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 expand $\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$ as $$-\sum_{m=1}^{\infty} \frac{(\frac{x^2\sin^2k}{2n})^m}{m}$$ which converges for $\frac{x^2\sin^2k}{2n} \le 1 \to x^2 \le 2n \to |x| \le \sqrt{2n}$.
Then the summation in your question would become $$-\sum_{k=1}^n\sum_{m=1}^{\infty} \frac{(\frac{x^2\sin^2k}{2n})^m}{m}$$
Switching the order of summation, I get $$-\sum_{m=1}^{\infty} \frac{x^{2m}}{2^m n^{m-1} m} \frac{1}{n}\sum_{k=1}^n \sin^{2m}(k)$$
The limit $ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \sin^{2m}(k)$ is equivalent to $$\frac{1}{\pi}\int_0^{\pi}\sin^{2m}(t)dt$$
So the sum can be rewritten as $$-\frac{n}{\pi} \int_0^{\pi} \sum_{m=1}^{\infty} \frac{\left(\frac{x^2 \sin^2(t)}{2n}\right)^m}{m} dt$$
The inside sum can be rewritten so that it ends up looking like the original $$\frac{n}{\pi} \int_0^{\pi} \ln\left( 1 - \frac{x^2 \sin^2(t)}{2n} \right) dt$$
This seems to match up with the original sum, although I feel like there is some easier way to convert the sum into this integral. We now want to find $$\lim_{n \to \infty} \frac{\int_0^{\pi} \ln\left( 1 - \frac{x^2 \sin^2(t)}{2n} \right) dt}{\frac{\pi}{n}}$$
This is a $\frac{0}{0}$ indeterminate form, so using L'Hôpital's rule, I get $$\lim_{n \to \infty} \frac{\int_0^{\pi} \frac{x^2 \sin^2(t)}{2n^2 \left( 1- \frac{x^2 \sin^2(t)}{2n} \right)} dt}{-\frac{\pi}{n^2}}$$
Simplifying, this becomes $$-\frac{x^{2}}{2\pi}\int_{0}^{\pi}\frac{\sin^{2}\left(t\right)}{1-\frac{x^{2}}{2n}\sin^{2}\left(t\right)}dt$$
As $n \to \infty$, $\frac{x^2}{n} \to 0$, so the final answer is $$-\frac{x^2}{2\pi} \int_0^{\pi} \sin^2(t) dt = -\frac{x^2}{2\pi} \frac{\pi}{2} = -\frac{x^2}{4}$$
