0
$\begingroup$

I wish to find a way to rearrange a series such that it diverges to infinity. If we take, for example, the alternating harmonic series: $$\sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{n}$$

How can we rearrange this particular conditionally convergent series such that it diverges to $\infty$? Is there any simple trick for doing this? Also, if you could explain the process you used to find this rearrangement, that would be extremely helpful.

Thanks, Lauren.

$\endgroup$
  • $\begingroup$ I think what you are looking for can be found here: en.wikipedia.org/wiki/… $\endgroup$ – 5xum Oct 19 '15 at 23:51
  • $\begingroup$ That link doesn't appear to show an example of how to rearrange to diverge to infinity, only converge to something finite. $\endgroup$ – Lauren Hayes Oct 19 '15 at 23:56
  • 1
    $\begingroup$ Basic recipe: let $a_n$ be the positive terms and $b_n$ be the negative terms. Take the first chunk of $a_n$ so that the sum of the chunk exceeds $2$. Then add one of the $b_n$, which is no larger in magnitude than $1$, so the total so far of the included $a_n$' s and $b_n$'s is at least $1$. Now add another chunk of $a_n$'s whose sum exceeds $2$. Then add the next $b_n$. Every time, we are adding as many $a_n$'s as needed to increase the sum by at least $2$, then adding a single $b_n$. After you add $b_N$, the partial sum of the $a_n$'s and $b_n$'s included so far will be at least $N$. $\endgroup$ – Bungo Oct 19 '15 at 23:58
0
$\begingroup$

Let $g(1)=1 .$ Let $a_{1,g(1)}=1 .$ Define $g(n+1)$, and $a_{j,g(n+1)}$ for $1\leq j \leq g(n+1)$ , recursively ,as follows : Given that $a_{n,g(n)}=1/(2 k_n-1)$ with $k_n\in Z^+ $, choose $g(n+1)$ large enough that $$\sum_{j=1}^{j=g(n+1)} (1/(2k_n+2j-1)>1 . $$ And let $a_{n+1,j}=1/(2k_n+2j-1)$ for $1\leq j\leq g(n+1) .$ Consider the series $$a_{1,g(1)}-1/2+a_{1,g(2)} +...+a_{g(2),g(2)}-1/4+a_{1,g(3)}+... +a_{g(3),g(3)} -1/6+a_{1,g(4)}+... $$ A conditionally convergent real series that is not absolutely convergent can actually be re-arranged into a series that sums to anything you want, including $\pm \infty$.

$\endgroup$

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.