Riemann Sum to show convergence help?

Question

I have the function $f(x)=\sin\left(\dfrac{\pi x}{2}\right)$ on the partition of $[0,1]$ given by $$P_{n}: 0 < \frac{1}{n} < \frac{2}{n} < ... < \frac{n-1}{n} < 1$$

I have shown that $$L(f,P_{n})= \sum^{n-1}_{j=0} \frac{1}{n}\sin\left(\frac{j \pi}{2n}\right)$$

However i don't know how to use this to show:$$\frac{1}{n}\left(\sin\left(\frac{\pi}{2n}\right)+\sin\left(\frac{2 \pi}{2n}\right)+...+\sin\left(\frac{(n-1)\pi}{2n}\right) \rightarrow \sin\left(\frac{2}{\pi}\right)\right)$$

Any help would be appreciated.

Answer 1

Hint. Here you may write

$$ \sum^{n-1}_{j=0} \frac{1}{n}\sin\left(\frac{j \pi}{2n}\right) \to \int_0^1 \sin \left( \frac{\pi x}{2}\right)dx=\frac2{\pi}. $$

Answer 2

It appears you need to compute the limit of the lower Riemann sum using some analytical tools -- as opposed to declaring simply that it converges to the integral.

The following identity is useful

$$\sum_{j=1}^n \sin (jx) = \frac{\sin\left( \frac{nx}{2}\right)\sin\left( \frac{(n+1)x}{2}\right)}{\sin\left( \frac{x}{2}\right)},$$

and is proved easily by taking the imaginary part of the geometric sum $\sum_{j=1}^n (e^{ix})^j.$

Consequently,

$$\begin{align}\frac{1}{n} \sum_{j=0}^{n-1}\sin\left(\frac{j \pi}{2n}\right) &= \frac{1}{n} \sum_{j=1}^{n-1}\sin\left(\frac{j \pi}{2n}\right) \\ &= \frac{\sin\left( \frac{(n-1)\pi}{4n}\right)\sin\left( \frac{n\pi}{4n}\right)}{n\sin\left( \frac{\pi}{4n}\right)} \\ &= \frac{4}{\pi} \left(\frac{\sin\left( \frac{\pi}{4n}\right) }{\frac{\pi}{4n}}\right)^{-1}\sin\left(\frac{\pi}{4}(1-1/n)\right)\sin\left(\frac{\pi}{4}\right)\end{align},$$

and

$$\lim_{n \to \infty}\frac{1}{n} \sum_{j=0}^{n-1}\sin\left(\frac{j \pi}{2n}\right) = \frac{4}{\pi}\lim_{n \to \infty}\left(\frac{\sin\left( \frac{\pi}{4n}\right) }{\frac{\pi}{4n}}\right)^{-1} \lim_{n \to \infty}\sin\left(\frac{\pi}{4}(1-1/n)\right) \sin\left(\frac{\pi}{4}\right) \\ = \frac{4}{\pi} \cdot 1 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \\ = \frac{2}{\pi}.$$