I want to solve this summation, however I have no idea where to start. Could any one help me find a good starting place?

$$ \sum_{i=1}^{n}\sin\left(i \over n\right)\frac{1}{n} $$ Any help would be greatly appreciated. Thank you very much.

The mathlab result is I attained is here.


closed as unclear what you're asking by Did, user91500, Claude Leibovici, user26857, Shailesh Oct 30 '16 at 9:18

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  • 2
    $\begingroup$ Hello, welcome to math.se. We typically expect some context for your question. As you might imagine, people sometimes post their homework questions here expecting someone else to do their work for them. To counter that suspicion, maybe you should post some thoughts you have had in attacking this problem and specifically point out where you are having trouble. $\endgroup$ – Ron Gordon Oct 29 '16 at 12:53
  • $\begingroup$ Hello, no buddy but i'm trying to check my code in matlab with it, someone asked this question to a forum and I just wanna to solve it, but I didn't know the matlab result is true or not. $\endgroup$ – Amin Qassemi Oct 29 '16 at 12:56
  • $\begingroup$ And... what was the MATLAB result for given $n$? $\endgroup$ – Parcly Taxel Oct 29 '16 at 13:00
  • $\begingroup$ I've attached my result in matlab but I don't have any idea to reach it. $\endgroup$ – Amin Qassemi Oct 29 '16 at 13:03
  • $\begingroup$ this is why I think the matlab result is wrong, thanks all, $\endgroup$ – Amin Qassemi Oct 29 '16 at 13:12

The sum is equal to

$$\begin{align}\frac1n\operatorname{Im} \sum_{k=1}^n e^{i k/n} &= \frac1n\operatorname{Im} \left (\frac{e^{i}-1}{1-e^{-i/n}} \right ) \\ &=\frac1{2 n}\operatorname{Im} \left (\frac{\left (e^i-1 \right ) \left (1-e^{i/n} \right )}{1-\cos{\frac1n}} \right )\\ &= \frac1{2 n \left (1-\cos{\frac1n} \right )} \operatorname{Im}{ \left (e^i +e^{i/n}-e^{i(1+1/n)} - 1 \right )} \\ &= \frac{\sin{1}+\sin{\frac1n}-\sin\left (1+\frac1n \right ){}}{2 n \left (1-\cos{\frac1n} \right )}\\ &=\frac{\sin{1} \left (1-\cos{\frac1n} \right )+\sin{\frac1n} (1-\cos{1})}{2 n \left (1-\cos{\frac1n} \right )} \\ &= \frac{\sin{1}}{2 n} + \frac{1-\cos{1}}{4 n \sin^2{\frac1{2 n}}} \sin{\frac1n}\end{align}$$


$$\frac1n \sum_{k=1}^n \sin{\frac{k}{n}} = \frac{\sin{1}}{2 n} + \frac{1-\cos{1}}{2 n \tan{\frac1{2 n}}} $$

As $n \to \infty$, the sum approaches $1-\cos{1}$, which is equal to $\int_0^1 dx \ \sin{x}$ as expected.


first, a cutie

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


It is evidente for $n=1$. Suppose it holds for some $n$. Then,

$$ \sum_{k=1}^{n+1} \sin (kx) = \frac{ \sin \left( \frac{n+1}{2} x \right)}{\sin \left( \frac{x}{2} \right) } \sin \left( \frac{ n x }{2} \right) + \sin[(n+1)x] = $$ $$\frac{ \sin \left( \frac{n+1}{2} x \right)}{\sin \left( \frac{x}{2} \right) } \sin \left( \frac{ n x }{2} \right) + 2 \sin \left( \frac{ (n+1) x }{2} \right)\cos \left( \frac{ (n+1) x }{2} \right) = \frac{ \sin \left( \frac{(n+2)x}{2} \right) }{\sin \left( \frac{x}{2} \right)} \sin \left( \frac{(n+1)x}{2} \right) $$

Now, with $x= 1/n$, we have

$$ \boxed{ \sum_{i=1}^n \sin \left( \frac{i}{n} \right) \frac{1}{n} = \frac{ \sin\left(\frac{n+1}{2n}\right)}{ \sin \left( \frac{1}{2n} \right)} \sin \left( \frac{1}{2} \right) \frac{1}{n} }$$

  • $\begingroup$ I just edited your answer. You had put x=n in place of x=1/n. $\endgroup$ – Rohan Oct 29 '16 at 13:06
  • $\begingroup$ yes, thanks i typed too fast $\endgroup$ – ILoveMath Oct 29 '16 at 13:06
  • $\begingroup$ @ILoveMath and Rohan thanks both of you guys. $\endgroup$ – Amin Qassemi Oct 29 '16 at 14:36

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