# Does $\sum_{i=1}^n \sin(i)$ converge? Prove it.

I know that $\sum_{i=1}^n \sin(i)$ doesn't converge. But I'm having trouble proving it.

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Can you show whatever work you have done? –  Calvin Lin Jan 13 '13 at 6:40
Note that each of the intervals $[0+\pi/6,\pi-\pi/6]$, $[2\pi+\pi/6, 3\pi-\pi/6]$, $\ldots$ contains an integer; and, on each of these intervals, the value of $\sin$ is at least $1/2$. –  David Mitra Jan 13 '13 at 12:26

If a summation $\sum_{n=1}^\infty a_n$ converges, then $a_n\rightarrow 0$. But $sin(n)$ does not go to zero as $n\rightarrow \infty$.

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Surely this is slightly incomplete. I know that $\sin(x)$ has no limit as the real variable $x$ goes to infinity. But you haven't justified that the sequence $\sin(n)$ evaluated at integers $n$ doesn't happen to converge. –  Hammerite Jan 13 '13 at 7:37
True. It suffies to argue, for example, that $\sin x$ and $\sin(x+1)$ can't both be close to $0$; in fact the function $\max\{|\sin x|,|\sin(x+1)|\}$ is always at least $\sin(\pi-1/2) > 0.479$. In particular the sequence $\{a_n\}$ is not Cauchy, hence can't converge to any limit, let alone $0$. –  Greg Martin Jan 13 '13 at 8:03

Hint: If $\sum_{i=1}^{\infty} a_n$ converges, then $a_n$ tends to 0.

Show that since $\frac {\pi}{1}$ is not rational, then by the pigeonhole principle, for any $N$, there exists $n > N$ such that $| \sin n | > \frac {1}{2}$.

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I am having some problem following everything. Could you may be elaborate what has irrationality of $\pi$ and pigeonhole principle anything to do here? –  007resu Jan 13 '13 at 7:29
@Freddy Show that there always exists some large $n$ which gets arbitrarily close to $(2k+1)\pi$, for some integer $k$. –  Calvin Lin Jan 13 '13 at 7:31

If $\sum a_n$ converges, $\lim a_n=0$.

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The easiest way is what Calvin, Vahid and user57101 have suggested. However, you can compute $a_n = \displaystyle \sum_{k=1}^n \sin(k)$ and see for yourself how the sequence behaves for large $n$. \begin{align} 2\sin(1/2) a_n & = 2\sum_{k=1}^n \sin(k) \sin(1/2) = \sum_{k=1}^n \left(\cos(k-1/2)-\cos(k+1/2) \right)\\ & = (\cos(1/2) - \cos(3/2)) + (\cos(3/2) - \cos(5/2)) + \cdots\\ & + (\cos(n-3/2) - \cos(n-1/2)) + (\cos(n-1/2) - \cos(n+1/2))\\ & = \cos(1/2) - \cos(n+1/2) \end{align} Hence, $$a_n = \dfrac{\cos(1/2) - \cos(n+1/2)}{2 \sin(1/2)} = \dfrac12 \cot(1/2) - \dfrac{\cos(n+1/2)}{2 \sin(1/2)}$$ which oscillates about $\dfrac12 \cot(1/2)$ as $n \to \infty$ and hence it doesn't converge. However, if you want to assign a regularized value to the series, then you can assign it a value $\dfrac12 \cot(1/2)$.

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