Show that the infinite series
$$\sum_{x = 1}^{\infty} \frac{\sin(x)}{x}$$ is convergent.
Please answer so that a Calculus student can understand.
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The Dirichlet Test says that if $$ \left|\,\sum_{k=0}^na_k\,\right|\le M $$ and $$ \lim_{k\to\infty}b_k=0 $$ where $$ 0\le b_k\le b_{k-1} $$ then $$ \sum_{k=0}^\infty a_kb_k $$ is convergent.
Since $$ \begin{align} \left|\,\sum_{k=0}^n\sin(k)\,\right| &=\left|\,\mathrm{Im}\left(\sum_{k=0}^ne^{ik}\right)\,\right|\\ &=\left|\,\mathrm{Im}\left(\frac{e^{i(n+1)}-1}{e^i-1}\right)\,\right|\\ &\le\frac1{\sin(1/2)} \end{align} $$ we get that $$ \sum_{k=1}^\infty\frac{\sin(k)}{k} $$ converges.
robjohn
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Can you explain the bound by 1/sin(1/2)? – Seth Jun 12 '14 at 02:26
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2$|e^{i(n+1)}-1|\le2$ and $|e^i-1|=\left|,2ie^{i/2}\frac{e^{i/2}-e^{-i/2}}{2i},\right|=2\sin(1/2)$. – robjohn Jun 12 '14 at 03:04
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This is the Dirichlet criteria, $\frac{1}{x}$ is monotone decreasing to zero and
$\sum\limits_{x=1}^m \sin x$ is bounded.
In response to user61527's question note further that the boundedness can be seen from the well known formula,
$$ \sum\limits_{i=1}^{n} \sin i\theta=\frac{\cos \frac{\theta}{2}- \cos \left( n+ \frac{1}{2} \right) \theta}{2 \sin \frac{\theta}{2} } $$ which gives a bound of
$$\frac{1}{ |\sin \frac{\theta}{2}| } $$
Rene Schipperus
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1How would you give a simple explanation that $\sum_{x = 1}^m \sin x$ is bounded? – Jun 12 '14 at 01:46
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@user61527, prove it, say by induction. You can also try the complex numbers approach robjohn writes about in his answer. – DonAntonio Jun 12 '14 at 03:45