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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.

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

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