Simple question, what is the rate that this sum converges to $\infty$ as $n\rightarrow\infty$ (the rate as a function of $n$):
$$\sum_{k=1}^n \cfrac{\sin^3(\theta_k^{(n)})\pi(n-1)}{n^2\cos^2((n-\frac{1}{2})\theta_k^{(n)}-\frac{\pi}{4})}\cfrac{1}{1-\cos(\theta_k^{(n)})}\ , $$
where $\theta_k^{(n)}=\pi\cfrac{4(n-k)+3}{4n+2}$.
For context, the first term in the sum is an asymptotic approximation for the weights of gaussian integration, and the second term is an asymptotic approximtion for $1/(1-\xi)$ evaluated at the gaussian integration points (so integrating $\int_{-1}^11/(1-\xi)\text{d}\xi$ using gaussian integration).
After simplifications $$a_k=\frac{ (n-1)\,\pi}{n^2}\,\tan \left(\frac{ (4 k-1)\pi}{4 (2 n+1)}\right)$$ $$S_n=\sum_{k=1}^n a_k \sim 2 \log (n)+(\gamma +\log (2))-\frac{\log (n)}{n}+\gamma\frac{\log(n)+3}{n^2}+\cdots$$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 1000 & 15.0790 & 15.0775 \\ 2000 & 16.4684 & 16.4667 \\ 3000 & 17.2804 & 17.2787 \\ 4000 & 17.8564 & 17.8547 \\ 5000 & 18.3030 & 18.3013 \\ 6000 & 18.6679 & 18.6662 \\ 7000 & 18.9764 & 18.9747 \\ 8000 & 19.2436 & 19.2419 \\ 9000 & 19.4793 & 19.4775 \\ 10000 & 19.6901 & 19.6884 \\ \end{array} \right)$$
Edit
If we compute the equivalent integral $$T_n=\frac{(n-1) (2 n+1) }{n^2}\log \left(\cot \left(\frac{3 \pi }{4(2n+1)}\right)\right)$$ and, with $R^2=0.9999999972$, gives $$S_n \sim T_n+\frac \pi 2$$
Expanded as a series $$T_n=2 \log (n)+2 \log \left(\frac{8}{3 \pi }\right)-\frac{\log (n)-\log \left(\frac{3 e \pi }{8}\right)}{n}+O\left(\frac{1}{n^2}\right)$$
Notice that $\frac \pi 2+2 \log \left(\frac{8}{3 \pi }\right)=1.24300$ while $\gamma+\log(2)=1.27036$.
Update
Using Euler-MacLaurin summation formula $$\frac {S_n}{n-1}=\frac{2 \log \left(\frac{8 n}{3 \pi }\right)+\frac{123265132}{68201595}}{n}+\frac{\log \left(\frac{8 n}{3 \pi}\right)+\frac{129834161}{68201595}}{n^2}+O\left(\frac{1}{n^3}\right)$$