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?


1 Answer 1


I agree with your confusion about part 2 of the exercise statement, which seems to say a whole lot of nothing.

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, $$|\sum_{k=1}^{n}b_k-S_{N_2}| \le \sum_{k\in A}|a_k|\le \sum_{k= N_2+1}^\infty |a_k|<\epsilon/2.$$


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