Evaluate the convergence of the following series. $$\sum_{n=1}^\infty\dfrac{\sin n}{n}$$ $$\sum_{n=1}^\infty\dfrac{\sin^2 n}{n}$$ $$\sum_{n=1}^\infty\dfrac{\sin^3 n}{n}$$ The first sum is the imaginary part of the sum $\sum_{n=1}^{\infty} \frac{e^{i n}}{n}$. It's known that the series $f(z) = \sum_{n=1}^{\infty} \frac{z^n}{n}$ has a radius of convergence of 1, and it is also known that it is convergent everywhere on the boundary except for $z=1$. Noting that $|e^i| = 1$ we see that the first sum converges. I have an inkling the other sums could be handled in a similar way. But I don't know exactly how. Please help!
Evaluating convergence of sums of different power of sines over n
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complex-analysis
trigonometry
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1http://math.stackexchange.com/questions/2160951/sin2-n-n-series-when-sin-is-defined-on-complex-numbers/2160964#2160964 ... SBA Answered the second one yesterday – 2017-02-27
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0@DonaldSplutterwit :P Thanks for the mention. – 2017-02-27
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0Thank you for the link. Very useful! ;-) – 2017-02-27
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0$\texttt{Odd powers}$ converge. $\quad\texttt{^{1}} \implies {\pi - 1 \over 2}\,,\quad \texttt{^{3}}\implies {\pi \over 4}$. – 2017-03-01
1 Answers
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Hint: the identities $$ \sin^2(x)=\frac{1-\cos(2x)}2 $$ and $$ \sin^3(x)=\frac{3\sin(x)-\sin(3x)}4 $$ should be useful.