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).

• Remember... $\ln(a)-\ln(b)=\ln(\frac ab)$. – manooooh May 22 at 17:03
• Welcome to math.SE!! – manooooh May 22 at 17:03

Use l'Hôpital (and some logarithm identities) to get:

$$\begin{equation} \begin{split} \lim_{n\to\infty} \frac{\ln(\frac34 \pi + 2n)-\ln(\frac\pi4+2n)}{\ln(2n+2)-\ln(2n)} &= \lim_{n\to\infty} \frac{\ln(3-\frac{16n}{8n+\pi})}{\ln(1+\frac1n)} \\ &\overset{\text{l'Hôpital}}= 16\pi \lim_{n\to\infty} \frac{n(n+1)}{(8n+\pi)(8n+3\pi)} \\ &\overset{\text{linearity of the limit}}=16\pi \bigg(\lim_{n\to\infty}\frac n{8n+\pi}\bigg)\bigg(\lim_{n\to\infty}\frac{n+1}{8n+3\pi}\bigg) \\ &= 16\pi\cdot\frac18\cdot\frac18=\frac\pi4. \end{split} \end{equation}$$

• That's nice! Thanks – user676375 May 22 at 17:54

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 $$\frac\pi4$$.

• Thank you! I like it – user676375 May 22 at 17:55

Using the rules of logarithm so write $$\frac{\ln\left(\frac{\frac{5\pi}{4}+2n}{\frac{\pi}{4}+2n}\right)}{\ln\left(\frac{2n+2}{2n}\right)}$$

$$a_n= \frac{\ln\left(\frac{3\pi}{4} + 2n\right)-\ln\left(\frac{\pi}{4}+2n\right)}{\ln(2n+2)-\ln(2n)}=\frac{\log \left(1+\frac{\pi }{4 n+\frac{\pi }{2}}\right) } {\log \left(1+\frac 1n \right)}$$ Using Taylor expansions $$a_n=\frac{\frac{\pi }{4 n}-\frac{\pi ^2}{16 n^2}+O\left(\frac{1}{n^3}\right) } {\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right) }=\frac{\pi }{4}-\frac{(\pi -2) \pi }{16 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.

This limit can be evaluated by the mean value theorem. Let $$f(x)=\ln x$$, then because $$f(x)$$is continuous and differentiable for all $$x$$, $$\frac{f(2n+2)-f(2n)}{2} = f'(2n+c_1), 0 and $$\frac{f(2n+\frac{3\pi}{4})-f(2n+\frac{\pi}{4}) }{\frac{\pi}{2}}=f'(2n+c_2), 0 Since $$f'(x)=\frac{1}{x}$$, $$\lim_{n\to\infty}{\frac{\frac{f(2n+2)-f(2n)}{2}}{\frac{f(2n+\frac{3\pi}{4})-f(2n+\frac{\pi}{4}) }{\frac{\pi}{2}}}\cdot \frac{\frac{\pi}{2}}{2}}=\lim_{n\to\infty}{\frac{2n+c_2}{2n+c_1}\cdot \frac{\pi}{4}}=\frac{\pi}{4}$$