Asymptotic behaviour of Lambert series How can I examine asymptotic behaviour of Lambert series $f(x) = \sum_{}^{} \frac{x^{n}}{1-x^{n}}$ when $x$ approaches $1^{-}$- for instance find function $ g(x)$ such that $\lim_{x\rightarrow 1^{-}}\frac{f(x)}{g(x)} = const$. I would really appreciate some hints or referalls to any similiar examples since I have no idea where to begin.
 A: An elementary approach is to note that the map $t \mapsto \frac{x^t}{1-x^t}$ is decreasing for $0 < x < 1$, so
$$
\int_1^\infty \frac{x^t}{1-x^t}\,dt \leq \sum_{n=1}^{\infty} \frac{x^n}{1-x^n} \leq \frac{x}{1-x} + \int_1^\infty \frac{x^t}{1-x^t}\,dt.
$$
Since
$$
\int_1^\infty \frac{x^t}{1-x^t}\,dt = \frac{\log(1-x)}{\log x} \sim \frac{\log(1-x)}{x-1}
$$
as $x \nearrow 1$, we conclude that
$$
\lim_{x \nearrow 1} \frac{x-1}{\log(1-x)}\sum_{n=1}^{\infty} \frac{x^n}{1-x^n} = 1.
$$
A: We can find an asymptotic formula using the residue theorem and the inverse Mellin transform : $$\begin{array}{ll}\displaystyle\sum_{n=1}^\infty \tau(n) e^{-nx} &=& \displaystyle\frac{1}{2i\pi}\int_{\sigma-i\infty}^{\sigma+i\infty} \Gamma(s) \zeta(s)^2 x^{-s} ds\\ &=& Res(\Gamma(s) \zeta(s)^2 x^{-s},1)+Res(\Gamma(s) \zeta(s)^2 x^{-s},0) +Res(\Gamma(s) \zeta(s)^2 x^{-s},-1) \\ && \qquad \qquad\displaystyle +\frac{1}{2i\pi}\int_{-3/2-i\infty}^{-3/2+i\infty} \Gamma(s) \zeta(s)^2 x^{-s} ds\\ &=& \displaystyle \frac{A\ln x+B}{x} +C+ \mathcal{O}(x) \qquad\qquad (x\to 0^+,\sigma > 1)\end{array}$$
Where $$Res(\Gamma(s) \zeta(s)^2 x^{-s},1)=Res((1-\gamma (s-1))(\frac{1}{s-1}+\gamma)^2 x^{-s},1)= \frac{-\ln x+\gamma}{x}$$ and $$Res(\Gamma(s) \zeta(s)^2 x^{-s},0) = Res(\frac{\zeta(0)^2}{s},0) = \frac{1}{4}$$
Therefore
$$\boxed{\sum_{n=1}^\infty \frac{y^n}{1-y^n}=\sum_{n=1}^\infty \tau(n) y^n = \frac{-\ln (-\ln y)+\gamma}{-\ln y} +\frac{1}{4}+ \mathcal{O}(\ln y) \qquad\qquad (y \to 1^-)}$$
A: I suppose this is
$$ f(x) = \sum_{n=1}^\infty \frac{x^n}{1-x^n}$$
Note that when $|x^n| < 1$,
$$ \frac{x^n}{1-x^n} = \sum_{k=1}^\infty x^{kn}$$
so $f(x)$ becomes
$$ \sum_{n=1}^\infty \sum_{k=1}^\infty x^{kn}  = \sum_{j=1}^\infty \tau(j) x^j$$
where $\tau(j)$ is the number of divisors of $j$.
