calculation $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{1}{k}\sin\frac{k\pi}{n+1}$. calculation$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{1}{k}\sin\frac{k\pi}{n+1}.$$
I try to use $|\sin x-x|\leqslant\frac{x^2}{2}$, but I don't feel like it's going to work. Is there any other way?
 A: This is essentially a Riemann sum of the function $$f(x) = \frac{\sin \pi x}{x}$$ over the unit interval $[0,1]$.
The partition seems to be $x_k = k/(n+1)$ for $0 \le k \le n+1$ but only $n$ points are being sampled ($x_1$ through $x_n$) instead of $n+1$. But the difference is negligible, and we have
$$ \sum_{k=1}^n f(x_k) \cdot \Delta x_k = \sum_{k=1}^n \frac{n+1}{k} \sin\left(\frac{\pi k}{n+1}\right) \cdot \frac{1}{n+1} \overset{n\to\infty}{\longrightarrow} \int_0^1 \frac{\sin \pi x}{x} \, dx = \mathrm{Si}(\pi) = 1.851\ldots$$
A: Let $ n\in\mathbb{N}^{*} $, we have the following : \begin{aligned}\sum_{k=1}^{n}{\frac{1}{k}\sin{\left(\frac{k\pi}{n+1}\right)}}&=\frac{2}{n+1}\int_{0}^{\frac{\pi}{2}}{\sum_{k=1}^{n}{\cos{\left(\frac{2kx}{n+1}\right)}}\,\mathrm{d}x}\\ &=\frac{2}{n+1}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}\sin{\left(\frac{nx}{n+1}\right)}}{\sin{\left(\frac{x}{n+1}\right)}}\,\mathrm{d}x}\end{aligned}
For all $ x\in\left(0,\frac{\pi}{2}\right] $ : \begin{aligned}\left\vert\frac{\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)\sin{\left(\frac{x}{n+1}\right)}}-\frac{\sin{x}}{x}\right\vert&=\left\vert\frac{\left(x-\left(n+1\right)\sin{\left(\frac{x}{n+1}\right)}\right)\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)x\sin{\left(\frac{x}{n+1}\right)}}+\frac{\sin{\left(\frac{nx}{n+1}\right)}-\sin{x}}{x}\right\vert\\ &\leq\frac{\sin{\left(\frac{nx}{n+1}\right)}}{x\sin{\left(\frac{x}{n+1}\right)}}\left\vert\frac{x}{n+1}-\sin{\left(\frac{x}{n+1}\right)}\right\vert+\frac{\left\vert\sin{x}-\sin{\left(\frac{nx}{n+1}\right)}\right\vert}{x}\\ &\leq\frac{x^{2}\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)^{3}\sin{\left(\frac{x}{n+1}\right)}}+\frac{2\sin{\left(\frac{x}{2\left(n+1\right)}\right)}\cos{\left(\frac{\left(2n+1\right)x}{2\left(n+1\right)}\right)}}{x}\end{aligned}
Combining the inequalities : $ \left(\forall y\in\left[0,\frac{\pi}{2}\right]\right),\ \frac{2}{\pi}x\leq\sin{x}\leq x $, $ \left(\forall y\in\mathbb{R}\right),\ \cos{y}\leq 1 $ and $ \left(\forall y\in\left[0,\frac{\pi}{2}\right]\right),\ \sin{y}\leq 1 $, we get : \begin{aligned}\left\vert\frac{\sin{\left(\frac{nx}{n+1}\right)}}{\left(n+1\right)\sin{\left(\frac{x}{n+1}\right)}}-\frac{\sin{x}}{x}\right\vert\leq\frac{x^{2}\times 1}{\left(n+1\right)^{3}\times\frac{2x}{\pi\left(n+1\right)}}+\frac{2\times\frac{x}{2\left(n+1\right)}\times 1}{x}&=\frac{\pi x}{2\left(n+1\right)^{2}}+\frac{1}{n+1}\\ &\leq\frac{n+1+\frac{\pi^{2}}{4}}{\left(n+1\right)^{2}}\end{aligned}
Thus : $$ \frac{1}{n+1}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}\sin{\left(\frac{nx}{n+1}\right)}}{\sin{\left(\frac{x}{n+1}\right)}}\,\mathrm{d}x}\underset{n\to +\infty}{\longrightarrow}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}\sin{x}}{x}\,\mathrm{d}x}=\frac{1}{2}\int_{0}^{\pi}{\frac{\sin{x}}{x}\,\mathrm{d}x} $$
Which means : $$ \lim_{n\to +\infty}{\sum_{k=1}^{n}{\frac{1}{k}\sin{\left(\frac{k\pi}{n+1}\right)}}}=\mathrm{Si}\left(\pi\right) $$
