# Show that $A_n=\sum\limits_{k=1}^n \sin k$ is bounded?

Let $A_n=\sum\limits_{k=1}^n \sin k$ , show that there exists $M>0$ , $|A_n|<M$ for every $n$ .

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What are your thoughts on the problem? What have you tried? Just so you know, it is also considered a bit rude to post in the imperative. It is quite alright to just ask a question if you have one instead of giving orders. –  JavaMan Aug 25 '11 at 2:09

Note that $$2(\sin k)(\sin(0.5))=\cos(k-0.5)-\cos(k+0.5).$$ This is obtained by using the ordinary expression for the cosine of a sum.

Add up, $k=1$ to $n$. On the right, there is mass cancellation. We get $$\cos(0.5)-\cos(n+0.5).$$ Thus our sum of sines is $$\frac{\cos(0.5)-\cos(n+0.5)}{2\sin(0.5)}.$$ We can now obtain the desired bound for $|A_n|$. For example, $2$ works, but not by much.

We could modify the appearance of the above formula by using the fact that $\cos(0.5)-\cos(n+0.5)=2\sin(n/2)\sin(n/2+0.5)$.

Generalization: The same idea can be used to find a closed form for $$\sum_{k=0}^{n-1} \sin(\alpha +k\delta).$$ Sums of cosines can be handled in a similar way.

Comment: This answer was written up because the OP, in a comment, asked for a solution that only uses real functions. However, summing complex exponentials, as in the solution by @Eric Naslund, is the right way to handle the problem.

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Hint: Since $$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$$ we can rewrite $$\sum_{n=1}^{K} \sin n = \frac {1}{2i}\sum_{n=1}^K (e^i)^n-\frac {1}{2i}\sum_{n=1}^K (e^{-i})^n$$ and both of these are geometric series.

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That's good. Can you show me a method just with real function. –  Leitingok Aug 25 '11 at 2:26
But $|| e^i|| = 1$, so it is not a geometric series! (I mean, it is but it does not converge).. Am I wrong? –  Ant Apr 8 '14 at 18:05
@Ant: It's a finite series so it does converge. It is true that the infinite series $\sum_{n=1}^\infty e^{inx}$ does converge, but that is not what appears here. –  Eric Naslund Apr 8 '14 at 18:10
of course.. but how does this solve the problem? I mean if we're talking about finite series every series converge. $\sum_0^k \sin(n)$ is a finite series hence converge. So why rewriting the partial sum of $\sin(n)$ that way? How does it help exactly? Thank you for your reply! :-) –  Ant Apr 8 '14 at 18:14
@Ant: The question asked to show that $\sum_{n=0}^k \sin(n)$ is bounded for all $k$. Boundedness is a less stringent condition than convergence. (All convergent sequences are bounded but not vice versa) –  Eric Naslund Apr 8 '14 at 18:36

A more general formula would be:

$$A_n=\sum_{k=1}^{n} \sin k\theta = \frac{\sin\theta+\sin n\theta-\sin(n+1)\theta}{2(1-\cos\theta)}$$ So $A_n$ is clearly bounded (just simply check the case where $\theta=1$).

The formula can be proved by induction using the trig identity: $\sin\alpha+\sin\beta=2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2})$.

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