calculate riemann sum of sin to proof limit proposition

Question

$$\lim_{n \to \infty}\frac1n\sum_{k=1}^n\sin(\frac{k\pi}{n})$$

I'm having trouble expressing $\sin(x)$ differently here in order to calculate the riemann sum. I want to show that this converges to $\frac{2}{\pi}$ so it equals to $\int_0^1 \sin(x\pi)$.

Is there any easy way to express $\sin(x)$ different here?

Added:

$$\frac{1}{2i}(\frac{\cos(\frac{(n+1)\pi}{n})+i\sin(\frac{(n+1)\pi}{n})-\cos(\frac{\pi}{n})+i\sin(\frac{\pi}{n})}{\cos(\frac{\pi}{n})+i\sin(\frac{\pi}{n})-1}-\\\frac{\cos(-\frac{(n+1)\pi}{n})+i\sin(-\frac{(n+1)\pi}{n})-\cos(-\frac{\pi}{n})+i\sin(-\frac{\pi}{n})}{\cos(-\frac{\pi}{n})+i\sin(-\frac{\pi}{n})-1})$$

Answer 1

Use that $$\sum_{k=1}^n \sin kx=\frac{\sin\dfrac{kx}2\sin (k+1)\dfrac x2}{\sin\dfrac{x}2}$$

ADD One can deduce the above in several ways. The first is to note it is the imaginary part of $$\sum_{k=0}^n e^{ikx}=\frac{e^{(n+1)ix}-1}{e^{ix}-1}$$

Another choice is to use $$\cos b-\cos a=2\sin\frac{b-a}2\sin\frac{b+a}2$$

Now let $$b=\left(n+\frac 1 2\right)x\\a=\left(n-\frac 1 2\right)x$$

Then you get

$$\cos \left(n+\frac 1 2\right)x-\cos \left(n-\frac 1 2\right)x=2\sin\frac{(n+1)x}2\sin\frac{x}2$$

Then sum and telescope.

Answer 2

The sum can have the closed form

$$\sum_{k=1}^n\sin\left(\frac{k\pi}{n}\right)= \frac{\sin \left( {\frac {\pi }{n}} \right)}{ \left( 1-\cos \left( {\frac { \pi }{n}} \right) \right)} .$$

Added: To prove the above identity, you need the two facts

1) $$ \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}, $$

2) $$ \sum_{k=1}^{n}x^m={\frac {{x}^{n+1}-x}{-1+x}}. $$

Answer 3

The Riemann sum can be written in closed form by reference to the following trig identity. List of Trig Identities (https://en.wikipedia.org/wiki/List_of_mathematical_series)

$$ \sum_{k=0}^{n-1}\sin(k\pi/N)=\cot(\pi/2N) $$

If you're curious about how that can be derived, you will need a bit of cleverness, using the difference of cosine identities.

The Riemann sum is $$ (1/N)\sum_{k=0}^{n-1}\sin(k\pi/N)=(1/N)\cot(\pi/2N) $$

to take the limit, as $N\rightarrow\infty$, of $\cot(\pi/2N)/N$, write as $\frac{\cos(\pi/2N)}{Nsin(\pi/2N)}=\frac{\cos(\pi/2N)}{sin(\pi/2N)/(1/N)}$ The numerator approaches 1 as $N\rightarrow\infty$ and the denominator approaches $\pi/2$ on account of the well known formula $\lim_{x\rightarrow 0}sin(\alpha*x)/x=\alpha$. Thus the answer is $1/(\pi/2)=2/\pi$ $\square$

The Trig Identity

As for the trig identity, the strategy is to rewrite each sine as a difference, which can then be used to telescope the series. If we multiply and divide each term in the series by $\sin(\theta/2)$, the numerator becomes $\sin(k\theta)\sin(\theta/2)$, which can then be written as a difference of cosines. The series then telescopes so only the first and last term remain. The conclusion is an exercise in trig identities, which I will omit for now. I may come back and edit this solution if there are still questions.

Answer 4

In general,

$\sum_{n=1}^p \sin n\theta=\Im (\sum_{n=1}^p e^{ i n\theta})=\Im (e^{i\theta}\frac{ 1-e^{i p\theta}}{1-e^{i \theta}}),$

which, for $1-e^{i n\theta}=e^{i0}-e^{i n\theta}=e^{i n\theta/2}(e^{-i n\theta/2}-e^{i n\theta/2})= e^{i n\theta/2}\cdot (-2i)\sin(n\theta/2)$, (we can use trigonometric formulas too, which one can notice is essentially the same) equals

$$\Im (e^{i\theta}\frac{ 1-e^{i p\theta}}{1-e^{i \theta}}) =\Im (e^{i\theta}\frac{e^{i p\theta/2}\cdot (-2i)\sin(p\theta/2)}{e^{i \theta/2}\cdot (-2i)\sin(\theta/2}))\\ =\Im (\frac{e^{i (p+1)\theta/2}\sin(p\theta/2)}{\sin(\theta/2)}) =\frac{\sin[(p+1)\theta/2]\sin(p\theta/2)}{\sin(\theta/2)}.$$

Therefore, $$\lim_{n \to \infty}\frac1n\sum_{k=1}^n\sin(\frac{k\pi}{n}) =\lim_{n \to \infty}\frac1n\frac{\sin[(n+1)\pi/2n]\sin(n\pi/2n)}{\sin(\pi/2n)} =\lim_{n \to \infty}\frac1n\frac{\cos (\pi/2n)}{\sin(\pi/2n)}=\frac{2}{\pi},$$ for $\lim_{x\to0}\frac{\sin x}{x}=1.$

This 'proves' $\int_{0}^1 \sin(\pi x) dx=\frac{2}{\pi}.$