# Series Rearrangement Theorem Proof - Help Needed!!!

In this write-up, I'd request you to do two things:

1. To verify the proof that I've written.
2. Answer the doubts that I've raised.

I know this is a long write-up, but kindly help me.

In this textbook I'm reading, the series rearrangement theorem is stated as follows:

If $$\sum_{n=1}^\infty a_n$$ converges absolutely, and $$b_1,b_2,\cdots,b_n,\cdots$$ is any arrangement of the sequence $$\{a_n\}$$, then $$b_n$$ converges absolutely and $$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty b_n$$

And then, we're given the exercise(with hints) to prove it. Here's how it goes (textbook statements in quotes):

1. Let $$\varepsilon$$ be a positive real number, let $$L=\sum_{n=1}^\infty a_n$$, and let $$S_k=\sum_{n=1}^k a_n$$. Show that for some index $$N_1$$ and for some index $$N_2\ge N_1$$, $$\sum_{n=N_1}^\infty |a_n| < \frac{\varepsilon}{2} \text{ and } |S_{N_2}-L| < \frac{\varepsilon}{2}$$

Since all the terms $$a_1,a_2,\cdots,a_{N_2}$$ appear somewhere in the sequence $$\{b_n\}$$, there is >an index $$N_3\ge N_2$$ such that if $$n\ge N_3$$, then $$\sum_{k=1}^nb_k-S_{N_2}$$ is at most a sum of terms $$a_m$$ with $$m\ge N_1$$. Therefore, if $$n\ge N_3$$, $$\left| \sum_{k=1}^nb_k-L \right| \le \left| \sum_{k=1}^nb_k-S_{N_2} \right| + |S_{N_2}-L| \le \sum_{k=N_1}^\infty|a_k|+|S_{N_2}-L|<\varepsilon$$

My Proof:
Since it is given that $$\sum_{n=1}^\infty a_n$$ converges absolutely, i.e, $$\sum_{n=1}^\infty |a_n|$$ converges, by the definition of convergence of series we have the following: There exists a number $$K>0$$ such that if $$n\ge K$$ there exists a positive real number $$\varepsilon/2$$, such that \begin{aligned} \left| \sum_{k=1}^n |a_k|-\sum_{k=1}^\infty |a_k| \right| &< \frac{\varepsilon}{2} \\ \implies \left| \sum_{k=1}^n |a_k|-\sum_{k=1}^n |a_k|-\sum_{k=n+1}^\infty |a_k| \right| &< \frac{\varepsilon}{2} \\ \implies \sum_{k=n+1}^\infty |a_k| &< \frac{\varepsilon}{2} \end{aligned} This is saying: for $$k\ge n+1>K$$, the sum could be made as small as we choose (or informally: one may take out any finite number of terms from a convergent series to make the sum as small as desired). Taking $$K+1=N_1$$, we get $$\sum_{k=N_1}^\infty |a_k| < \frac{\varepsilon}{2} \tag{1}$$ Next we have \begin{aligned} |S_{N_2}-L| &= \left| \sum_{n=1}^{N_2}a_n-\sum_{n=1}^\infty a_n \right| \\ &= \left| \sum_{n=1}^{N_2}a_n-\sum_{n=1}^{N_2}a_n-\sum_{n=N_2+1}^\infty a_n \right| \\ &= \left| \sum_{n=N_2+1}^\infty a_n \right| \\ &\le \sum_{n=N_2+1}^\infty |a_n| \text{ (triangle inequality)} \\ &= \sum_{n=N_1}^{N_2} |a_n|+\sum_{n=N_2+1}^\infty |a_n|-\sum_{n=N_1}^{N_2} |a_n| \\ &= \sum_{n=N_1}^\infty |a_n|-\sum_{n=N_1}^{N_2} |a_n| \\ &< \sum_{n=N_1}^\infty |a_n| \\ &< \frac{\varepsilon}{2} \text{ (from (1))} \end{aligned} There we go. We have $$|S_{N_2}-L| < \sum_{n=N_1}^\infty |a_n| < \frac{\varepsilon}{2}$$

