# Finding a rearrangement of a series such that the rearrangement diverges to infinity

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.

• I think what you are looking for can be found here: en.wikipedia.org/wiki/…
– 5xum
Oct 19, 2015 at 23:51
• That link doesn't appear to show an example of how to rearrange to diverge to infinity, only converge to something finite. Oct 19, 2015 at 23:56
• 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 $N$.
– user169852
Oct 19, 2015 at 23:58

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 $\pm \infty$.