4
$\begingroup$

The theorem is

If $\sum a_n$ is a series of complex numbers which converges absolutely then every rearrangement of $\sum a_n$ converges, and they all converge to the same value.

The proof given is as follows:

Let $\sum a_n'$ be a rearrangement, with partial sums $s_n'$. Given $\epsilon > 0$, there exists an integer $N$ such that $m \ge n \ge N$ implies

\begin{equation} \sum_{i = n}^m |a_i| \le \epsilon. \end{equation}

Now choose $p$ such that the integers $1,2, \cdots, N$ are all contained in $k_1, k_2, \cdots, k_p$ where $\{k_n\}$ is a bijective function from $\mathbb{N} \rightarrow \mathbb{N}$. Furthermore, $s_n' = s_{k_n}$ Then if $n>p$, the numbers $a_1, \cdots, a_N$ will cancel in the difference $s_n - s'_n$, so that

\begin{equation} |s_n - s_n'| \le \epsilon \end{equation} from the above inequality. Hence $\{s_n'\}$ converges to the same sum as $\{s_n\}$.

I am having trouble seeing why the second inequality follows from the first. Thanks.

$\endgroup$
  • $\begingroup$ What do you mean by first and second inequality? $\endgroup$ – Snufsan Sep 14 '14 at 8:15
2
$\begingroup$

Your first inequalty

\begin{equation} \sum_{i = n}^m |a_i| \le \epsilon. \end{equation}

says that no matter how many members of the tail of the sum (where the tail in this case, are all members after the $N$-th) you sum up absolutely, their sum will always be smaller than $\epsilon$.

The difference of sums $|s_n - s_n'|$ contains only elements in the tail of the sequence, finitely many, that means we can find the element with the largest index (from the original sequence $a_n$), lets say its $a_q$ (q is the largest index). This means that

$$|s_n - s_n'| \leq \left| \sum_{i=N+1}^{q}a_i \right| \leq \sum_{i = N+1}^q |a_i| \le \epsilon.$$

We bound $|s_n - s_n'|$ with it's maximum value $\left| \sum_{i=N+1}^{q}a_i \right|$, and then use the generalized triangle inequality to bound it with $\sum_{i=N+1}^q |a_i|$

$\endgroup$
  • $\begingroup$ Oh I see! I just got a little confused because the author switched up the indices when looking at the tail sum. Thanks. $\endgroup$ – Sandeep Silwal Sep 14 '14 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.