# Evaluating the sum after reordering an infinite series $1-\dfrac {1} {2}+\dfrac {1} {3}-\dfrac {1} {4}+\ldots$

If in the series $1-\dfrac {1} {2}+\dfrac {1} {3}-\dfrac {1} {4}+\ldots$ the order of the terms be altered, so that the ratio of the number of positive terms to the number of negative terms in the first $n$ terms is ultimately $a^{2}$, show that the sum of the series will become $\log \left( 2a\right)$.

Solution attempt. Let $p$ be the number of positive terms in the first $n$ terms of the reordered series so based on the question we are allowed $a^{2}=\dfrac {p} {n-p}$. Now solving for p we get $p=\dfrac {a^{2}n} {\left( 1+a^{2}\right) }$ and $n-p=\dfrac {n } {\left( 1+a^{2}\right) }$.

We also observe that in the original series only the odd terms are positive and only the even terms are negative.

Let's define $S_{odd}=1+\dfrac {1} {3}+\dfrac {1} {5}+.\ldots +\dfrac {1} {2n-1}$ and $S_{even}=\dfrac {-1} {2}\dfrac {-1} {4}\ldots -\dfrac {1} {2n}$

$S_{Reordered}=S_{odd_{p}}+S_{even_{n-p}}$

$S_{Reordered}=\sum _{t=1}^{t=P}\dfrac {1} {2t-1}+\sum _{t=1}^{t=(n-P)}\dfrac {-1} {2t}$
$S_{Reordered}=\sum _{t=1}^{t=\dfrac {a^{2}n} {\left( 1+a^{2}\right) }}\dfrac {1} {2t-1}+\sum _{t=1}^{t=\dfrac {n } {\left( 1+a^{2}\right) }}\dfrac {-1} {2t}$

In order to extend $S_{Reordered}$ from n terms to an infinite length. I guess i should take the limit of the $S_{Reordered}$ as $n\rightarrow \infty$ which would make the upper values of $t$ (Not sure of technical term here) to be $\infty$ giving me

$S_{Reordered}=\sum _{t=1}^{t=\infty}\dfrac {1} {2t-1}+\sum _{t=1}^{t=\infty}\dfrac {-1} {2t}$

and i have lost the $a$ from the expression. I guess i am stuck i need to perform some step to capture a before taking the limit. Any help would be much appreciated.

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Wait, what exactly are you doing? You are reordering the terms without changing them, and get a different result depending on the ordering? (Could you give an example of the series for some $a$?) –  TMM Mar 3 '12 at 20:33
(That also explains your final result: Only reordering will always give you the same result, which does not depend on $a$.) –  TMM Mar 3 '12 at 20:34
I believe it is well known result that reordering the terms without changing them in a non absolute series changes there sum. I quote "We say that the series $\sum _{n=1}^{n=\infty }v_{n}$ consists of the terms of $\sum _{n=1}^{n=\infty }u_{n}$ in a different order if a law is given by which corresponding to each positive integer $p$ we can find one(and only one) integer $q$ and vice versa, and $v_{q}$ is taken equal to $u_{p}$." I am afraid i do not have an example of such a series for a value of $a$. I presented all the info provided by the question. –  Hardy Mar 3 '12 at 20:41

I think you will find this result useful:

Fix $p$ and $q$, with $p\geq q\geq 1$. Then

$$\lim_{n \to \infty} \sum_{k=qn}^{pn} \frac{1}{k} = \log \frac{p}{q}$$

Usually, if you take the altered harmonic series and sum $p$ positive terms and then add $q$ negative terms, you'll get

$$\log 2 + \frac{1}{2} \log \frac{p}{q}$$

For example,

$$1+1/3+1/5-1/2-1/4+1/7+1/9+1/11-1/6-1/8+++--\cdots$$

will give

$$\log 2 + \frac{1}{2} \log \frac{3}{2}$$

So you now choose $p/q=a^2$

$$\log 2 + \frac{1}{2} \log a^2 =\log 2 + \log a = \log {2a}$$

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@Hardy For more info on this check Apostol's Calculus, Ch 10, 10.21 –  Pedro Tamaroff Mar 5 '12 at 20:49

Here is the germ of an idea: $$\frac12+\frac14+\frac14+\cdots+\frac1{2n}=\frac12\ln n+\frac\gamma2+o(1)$$ where $\gamma$ is Euler's constant, while the corresponding sum of odd reciprocals differs from the above by $\ln 2+o(1)$. I rather suspect that will do the trick.

By the way, I don't think the result is true if you allow reordering of the positive elements among themselves, or the negative ones. You should be able to reorder the harmonic series to diverge as slowly as you might wish by starting with $\sum 1/2^n$ and then inserting the reciprocals of non-powers of two at extremely rare intervals.

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