 18 '13 at 21:57
• Your last sentence is false, and it is impossible to understand conditional convergence as long as you think that is true. The fact that $\sum_np_n$ and $\sum_nq_n$ can both diverge to $\infty$ even though $\sum_n(p_n-q_n)$ converges is the point of the concept. – Michael Hardy Aug 18 '13 at 23:21
• So can't that happen too if one of them is bounded? – Hawk Aug 19 '13 at 2:09

Note that $\sum p_n$ diverges to $\infty$ but $\sum q_n$ diverges to $-\infty$ in the case where both diverge. An expression of the form $\infty - \infty$ (to speak loosely) doesn't necessarily diverge.
Example : $1 - \frac{1}{2} + \frac{1}{3}-\frac{1}{4}+ \cdots$ converges to $\ln 2$. But the the sum of positive terms and negative terms diverge to $\infty$ and $-\infty$ respectively.
• But then aren't you still "adding" the two divergent sums together (with rearrangement) to get $\ln 2$ – Hawk Aug 18 '13 at 22:02
Basically we have $a_n = p_n + q_n$ and $|a_n| = p_n - q_n$.
Writing it as $q_n = a_n - p_n$. If $\sum q_n$ diverges, one of $\sum p_n$ or $\sum a_n$ must diverge.
If $\sum q_n$ converges, both $\sum a_n$ and $\sum p_n$ must converge. But if $\sum a_n$ converges (which we know it is) and $\sum q_n$ diverges, this leaves only $\sum p_n$ to diverge.