# A limit $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{k\sin\frac{k\pi}{n}}{1+(\cos\frac{k\pi}{n})^2}$ maybe relate to riemann sum

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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May be you made a typo. The limit as it is equals to $\infty$ – Norbert Nov 11 '12 at 3:45
What I can't deal with in this case is the lack of some expression representing the length of the subintervals in which some interval is divided, in order to construct Riemann sums...are you sure you copied correctly the sum? – DonAntonio Nov 11 '12 at 3:45
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 $\infty$ – Laura Nov 11 '12 at 5:03

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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It will be too easy for you this limit will evaluate to riemann sum :) – Norbert Nov 11 '12 at 3:50
thanks for your answer in detail, and in your answer. I don't know in this inequality: $\liminf_{n\to\infty} \sum_{k=1}^{n} \frac{k \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}} \geq \lim_{n\to\infty} \sum_{k=n-m}^{n-1} \frac{k \sin \frac{k \pi}{n}}{1 + \cos^2 \frac{k \pi}{n}}$ why use $\liminf$ ? – Laura Nov 11 '12 at 6:20
@R_bot, I used $\liminf$ because we know nothing on the convergence of the original limit at that point. – Sangchul Lee Nov 11 '12 at 6:29
@sos440: hmm, nice. (+1) – user 1618033 Dec 10 '12 at 21:20