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? 
 A: Since $\cos$ is even, you may assume WLOG that $\theta \geq 0$. 
Note that 
$\displaystyle \sum_{k=1}^n\frac{-\theta^2}{2k\ln n} = \frac{-\theta^2}{2}\frac{1}{\ln n}\sum_{k=1}^n \frac 1k \xrightarrow[n\to \infty]{}\frac{-\theta^2}{2}$ and 
$$\left|\sum_{k=1}^n \ln \Big(1 - \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big)\Big) - \sum_{k=1}^n\frac{-\theta^2}{2k\ln n} \right|
\leq A_n + B_n
$$
where $\displaystyle A_n = \left|\sum_{k=1}^n \ln \Big(1 - \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big)\Big) - \left(- \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big) \right)  \right| $ 
and 
$\displaystyle B_n = \left|\sum_{k=1}^n - \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big) - \frac{-\theta^2}{2k\ln n} \right|$
Since $\ln(1+x) = x+O(x^2)$, there is some $K>0$ and $\epsilon$ such that $$|x|\leq \epsilon \implies |\ln(1+x)-x|\leq Kx^2$$
Since $\cos(x) = 1-\frac{x^2}2+O(x^4)$, there is some $K'>0$ and $\epsilon'$ such that $$|x|\leq \epsilon' \implies |\cos(x)-1+\frac{x^2}2|\leq K'x^4$$
For $n\geq \max\left(\exp\left(\frac{\theta}{\arccos(1-\epsilon)} \right), \exp\left(\frac{\theta}{\epsilon'} \right) \right)$,
$$\begin{align}A_n+B_n&\leq K\left(\cos(\frac{\theta}{\sqrt{\ln n}})-1 \right)^2\sum_{k=1}^n \frac 1{k^2} + K' \frac{\theta^4}{\ln^2 n} \sum_{k=1}^n \frac 1k \\&= O\left( \left(\cos(\frac{\theta}{\sqrt{\ln n}})-1 \right)^2 \right) + O\left(\frac{1}{\ln n} \right)
\\ &=o(1)
 \end{align}$$
Finally,
$$\sum_{k=1}^n \ln \Big(1 - \frac{1}{k} + \frac{1}{k}\cos\Big(\frac{\theta}{\sqrt{\ln n}}\Big)\Big)\xrightarrow[n\to \infty]{}\frac{-\theta^2}{2}$$
A: 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}
$
