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Please solve this, I am confused how to solve it, what will be its limit?

$$\lim_{n \rightarrow \infty} \left[\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{2\pi}{n}\right)+\dots+\sin\left(\frac{n\pi}{n}\right)\right]$$

Equals

1) $\large\frac{1}{\pi}$

2) $\large \frac{2}{\pi}$

3) $\large \frac{3}{\pi}$

4) none

projectilemotion
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Mrutyunjay Meher
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  • See http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro – lab bhattacharjee Mar 15 '17 at 17:54
  • Looks pretty close to a Riemann sum for $\int_0^1 \sin(\pi x),dx$, but without the factor of $\frac{\pi}{n}$. A geometric series might work. – Matthew Leingang Mar 15 '17 at 17:56
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    @MatthewLeingang I would write it as $n/\pi$ times the Riemann sum you mentioned. Thus the limit is $\infty.$ – zhw. Mar 15 '17 at 18:16

3 Answers3

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Let $f\left( x\right):=x\sin \pi x$ so $$\lim_{n\to\infty}\sum_{k=1}^n\sin\frac{k\pi}{n}=\lim_{n\to\infty}\sum_{k=1}^n\frac{f\left( k/n\right)}{k/n}.$$The large-$n$ approximation $$\sum_{k=1}^n\frac{f\left( k/n\right)}{k/n}\approx\int_1^n\frac{f\left( x/n\right)}{x/n}dx=n\int_{1/n}^1\frac{f\left( y\right)}{y}dy\approx n\int_0^1\frac{f\left( y\right)}{y}dy=\frac{2n}{\pi}$$then gives us$$\lim_{n\to\infty}\sum_{k=1}^n\sin\frac{k\pi}{n}=\lim_{n\to\infty}\frac{2n}{\pi}=\infty.$$

J.G.
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$$\sin \pi/n +\sin 2\pi/n+...+\sin n\pi/n$$ $$=\frac{sin\frac{(\pi/n+\pi)}{2}\sin (n\pi/2n)}{\sin\pi/2n}$$ $$=\sin (π/2)\frac{\cos\pi/2n}{\sin\pi/2n}$$ $$=cot(π/2n)$$ $$=cotx$$

where $x=π/2n$

Now as $n\rightarrow \infty, x \rightarrow 0 , \cot x \rightarrow \infty$

Used $$sina+sin(a+b)+sin(a+2b)+...+sin[a+(n-1)b]=\frac{\frac{sin[2a+(n-1)b]}{2}sin(nb/2)}{sin(b/2)}$$

Rayees Ahmad
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Try to prove the following the first equality either by induction or with some complex analysis :

$$\sum_{k=1}^n\sin\frac{k\pi}n=\frac{\cos\frac\pi{2n}-\cos\left[\left(n+\frac{1}2\right)\frac\pi n\right]}{2\sin\frac\pi{2n}}\xrightarrow[n\to\infty]{}\infty$$

DonAntonio
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