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 $\sin$

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