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.