# Evaluate $\lim_{n\rightarrow\infty} \sum_{k=1}^n \frac{1}{k}\tan(\frac{k\pi}{4n + 4})$

My approach:

$$\lim_{n\rightarrow\infty} \sum_{k=1}^n \frac{1}{k}\tan(\frac{k\pi}{4n + 4}) = \lim_{n\rightarrow\infty} \sum_{k=1}^{n+1} \frac{1}{k}\frac{n+1}{n+1}\tan(\frac{k\pi}{4n + 4}) = \lim_{n\rightarrow\infty} \sum_{k=1}^{n+1} \frac{1}{k}\frac{n+1}{n+1}\tan(\frac{\pi k}{4(n + 1)}) = \lim_{n\rightarrow\infty} \sum_{k=1}^{n+1} \frac{1}{n+1}\frac{n+1}{k}\tan(\frac{\pi}{4}\frac{k}{n+1}) = \int_0^1 \frac{\tan(\frac{\pi}{4}x)}{x}dx = \int_0^{\pi/4} \frac{\tan(x)}{x}dx$$

I've found out after googling that this integral has no solution in elementary functions which has me a bit stumped since our teacher told us this limit has a solution.

EDIT: I tried approximating the sum itself and at $$n = 1 000 000$$ it evaluates to about $$0.848967...$$ which appears to be really close to the value that the integral evaluates to according to Wolfram Alpha.

• It looks as if your computation is correct. This is probably not the limit your instructor was thinking about.
– robjohn
Mar 16 at 13:47