How many ways to show $\sum_{n=1}^{\infty}\ln \left|1-\frac{x^2}{n^2\pi^2}\right|$ is pointwise convergence? $\sum_{n=1}^{\infty}\ln \left|1-\frac{x^2}{n^2\pi^2}\right|$. where $x\not= k\pi, k\in \mathbb{Z}$ is pointwise convergence.

Assume $n$ is sufficiently large($\geq N$),for a fixed point $x_0$,$-\ln \left|1-\frac{x_0^2}{n^2\pi^2}\right|=\frac{x_0^2}{n^2\pi^2}+\frac{1}{2}\frac{x_0^4}{n^4\pi^4}+\cdots+\frac{1}{n}\left(\frac{x_0^2}{n^2\pi^2}\right)^n+\cdots < 
\frac{x_0^2}{n^2\pi^2}+\left(\frac{x_0^2}{n^2\pi^2}\right)^2+\left(\frac{x_0^2}{n^2\pi^2}\right)^3+\cdots+\left(\frac{x_0^2}{n^2\pi^2}\right)^n+\cdots=\frac{x_0^2}{n^2\pi^2-x_0^2}$
then $\sum_{n\geq N}\ln \left|1-\frac{x^2}{n^2\pi^2}\right| \leq \sum_{n \geq N}\frac{x_0^2}{n^2\pi^2-x_0^2} \Rightarrow 0$

Is there any other way to show this? because my textbook omits this process.Maybe there is a straightforward way to solve this.
 A: I think that most (if not all) ways of showing the convergence rely on comparison with $\sum_{n\ge N}Cn^{-2}$. One way of avoiding the use of the Taylor expansion of $\ln(1-t)$ is the following. Assume that $n$ is large enough, more specifically, assume that $n>2|x|/\pi$. Then $t=x^2/(n^2\pi^2)<1/4$.
In the interval $t\in(0,1/4)$ the derivative $|D\ln(1-t)|=1/|1-t|$ is bounded from above by $1/(1-\frac14)=\dfrac43$. Therefore, by the mean value theorem, we have the estimate, valid for  all $t\in(0,1/4)$,
$$
|\ln(1-t)|=|\ln(1-t)-\ln 1|=t\cdot \frac1{|1-\xi|}\le \frac{4t}3,
$$
where $\xi\in(0,t)$. Hence the terms of your series have the upper bound
$$
\left|\ln(1-\frac{x^2}{n^2\pi^2})\right|\le\frac{4x^2}{3\pi^2n^2}
$$
for all $n\ge 2|x|/\pi$.
Convergence of your series follows from this. Pointwise for all $x$ and uniformly in any bounded interval (save for the points $x=k\pi$ for some $k\in\Bbb{Z}$).
A: Another way for math freaks: using the Euler formula for the sine
$$
\sum_{n=1}^\infty\ln\left|1-\frac{x^2}{n^2\pi^2}\right|=
\ln\prod_{k=1}^\infty \left|1-\frac{x^2}{n^2\pi^2}\right|=\ln\left|\frac{\sin x}{x}\right|.
$$