Now, I've got a problem with what's stated in the second part of the problem statement. Since $$n\ge N_3\ge N_2\ge N_1$$, if $$S_{N_2}$$ is removed from $$\sum_{k=1}^nb_k$$, then we're removing terms $$a_1$$ to $$a_{N_2}$$ from $$\sum_{k=1}^nb_k$$ (since we've chosen an index $$N_3$$ such that terms in $$\{b_n\}$$ below $$N_3$$ will have all $$a_1$$ till $$a_{N_2}$$). Then we'd be left with terms in $$\{b_n\}$$ from $$a_{N_2+1}$$ till $$a_{N_3}$$ or beyond. So, shouldn't $$m\ge N_2+1$$ instead of $$m>N_1$$?

If my understanding is correct, then \begin{aligned} \left| \sum_{k=1}^nb_k-L \right| &= \left| \sum_{k=1}^nb_k-S_{N_2}+S_{N_2}-L \right| \\ &\le \left| \sum_{k=1}^nb_k-S_{N_2} \right| + \left| S_{N_2}-L \right| \\ &= \left| \sum_{k=N_2+1}^n a_n \right| + \left| S_{N_2}-L \right| \\ &\le \sum_{k=N_2+1}^\infty |a_n| + \left| S_{N_2}-L \right| \\ &< \sum_{k=N_1}^\infty |a_n| + \left| S_{N_2}-L \right| \\ &< \frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\ &= \varepsilon \end{aligned} Therefore $$\left| \sum_{k=1}^{n\ge N_3}b_k-L \right| < \varepsilon \tag{2}$$

Coming to the second part of the exercise statement:

1. The argument in part 1 shows that if $$\sum_{n=1}^\infty a_n$$ converges absolutely then $$\sum_{n=1}^\infty b_n$$ converges and $$\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty a_n$$. Now show that because $$\sum_{n=1}^\infty a_n$$ converges, $$\sum_{n=1}^\infty b_n$$ converges to $$\sum_{n=1}^\infty a_n$$.

I don't understand this statement - from $$(2)$$ doesn't it follow that $$\sum b_n$$ converges to $$L$$, the same limit that $$\sum a_n$$ converges to? Isn't the request of the second statement answered in the first statement itself?

I can't follow the argument you make about $$|S_{N_2}-L|<\epsilon/2,$$ but even if it is right, it seems overcomplicated and to miss the point. The point is that $$S_n\to L$$ (this is just what $$\sum_n a_n=L$$ means) and so for sufficiently large $$n,$$ $$|S_n-L|<\epsilon/2.$$ Thus there is some $$N_2\ge N_1$$ such that $$|S_{N_2}-L|<\epsilon/2.$$
Your understanding of the third part is not correct. It is not the case that $$\sum_{k=1}^{n}b_k-S_{N_2} = \sum_{k=N_2+1}^n a_k.$$ We just have $$|\sum_{k=1}^{n}b_k-S_{N_2}| \le \sum_{k=N_2+1}^\infty |a_k|.$$ This is because the thing inside the absolute value on the left-hand-side can be written as a sum of some of the $$a_k$$ for $$k\ge N_2,$$ since all of the $$a_k$$ with $$k\le N_2$$ are present in $$\{b_1,\ldots, b_n\},$$ but then subtracted out when you subtract $$S_{N_2}.$$ There's no guarantee that all the $$a_k$$ that in the remainder have index $$\le n$$ and neither is there guarantee that all of the $$a_k$$ for $$N_2+1\le k \le n$$ occur in the remainder.
More concretely, let $$A = \{b_1,\ldots, b_n\}\setminus \{a_1,\ldots,a_{N_2}\}.$$ Then, because of how $$n$$ was chosen, we have $$A\subseteq \{a_{N_2+1},a_{N_2+1},\ldots\}$$ and $$\sum_{k=1}^{n}b_k-S_{N_2} = \sum_{k\in A}a_k.$$ And therefore, $$|\sum_{k=1}^{n}b_k-S_{N_2}| \le \sum_{k\in A}|a_k|\le \sum_{k= N_2+1}^\infty |a_k|<\epsilon/2.$$