Finding the asymptotic behavior of a function series

I solved the shape of an elastic sheet annulus clamped on the inner and outer circle with a point load (the figure below shows the cross section of an example): Solve Green function of an annulus to calculate the shape of a clamped elastic sheet

I found the solution at the point load is actually divergent though any other point is convergent (just like the field of an electrical point charge).

By non-dimensionalization, now the problem to look at is a function series $$f(r)=\sum_{n=1}^{\infty} \frac{1}{n} \cdot \frac{\left((2r)^n-(2r)^{-n}\right)\cdot (2^n -2^{-n})}{4^n -4^{-n}}$$

Numerical tests show the terms in the series decays slower than $1/n$ for $r=1$ and thus divergent: $f(1)=\infty$ while $f(r)$ is convergent on $[\frac{1}{2},1)$.

My ideal target is to approximate the behavior near the singularity, maybe in the form $f(r)=C_1 + C_2 \cdot (1-r)^{-\alpha}$.

I'd really appreciate your kind help.

Follow @Delta-u 's solution, here is the plot of $\ln{(1-r)} - f(r)$:

The black solid line is $f(r)=r$.

$$g(r)=\sum_{n=1}^{\infty} \frac{r^n +r^{-n}}{n \cdot (4^n +1)}$$ $$g(1)=\sum_{n=1}^{\infty} \frac{2}{n \cdot (4^n +1)} = 0.4715$$

$g(1)$ matches the numerical result.

• It seems that rather than $f(r)=C_1 + C_2 \cdot (1-r)^{-\alpha}$ the series look like $f(r)=C_1 + C_2 \ln(1-r)$ – Delta-u Jul 13 '18 at 15:18

You have: \begin{align} f(r)&=\sum_{n=1}^{\infty} \frac{1}{n} \cdot \frac{\left((2r)^n-(2r)^{-n}\right)\cdot (2^n -2^{-n})}{4^n -4^{-n}}\\ &=\sum_{n=1}^{\infty} \frac{1}{n} \cdot \frac{(2r)^n\cdot 2^n }{4^n}+g(r) \end{align} where \begin{align} g(r)=\sum_{n=1}^{\infty} \frac{1}{n} \cdot \frac{-(2r)^{-n}\cdot (2^n -2^{-n})}{4^n -4^{-n}}+\sum_{n=1}^{\infty} \frac{1}{n} \cdot \frac{(2r)^n\cdot (8^n -2^n-8^n+2^{-n})}{16^n -1} \end{align} so $g$ is well defined and continuous up to $r=1$.
As $\sum_n \frac{z^n}{n} = -\ln(1-z)$ you obtain: $$f(r)=-\ln(1-r)+g(1)+o(1) \text{ near } r=1$$