This question may be duplicate with this answer_1 and here I referred to the same paper by Hardy, G. H.; Ramanujan, S. referred to by wikipedia which is referred to in answer_1.
But here I focused on the different part of the paper.
I can understand most of the proof if assuming that Theorem C in the paper is already proven. But I have some small problems.
In the paper p13, it says
$\log g(x)=\sum_{2}^{\infty}\log \frac{1}{1-x^{\mu}}=\sum_{1}^{\infty}\frac{1}{\nu}\frac{x^{2\nu}}{1-x^{\nu}}\sim \frac{1}{1-x}\sum_{1}^{\infty}\frac{1}{\nu^2}$
...
$\nu x^{\nu−1}(1 − x) < 1 − x^{\nu} < \nu(1 − x)$
$\frac{1}{1-x}\sum \frac{x^{2\nu}}{\nu^2}<\log g(x)<\frac{1}{1-x}\sum \frac{x^{\nu+1}}{\nu^2}$
The first equation can be got by letting $f(x)=x^v$, then $f'(x)<\frac{1-f(x)}{1-x},x\to 1^{-}$. The situation where $x\to 1^{+}$ is similar.
But for the 2nd, $\sum_{2}^{\infty}\log \frac{1}{\nu(1-x)}<\log g(x)=\sum_{2}^{\infty}\log \frac{1}{1-x^{\nu}}$
Q:
How does $\sum_{2}^{\infty}\log \frac{1}{\nu(1-x)}$ becomes $\frac{1}{1-x}\sum_{\nu=1}^{\infty}\frac{x^{2\nu}}{\nu^2}$ by removing the $\log()$ function (Here I think the subscript is same as $\sum_{1}^{\infty}\frac{1}{\nu}\frac{x^{2\nu}}{1-x^{\nu}}$ although the paper doesn't explicitly show it)?
Edited:
By Anne Bauval's comment, the removal of $\log()$ occurs when $\sum_{2}^{\infty}\log \frac{1}{1-x^{\mu}}=\sum_{1}^{\infty}\frac{1}{\nu}\frac{x^{2\nu}}{1-x^{\nu}}$. This is where my doubt exists.
