Show that if the series A is absolutely convergent then rearrangement of the series is also convergent and converges to the same limit.
I can't think of any way to solve. Please help
[Math] Series Rearrangement
sequences-and-series
Related Solutions
This is a particular case of the astonishing Riemann Series Theorem . Note that you can rearrange a conditional convergent series in such a way as to make the rearranged series converge to whatever you want.
If you have a finite sum, then rearrangements are trivial. It's when you get into the infinite that you begin to see some strange things happen.
For absolutely convergent series, we get back a series that, again, acts the way we would like. It's only when we go with a rather loose concept of convergence, that is, conditional convergence, that we see strange things happen.
I think the typical example is the alternating harmonic series, $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n}$. In this original arrangement, the sum is $\ln2$. However, we may rearrange it for another sum as follows:
$$\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac{1}{5}-\frac{1}{10}\right)-\frac{1}{12}+\cdots=$$ $$\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots=$$ $$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{2n}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n}=\frac{1}{2}\ln2.$$
What we did was move a few terms over and see that we somehow got half of our original sum. This is the Riemann Rearrangement Theorem. In fact, any conditionally convergent series can be made to become anything, essentially. We can obtain any real number or even make a divergent rearrangement.
Here's a nonrigorous reason for why this is true. Note that a conditionally convergent series is that way because the absolute value of the $n$th term isn't going to 0 fast enough, and so both the positive and negative terms in such a series, on their own, would be infinite (so we can add only positive or only negative to pass any value in a finite number of terms). So, when we rearrange to get some real number $x$, which we will assume is positive without loss in generality (if negative, swap the words positive/above and negative/below in the next bit), we can add the positive terms up until we get to or above $x$, and then we add negative terms until we get below, and continue in that fashion. Since the $n$th term is going to zero eventually, our difference from $x$ gets smaller the further we take this process so that, in the limit, we get $x$.
If we want a divergent sum, we add a bunch of positive values up and then a few negative values, then go back to positive, and return to negative for a short time. I'm being very vague here, but since the positive and negative terms, on their own, diverge (conditional convergence), we can add the terms in such a way that the positive terms are dominating the few negative terms that we put in at a time so that the limit is infinite. We can, of course, do the same the other way around.
The reason that this doesn't work with absolute convergent series is due to the fact that the $n$th term is going to 0 fast enough. We could never add the positive terms to get an arbitrarily large number, and the negative terms that we do add, no matter how cleverly we may think that we're adding them, will be enough to put our sum back in it's place. The $n$th term of an absolutely convergent series is simply too weak to move the series away from it's original sum.
Best Answer
Suppose $A = \sum_{n=1}^{\infty} a_n$, where the convergence is absolute. Let $b : \mathbb{N} \to \mathbb{N}$ be a bijection.
Let $\epsilon > 0$. Since the series converges absolutely to $A$, there exists $N \in \mathbb{N}$ such that $$\left|\sum_{n=1}^{M} a_n - A\right| < \epsilon/2$$ and $$\sum_{n=M}^{\infty} |a_n| < \epsilon/2$$ for all $M \geq N$.
Let $M$ be large enough that $C_N = \{1,2,\ldots,N\} \subseteq \{b(1),b(2),\ldots,b(M)\} = B_M$. Then $$\begin{aligned} \left|\sum_{n=1}^{M} a_{b(n)} - A\right| &= \left|\sum_{n \in B_M} a_{n} - A\right| \\ &= \left|\sum_{n \in C_N} a_n + \sum_{n \in B_M \setminus C_N}a_n - A \right|\\ &\leq \left| \sum_{n \in C_N} a_n - A\right| + \left| \sum_{n \in B_M \setminus C_N}a_n \right| \\ &= \left| \sum_{n=1}^{N} a_n - A \right| + \left| \sum_{n \in B_M \setminus C_N} a_n \right| \\ & \leq \epsilon/2 + \left| \sum_{n \in B_M \setminus C_N} a_n \right| \\ & \leq \epsilon/2 + \sum_{n \in B_M \setminus C_N} |a_n| \\ & \leq \epsilon/2 + \sum_{n=N+1}^{\infty}|a_n| \\ & \leq \epsilon \end{aligned}$$