I'm making my term paper and I need to prove some results, my question is:
If I know that (I've proved this) $\sin(x) = x \displaystyle \prod_{n=1}^{\infty}\left(1 - \frac{x²}{(\pi n)^2 } \right)$ then applying $\ln(x)$ we have
$\ln(\sin(x)) = \ln\left(x \displaystyle \prod_{n=1}^{\infty}\left(1 - \frac{x²}{(\pi n)^2 } \right)\right)$ .
Now, can I assume that $\ln(\sin(x)) = \ln x + \displaystyle \sum_{n=1}^{\infty} \ln\left(1 - \frac{x²}{(\pi n)^2 } \right) $??
I need Uniform convergence of $\ln(x)$'series to assume that?
Assuming that $0 < x < \pi$ and that it has already been proved that
$$\sin x = x \prod_{n=1}^{\infty}\left(1 - \frac{x^2}{\pi^2n^2 } \right) = x \lim_{m \to \infty}\prod_{n=1}^{m}\left(1 - \frac{x^2}{\pi^2n^2 } \right), $$
We have
$$\tag{*}\ln(\sin x) = \ln x + \ln \left[\lim_{m \to \infty}\prod_{n=1}^{m}\left(1 - \frac{x^2}{\pi^2n^2 } \right) \right]$$
Here we have $a_n = -x^2/(\pi n)^2 > -1 $ for all $n \in \mathbb{N}$ and
$$\sum_{n = 1}^\infty|a_n| = \frac{x^2}{\pi^2}\sum_{n=1}^\infty\frac{1}{n^2} < \infty$$
It is not difficult to prove that convergence of the series $\sum|a_n|$ implies convergence of $\sum |\ln(1 +a_n)|$ if $a_n > -1$. This, in turn, implies convergence of $\sum \ln(1+a_n)$ by the comparison test, and so we have convergence of
$$\lim_{m \to \infty}\ln \left[\prod_{n=1}^{m}\left(1 - \frac{x^2}{\pi^2n^2 } \right)\right] = \lim_{m \to \infty}\sum_{n=1}^m\ln\left(1 - \frac{x^2}{\pi^2n^2 } \right) $$
By continuity of the logarithmic function we can write (*) as
$$\ln(\sin x) = \ln x + \lim_{m \to \infty}\sum_{n=1}^{m}\ln \left(1 - \frac{x^2}{\pi^2n^2 } \right) $$
Addendum: $\sum |a_n| < \infty \implies \sum |\ln(1+a_n)| < \infty$ when $-1 < a_n < 0$
If $\sum|a_n|$ converges then $|a_n| \to 0$ as $n \to \infty$ and for all sufficiently large $n$ , $|a_n| < 1/2$. Using the inequality $\ln(1+x) \leqslant x$, we have
$$|\ln(1+a_n)| = \ln \left( \frac{1}{1+a_n}\right) = \ln \left(1 + \frac{|a_n|}{1- |a_n|} \right) \leqslant \frac{|a_n|}{1 - |a_n|}< 2 |a_n|$$
Thus, $\sum|\ln(1+a_n)|$ converges by the comparison test.