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.

Another way for math freaks: using the Euler formula for the sine
$$
\sum_{n=1}^\infty\ln\left|1-\frac{x^2}{n^2\pi^2}\right|=
\ln\prod_{k=1}^\infty \left|1-\frac{x^2}{n^2\pi^2}\right|=\ln\left|\frac{\sin x}{x}\right|.
$$
