What does $\sum_{m=1}^n \ln (1 -\frac{\theta^2}{n}+ \frac{\theta^2}{2n m(\log m)^2} )$ converge to?

Question

What does the following sum converge to as $n\rightarrow\infty$?

$$\sum_{m=2}^n \ln (1 -\frac{\theta^2}{n}+ \frac{\theta^2}{2n m(\log m)^2} )$$

I think it converges to $-\theta^2/2$, but I am not sure how to show this?

Answer 1

$$\sum_{m=2}^{n}\ln\left(1-\frac{\theta^2}{n}+\frac{\theta^2}{2nm\ln^2m}\right)=-\frac{\theta^2}{n}\sum_{m=2}^{n}\left(1-\frac{1}{2m\ln^2m}\right)+\mathcal{O}\left(\frac{1}{n}\right)$$ and, since $\sum_{m=2}^\infty\frac{1}{m\ln^2m}$ converges, the given sum converges to $-\theta^2$.