How to rearrange alternating harmonic series to get sum of 1

Question

How can we rearrange $$ \sum \frac{(-1)^n}{n} $$ to get the sum 1 I know we know the alternation of sum by$$\frac{1}{2}\log{k}$$ Edit:sorry forgot to include $$k=a/b $$ a is number of positive terms taken

b is number of negative terms taken

If we use above formula $$ 1=\log(2)+(1/2)\log(k)$$ We get $k=e^2/4$ I am not sure what taking e^2 terms mean

It's a bit unclear what you're asking. What is $k$? What do you mean by "rearrange to get sum $1$"? — dgeyfman, Jul 26 '23 at 17:22
Are you familiar with the Riemann rearrangement theorem? Do you know the proof? The proof I know is rather constructive... — Xander Henderson, Jul 26 '23 at 17:23
Does this answer your question? Series rearrangement and Riemann's theorem — Robert Israel, Jul 26 '23 at 17:33
Pick a sequence $a_n/b_n$ that tends to $e^2/4$. Put, in the natural order, $a_n$ positive terms, then $b_n$ negative terms, and so on. — NDB, Jul 26 '23 at 17:37
"I know we know the alternation of sim by $\frac12{\log k}$." What does that mean? — Thomas Andrews, Jul 26 '23 at 17:51
@DanielGeyfman k is ratio of positive and negative terms taken sequentially,I have edited the question please check — unknown user, Jul 26 '23 at 18:16
@ThomasAndrews k is ratio of positive and negative terms taken sequentially,I have edited the question please check — unknown user, Jul 26 '23 at 18:17
"I know we get the alternation of sums" is the meaningless part. Why do you assume $1=\log 2 +\frac12\log k?$ — Thomas Andrews, Jul 26 '23 at 18:42
If you mean $a_n,b_n$ is the number of positive and negative terms in the $n$th partial sum, and $k$ is the limit of $a_n/b_n,$ then it isn't even clear that the limit exists. — Thomas Andrews, Jul 26 '23 at 18:46

score 0 · Answer 1 · answered Jul 26 '23 at 17:26

0

Here's how you can get any desired sum $S$:

Add up enough odd terms until the first time you exceed $S$
Subtract enough even terms until you go below $S$
Add more odd terms until you exceed $S$ again
Subtract more even terms until you go below $S$ again

If you want $S =1$ specifically, here's what that looks like:

$1 + \frac{1}{3} > 1$
$1 + \frac{1}{3} - \frac{1}{2} < 1$
$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} > 1$
$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} - \frac{1}{4} < 1,$

and so on forever.

answered Jul 26 '23 at 17:26

Rivers McForge

5,787

Does it follow any certain pattern for terms? – unknown user Jul 27 '23 at 11:37
A pattern could be constructed based on the algorithm given above, but it would most likely be somewhat messy and unintuitive. Additionally, this rearrangement is not unique; there are many different ways to change the order of the positive and negative terms to get a sum of $S = 1$ (or any other desired sum). – Rivers McForge Jul 28 '23 at 15:34

How to rearrange alternating harmonic series to get sum of 1

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