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I have a series: $$\sum_{n=1}^\infty(-1)^n\frac{\sin^4n}{\sqrt n}.$$ How can we prove that it converges?


Usually, with $\sin^4n$ we would use Comparison Test, but it only applies when the terms are nonnegative.

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    $\begingroup$ tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx $\endgroup$ – TomGrubb Dec 31 '15 at 14:35
  • $\begingroup$ @bburGsamohT Is that function strictly decreasing? You need that for the AST. $\endgroup$ – Gregory Grant Dec 31 '15 at 14:36
  • $\begingroup$ @GregoryGrant Compare it with the series $\sum(-1)^n\frac{1}{\sqrt{n}}$ $\endgroup$ – TomGrubb Dec 31 '15 at 14:37
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    $\begingroup$ If $\alpha$ is not a multiple of $2\pi$, then the partial sums $\sum_{n = 1}^N e^{i\alpha n}$ are bounded. $(-1)^n\sin^4 n = \frac{e^{i\pi n}}{16}(e^{in} - e^{-in})^4$. Ask Dirichlet. $\endgroup$ – Daniel Fischer Dec 31 '15 at 14:43
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    $\begingroup$ @bburGsamohT we can compare series that have only non-negative terms. $\endgroup$ – niar_q Dec 31 '15 at 14:47
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Hint: Noting that $$ \sin^4n=\frac{1}{8}(3-4\cos(2n)+\cos(4n)) $$ you have \begin{eqnarray} \sum_{n=1}^\infty(-1)^n\frac{\sin^4n}{\sqrt n}&=&\frac{3}{8}\sum_{n=1}^\infty(-1)^n\frac{1}{\sqrt n}-\frac{1}{2}\sum_{n=1}^\infty(-1)^n\frac{\cos(2n)}{\sqrt n}+\frac{1}{8}\sum_{n=1}^\infty(-1)^n\frac{\cos(4n)}{\sqrt n}. \end{eqnarray} Now you can do the rest to show that $\sum_{n=1}^\infty(-1)^n\frac{\cos(2n)}{\sqrt n}$ and $\sum_{n=1}^\infty(-1)^n\frac{\cos(4n)}{\sqrt n}$ are convergent.

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