Determine this limit $\lim_{n\to\infty} \frac{\ln\left(\frac{3\pi}{4} + 2n\right)-\ln\left(\frac{\pi}{4}+2n\right)}{\ln(2n+2)-\ln(2n)}.$ how can I determine the following limit? $$\lim_{n\to\infty} \frac{\ln\left(\frac{3\pi}{4} + 2n\right)-\ln\left(\frac{\pi}{4}+2n\right)}{\ln(2n+2)-\ln(2n)}.$$
This question stems from this question. The proof presented there is incorrect, and it would be trivial to show that the mentioned integral diverges if the above limit is $>0$ by using the comparison test (anyone feels free to do this by the way).

Let $f(x)=\ln\left(\frac{3\pi}3+x\right)-\ln\left(\frac\pi3+x\right)$ and $g(x)=\ln(x+2)-\ln x$. Then\begin{align}\lim_{x\to\infty}\frac{f(x)}{g(x)}&=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}\\&=\lim_{x\to\infty}\frac{-\frac{8 \pi }{(4 x+\pi ) (4 x+3 \pi )}}{-\frac{2}{x^2+2 x}}\\&=\lim_{x\to\infty}\frac{4\pi(x^2+2x)}{(4 x+\pi ) (4 x+3 \pi )}\\&=\frac\pi4.\end{align}So, your limit is $\frac\pi4$.
