Does $\lim_{n\to \infty} \sum_{k=1}^n\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$ exist? Let $x \in \mathbb{R}.$ Is is true that the following limit exists : $$\lim_{n \to \infty} \sum_{k=1}^n\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$$ What is the value of this limit?
I tried the Integral test for convergence, but nothing came out.
Any suggestions?

We can write
$$
\sum\limits_{k = 1}^n {\log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)}  =  - \frac{{x^2 }}{{2n}}\sum\limits_{k = 1}^n {\sin ^2 k}  + \sum\limits_{k = 1}^n {\left[ {\frac{{x^2 \sin ^2 k}}{{2n}} + \log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)} \right]} .
$$
Here
$$
\sum\limits_{k = 1}^n {\sin ^2 k}  = \frac{n}{2} + \mathcal{O}(1).
$$
Suppose that $n$ is so large that $x^2  \le n$. Then
$$
\left| {\frac{{x^2 \sin ^2 k}}{{2n}} + \log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)} \right|  \le \frac{{x^4 \sin ^4 k}}{{4n^2 }}\le \frac{{x^4}}{{4n^2 }}.
$$
Hence,
$$
\left| {\sum\limits_{k = 1}^n {\left[ {\frac{{x^2 \sin ^2 k}}{{2n}} + \log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)} \right]} } \right| \le \frac{{x^4 }}{{4n }}.
$$
From these estimates, we can see that
$$
\sum\limits_{k = 1}^n {\log \left( {1 - \frac{{x^2 \sin ^2 k}}{{2n}}} \right)}  \to  - \frac{{x^2 }}{4}
$$
uniformly on compact subsets of $\mathbb{R}$.
