Check if $\lim_{n\rightarrow\infty}\sum_{k=1}^n\ln\Big(1-\frac{1}{k}+\frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big)\Big)$ converges. How can I check if the following expression converges as $n \rightarrow\infty$? I am confused because $n$ appears twice... 
$$\lim_{n\rightarrow\infty} \sum_{k=1}^n \ln \Big(1 - \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big)\Big)$$
My initial thoughts are that $\cos(\frac{\theta}{\sqrt{\ln n}}) \rightarrow 0$ and so we are essentially summing $\ln(1) = 0$ infinitely many times, so it converges to 0. But is this correct? 

Since
$1-\cos(x)
=2\sin^2(x/2)
$,
and writing 
$t$ for $\theta$,
$\begin{array}\\
s(t, n)
&=\sum_{k=1}^n \ln \Big(1 - \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{t}{\sqrt{\ln n}}\Big)\Big)\\
&=\sum_{k=1}^n \ln \Big(1 - \frac{1}{k}\left(1- \cos\Big(\frac{t}{\sqrt{\ln n}}\Big)\right)\Big)\\
&=\sum_{k=1}^n \ln \Big(1 - \frac{1}{k}\left(2\sin^2\Big(\frac{t}{2\sqrt{\ln n}}\Big)\right)\Big)\\
&=\sum_{k=1}^n \ln \Big(1 - \frac{r}{k}\Big)
\qquad r = 2\sin^2\Big(\frac{t}{2\sqrt{\ln n}}\Big)\\
&\approx\sum_{k=1}^n (-\frac{r}{k}\Big)
\qquad r \approx \frac{t^2}{2\ln n} \text{ for large } n\\
&=-r(\ln(n)+O(1))\\
&\approx-\dfrac{t^2(\ln(n)+O(1))}{2\ln(n)}\\
&=-\dfrac{t^2}{2}+O(\dfrac{1}{\ln(n)})\\
\end{array}
$
