Here's Example 3.53 in the book Principles of Mathematical Analysis by Walter Rudin, third edition.

Consider the convergent series $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \ldots$$ and one of its rearrangements $$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} + \ldots$$ in which two positive terms are always followed by one negative. If $s$ is the sum of the original series, then $$s < 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6}.$$ Since $$\frac{1}{4k-3} + \frac{1}{4k-1} - \frac{1}{2k} = \frac{8k-4}{(4k-1)(4k-3)} - \frac{1}{2k} = \frac{2k(8k-4) - (4k-1)(4k-3)}{2k(4k-1)(4k-3)} = \frac{8k-3}{2k(4k-1)(4k-3)} > 0$$ for $k \geq 1$, we see that $$s^\prime_3 < s^\prime_6 < s^\prime_9 < \ldots,$$ where $s^\prime_n$ is the $n$th partial sum of the series after the rearrangement. Hence $$\lim_{n\to\infty}\sup s^\prime_n > s^\prime_3 = \frac{5}{6},$$ so that the rearranged series certainly does not converge to $s$.

Now here's my question:

How to determine, using the machinery developed by Rudin upto this point in the book, if the new (or rearranged) series converges at all? Rudin leaves it to the reader to check that the new series does converge. How to prove this convergence?

I would like to have answers that use only the results that Rudin has discussed so far in the book.

  • $\begingroup$ Does the Leibniz test happen to be among those tools at your disposal? $\endgroup$ Nov 19 '14 at 14:47
  • $\begingroup$ What is that? I can't recall it using this name, I'm afraid. $\endgroup$ Nov 19 '14 at 14:49
  • $\begingroup$ Also known as the alternating series test. (Doesn't seem applicable here, however.) $\endgroup$ Nov 19 '14 at 14:49
  • $\begingroup$ Oh yes, it is Theorem 3.43 in Baby Rudin. Then what? $\endgroup$ Nov 19 '14 at 14:52
  • $\begingroup$ Not directly applicable, but unless I'm very much mistaken here, from there it's a very small step to prove the convergence - I believe that's what the answer below means to tell you. $\endgroup$ Nov 19 '14 at 14:52

We can show that the series converges, and find its sum, as follows:

$\hspace{.3 in}1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots=\ln 2$ $\;\;\;$so

$\hspace{.27 in}\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots=\frac{1}{2}\ln 2$. $\;\;\;$Inserting zeros, we get

$\hspace{.26 in}0+\frac{1}{2}+0-\frac{1}{4}+0+\frac{1}{6}+0-\frac{1}{8}+0+\frac{1}{10}+\cdots=\frac{1}{2}\ln 2$.

Adding this to the original series gives

$\hspace{.26 in}1+0+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+0+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+0+\cdots=\frac{3}{2}\ln 2$, $\;\;$ so

$\hspace{.25 in} 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\cdots=\frac{3}{2}\ln 2$.

  • $\begingroup$ Wow! What a great, clear answer! $\endgroup$
    – ignoramus
    Nov 22 '14 at 3:38
  • $\begingroup$ @ignoramus Thanks! $\endgroup$
    – user84413
    Nov 22 '14 at 16:37
  • $\begingroup$ @user84413 how did you figure it out yourself? Can you please enlighten me too as to how this insight came to you? $\endgroup$ Oct 23 '16 at 7:12
  • $\begingroup$ @SaaqibMahmuud I had seen this argument in some textbook -- it is not something I thought of myself. $\endgroup$
    – user84413
    Oct 27 '16 at 17:11
  • $\begingroup$ This is all so good $\endgroup$
    – Henry Choi
    Oct 3 '20 at 4:59

Hint: Notice that $s_{3k}'$ converge as $k \to \infty$. How far can $s_{3k+1}$ and $s_{3k+2}$ be from $s_{3k}$?

  • $\begingroup$ @D Poole, how do we know that $s_{3k}^\prime$ converges? And, how to use this information toward establishing the convergence of $s_n^\prime$? $\endgroup$ Nov 19 '14 at 14:57
  • $\begingroup$ For $s'_{3k}$ take a good look at the alternating series test. You have $(1+\frac{1}{3})-(\frac{1}{2})+(\frac{1}{5}+\frac{1}{7})-(\frac{1}{4}) \pm \cdots$ - now can you see how to separate this series and apply the test? $\endgroup$ Nov 19 '14 at 15:01
  • 1
    $\begingroup$ I don't think that's quite what you want. Try to think of the following: Define the sequence $a_{2n-1}:=\frac{1}{4n-3}+\frac{1}{4n-1}$, $a_{2n}:=\frac{1}{n}$ for $n\geq 1$. Now, what does the alternating series test say about the series $\sum_{i=1}^\infty{(-1)^ia_i}$? And, how does that help you? $\endgroup$ Nov 19 '14 at 15:13
  • 1
    $\begingroup$ To see that $s_{3k}'$ converge, look at the powers of $k$ in the fraction in your formula for $a_{3k-2}'+a_{3k-1}'+a_{3k}'$. $\endgroup$
    – D Poole
    Nov 19 '14 at 15:40
  • 1
    $\begingroup$ @SaaqibMahmuud You do not need any results on rearrangements of infinite series for this method. It is an $\epsilon/2$ argument. Do you agree that you have the techniques to show that $s'_{3k}$ converges? If so, what is the largest that $|s'_{3k+1}-s'_{3k}|$ and $|s'_{3k+2}-s'_{3k}|$ can be? [For all sufficiently large $N$, you have that $|s'_{N} - s'_{\lfloor N/3 \rfloor}|< \epsilon/2$. Further, since $s'_{3k}$ converges to $s',$ for instance, then for all $N'$ large enough, $|s'_{3N'}-s'|<\epsilon/2$. Putting these together $ \ldots$] $\endgroup$
    – D Poole
    May 3 '16 at 12:07

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.