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Find $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{k\sin\frac{k\pi}{n}}{1+(\cos\frac{k\pi}{n})^2}$

I think this maybe relate to Riemann sum. but I can't deal with $k$ before $\sin$

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    a man in other forum post a lot of questions, include this. and i can't figure out, so i duplicate it here. he must make a typo. at first i think the fraction maybe $\frac{\frac{k}{n^2} \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}}$. but it is simple to work out.he maybe not mean this. but i had not found it equal to $\infty$2012-11-11

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If there is no typo, then the answer is $\infty$. Indeed, let $m$ be any fixed positive integer and consider the final $m$ consecutive terms:

$ \sum_{k=n-m}^{n-1} \frac{k \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}} = \sum_{k=1}^{m} \frac{(n-k) \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}}. $

As $n \to \infty$, each term converges to $k \pi$, in view of the substitution $x = \frac{k\pi}{n}$ and the following limit

$ \lim_{x\to 0}\frac{\sin x}{x(1 + \cos^2 x)} = 1. $

Thus

$ \liminf_{n\to\infty} \sum_{k=1}^{n} \frac{k \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}} \geq \lim_{n\to\infty} \sum_{k=n-m}^{n-1} \frac{k \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}} = \sum_{k=1}^{m} k \pi = \frac{m(m+1)}{2}\pi. $

Now letting $m \to \infty$, we obtain the desired result.


Indeed, we have

$ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{\frac{k}{n} \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}} \frac{1}{n} = \frac{1}{\pi^2} \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx = \frac{1}{4}. $

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    @sos440: hmm, nice. (+1)2012-12-10