# absolutely convergent series and conditionally convergent series rearrangement

I don't understand that why the terms of an absolutely convergent series can be rearranged in any order and all such rearranged series converge to the same sum.

my textbook gives me an example that it is not so for conditionally convergent series

$1-\frac12+\frac13-\frac14+\ldots=\ln2$
$1+\frac13-\frac12+\frac15+\frac17-\frac14+\frac19+\frac1{11}-\frac16...=\frac32\ln2$

which I also don't know how to prove

• (Regarding your first paragraph) Well, that's a theorem (of Weierstrass). So Step 1 is to find a proof of the theorem and try to understand it. Have you done that already? If not, try $\S$ 14.2.2 of math.uga.edu/~pete/2400full.pdf. – Pete L. Clark Apr 10 '13 at 15:39

Here is something wild: not only do rearrangements of conditionally convergent series give different answers, we can rearrange them to make them equal anything we want!

Let $\sum a_n$ be a conditionally convergent series, and define two new sequences:

$$a^+_n=\frac{a_n+|a_n|}{2}=\begin{cases}a_n&a_n\ge0\\0&a_n<0\end{cases}\\a^-_n=\frac{a_n-|a_n|}{2}=\begin{cases}a_n&a_n\le0\\0&a_n>0\end{cases}$$

By the divergence of $\sum|a_n|$, it follows that both $\sum a^+_n$ and $\sum a_n^-$ diverge, but we have $\lim_{n\to\infty}a_n^+=\lim_{n\to\infty}a_n^-=0$. Now, choose any $r\in\mathbb{R}$. We assume $r\ge 0$, but the proof is similar in the case $r\le 0$.

Without being too explicit, here is how to rearrange the terms of $\sum a_n$ so that the sum of the rearrangement is equal to $r$. Take the first terms to be the first terms of $a_n^+$ until the partial sum is greater than $r$. Then, allow the next terms to come from $a_n^-$ until the partial sum is less than $r$. Continue this process to obtain the desired rearrangement.

Two things we must note. First, the divergence of the positive and negative sums guarantees we will always be able to jump back and forth across $r$. Second, since the limit of the sequences go to zero, the jumping back and forth across $r$ will eventually get more and more precise, so that the limit will approach $r$.

Fascinating.

The first one equals to $$\sum_{i=0}^n (-1)^n(1/i) = \ln 2$$ The second one equals to $$[1+1/2+……+1/4n]-[1/2+1/4+……+1/2n]-[1/2+1/4+……+1/4n] =\ln4n-(1/2)\ln n-(1/2)\ln 2n=(3/2)\ln 2$$