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I've encountered this limit :

$$\lim_{n\to\infty} \frac1n \left(\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{2\pi}{n}\right)+\cdots+\sin{\frac{(n-1)\pi}{n}}\right)$$

Wolfram gives the value: $2\over \pi$.

We did something similar in class. If I consider the function: $\sin(x\pi)$ and the equidistant partition: $j\over n$, I can somehow transform this problem into an integral.

Could someone please give me some advices and\or guidance on how to proceed in this problem.

Thanks for any suggestions in advance.

edit: Thanks a lot guys.

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  • $\begingroup$ @user_of_math We havent't covered that yet. I believe there's and alternative solution. Thanks though. $\endgroup$ – David Oct 22 '14 at 16:48
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It's a Riemann sum for the integral $$ \int_0^1 \sin(\pi x)\,dx. $$ The partition is $0,\dfrac1n,\dfrac2n,\dfrac3n,\ldots\dfrac n n$, so that $\Delta x$ is $1/n$ in each case. Take the value of the function $x\mapsto\sin(\pi x)$ at the left endpoint of each subinterval.

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The expression whose limit you want to compute is a Riemann sum for the integral $$ \int_0^1\sin(\pi\,x)\,dx. $$

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Hint:-

$\displaystyle \lim_{n \to \infty}\dfrac{1}{n}\displaystyle\sum_{i=0}^{n-1}\sin \left(\dfrac{i\pi}{n}\right)=\displaystyle\int_{0}^1\sin (\pi x) \ dx$

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