重排绝对收敛级数

Remember that a series $\sum a_n$ absolutely converges if the sum $\sum |a_n|$ converges. As an example, let's look at a the geometric series formed by $a_n=\frac{1}{2^n}$, which we know converges absolutely.

By the infinite geometric series formula, $$\sum_{n=0}^{\infty} \frac{1}{2^n}=1+\frac{1}{2}+\frac{1}{4}+\cdots=\frac{1}{1-\frac{1}{2}}=2.$$ Now, what happens if we rearrange the terms? Well, one convenient way to rearrange this series would be $$\sum_{n=0}^{\infty} \frac{1}{2^n}=\left(1+\frac{1}{4}+\frac{1}{16}+\cdots\right)+\left(\frac{1}{2}+\frac{1}{8}+\cdots\right)=\sum_{n=0}^{\infty} \frac{1}{2^{2n}}+\sum_{n=0}^{\infty} \frac{1}{2^{2n+1}}.$$

Note that for both sums, the geometric ratio is $\frac{1}{4}$. So by using the infinite geometric sequence formula, $$\sum_{n=0}^{\infty} \frac{1}{2^{2n}}+\sum_{n=0}^{\infty} \frac{1}{2^{2n+1}}=\frac{1}{1-\frac{1}{4}}+\frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3}+\frac{2}{3}=2.$$

Hm, is this a coincidence? Let's change the series around one more time: $$\sum_{n=0}^{\infty} \frac{1}{2^n}=\left(1+\frac{1}{8}+\frac{1}{64}+\cdots\right)+\left(\frac{1}{2}+\frac{1}{16}+\cdots\right)+\left(\frac{1}{4}+\frac{1}{32}+\cdots\right)$$ $$=\sum_{n=0}^{\infty} \frac{1}{2^{3n}}+\sum_{n=0}^{\infty} \frac{1}{2^{3n+1}}+\sum_{n=0}^{\infty} \frac{1}{2^{3n+2}}.$$

This time, the geometric ratio for each sum is $\frac{1}{8}$, and so we have $$\sum_{n=0}^{\infty} \frac{1}{2^{3n}}+\sum_{n=0}^{\infty} \frac{1}{2^{3n+1}}+\sum_{n=0}^{\infty} \frac{1}{2^{3n+2}}\\=\frac{1}{1-\frac{1}{8}}+\frac{\frac{1}{2}}{1-\frac{1}{8}}+\frac{\frac{1}{4}}{1-\frac{1}{8}}=\frac{8}{7}+\frac{4}{7}+\frac{2}{7}=2.$$ Wow! The sum comes out to $2$ again! Do you think this is just a coincidence?

Nope! In fact, here we'll prove the following theorem:

If the sum $\sum a_n$ converges absolutely and equals $S$, then any rearrangement of $\sum a_n$ will also converge absolutely and equal $S$.

To prove this, suppose $f: \mathbb{N}\to\mathbb{N}$ is a bijection. We want to prove that $\sum a_{f(n)}$ converges absolutely and equals $S$.

Define the following terms: $$s_n=\sum_{i=0}^n a_i\text{ and } t_n=\sum_{i=1}^n a_{f(i)}.$$ We know by the Triangle Inequality that for any $m,n\in\mathbb{N}$, $$|t_m-S|=|t_m-s_n+s_n-S|\le |t_m-s_n|+|s_n-S|.$$ Since we're given that $\sum s_n$ converges to $S$, we know that we can make the quantity $|s_n-S|$ arbitrarily small, by finding a large enough value of $n$.

So, fix a positive quantity $\epsilon>0$. Then we know that there exists $N_1$ such that for any $n\ge N_1$, $$|s_n-S| < \frac{\epsilon}{2}.$$ Moreover, since $\sum a_n$ converges absolutely, there exists $N_2$ such that for any $c\ge N_2$, $$\sum_{j=N_1+1}^{c} |a_j|<\frac{\epsilon}{2}.$$ So, fix some natural number $N\ge N_1,N_2$. Since $N$ is finite, there exists $M$ such that for any $m\ge M$, $$\{1,2,\ldots, N\}\subseteq\{f(1),f(2),\ldots, f(m)\}.$$ (For instance, we can set $m=\max{\{f^{-1}(1),f^{-1}(2),\ldots, f^{-1}(N)\}}$.) So, the sum $t_m-s_N$ is a finite sum of terms $a_k$ with $k>N$, which means that there exists $q>N$ such that $$|t_m-s_N|=\bigg|\sum_{i=N+1}^q a_i\bigg|\le \sum_{i=N+1}^q |a_i|<\frac{\epsilon}{2}.$$ Therefore, for $m\ge M$, we have $$|t_m-S|\le |t_m-s_n|+|s_n-S|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$ As $\epsilon\to 0$, by the definition of convergence we see that $\sum a_{f(m)}$ converges absolutely to $S$. This completes the proof.

So, we see that the above examples were not coincidental! However, rearrangements of conditionally convergent series are part of another story...