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?
 A: First, maybe we need to be a little more precise because if $x$ is very large then for small $n$’s we would have logarithms of negative numbers! So, I will alter and generalize the sequence as folows:
For a fixed $0<t<\pi$, a real $x$ we will consider the sequence $(A_n)_{n\ge n(x)}$ where $n(x)=1+\lfloor x^2/2\rfloor$ and
$$A_n=\sum_{k=1}^n\ln\left(1-\frac{x^2}{2n}\sin^2(kt)\right)$$
We will prove that
$$ \forall\,t\in(0,\pi),\qquad \lim_{n\to\infty}A_n=-\frac{x^2}{4}\tag{$*$}$$
To this end we will use the next lemma.
Lemma 1. for $u\in[0,1/2]$ we have
$0\le -u-\ln(1-u)\le u^2.$$
Proof. Indeed, for $0\le u\le 1/2$ we have
$$-u-\ln(1-u)=\int_0^u\frac{t}{1-t}\,dt.$$
But if $0\le t\le 1/2$ then $0\le t/(1-t)\le 2t$ hence
$$0\le -u-\ln(1-u)=\int_0^u\frac{t}{1-t}dt\le \int_0^u2tdt=u^2.\qquad\qquad\square$$
We will also need the following result.
Lemma 2. For all $t\in (0,\pi)$ we have
$$\lim_{n\to\infty}\frac1n \sum_{k=1}^n\sin^2(kt)=\frac12\tag2$$
Proof. Because
$$\eqalign{\sum_{k=1}^n \sin^2(kt)&=\frac{1}{2}
\sum_{k=1}^n(1-\cos(2kt))\cr
&=\frac{n}{2}-\frac{1}{2}\Re\sum_{k=1}^{n}e^{2ikt}
\cr
&=\frac{n}{2}-\frac12\Re\frac{e^{2(n+1)it}-e^{2it}}{e^{2it}-1}\cr
&=\frac{n}{2}- \frac{\sin((2n+1)t)-\sin(t)}{4\sin t} 
}$$
In particular,
$$\lim_{n\to\infty}\frac1n \sum_{k=1}^n\sin^2(kt)=\frac12\qquad\qquad\square$$
Now consider a real number $x$, and let $n$ be a positive integer such that $n>x^2$. Using Lemma 1. With $u=x^2\sin^2(kt)/(2n)$ we get
$$0\le -\frac{x^2}{2n}\sin^2(kt)-\ln\left(1-\frac{x^2}{2n}\sin^2(kt)\right)\le \frac{x^4}{4n^2}$$
Adding these inequalities as $k$ varies from $1$ to $n$ we obtain
$$0\le-\frac{x^2}{2n}\sum_{k=1}^n\sin^2(kt)-A_n\le \frac{x^4}{4n}$$
We conclude that
$$\lim_{n\to\infty}\left( \frac{x^2}{2n}\sum_{k=1}^n\sin^2(kt)+A_n\right)= 0.$$
Now, using Lemma 2 we find that
$$\lim_{n\to\infty}\left(\frac{x^2}{4}+A_n\right)=0$$
and $(*)$ is proved. $\qquad\square$
Remark 1. Note that the limit does not depend on $t\in(0,\pi)$.
Remark 2. It is clear from the proof that we have uniform convergence on compact sets with respect to $x$.
A: We can write
$$
\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 $\mathbb{R}$.
A: This is a comment, not an answer, but there's no way to put it in a comment box.  Numerically, it certainly seems like the sequence converges.  In this graph

the blue curve represents the values for $n=100$, the red curve represents the values for $n=200$, and the green curve represents the values for $n=300$.  Matplotlib has its own ideas about what the scales should be. Actually, $0\leq x\leq 1.4$ and $y$ runs between $0$ and approximately $-1$.
A: 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 $$-\frac{x^2}{2\pi} \int_0^{\pi} \sin^2(t) dt = -\frac{x^2}{2\pi} \frac{\pi}{2} = -\frac{x^2}{4}$$
