# Convergence / divergence examination <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeX

Convergence / divergence examination $\sum _{k=1}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{k}\left(a\right)}{k}$
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talhekh
If $\mathrm{sin}\left(a\right)=0$, then the series obviously converges.
If $\mathrm{sin}\left(a\right)=1$, then the series rewrites $\sum \frac{1}{k}$ which diverges.
If $\mathrm{sin}\left(a\right)=-1$, then the series rewrites $\sum \frac{\left(-1{\right)}^{k}}{k}$ which converges by Alternating Test.
If $|\mathrm{sin}\left(a\right)|<1$ and $\mathrm{sin}\left(a\right)\ne 0$, then one has
$|\frac{{\mathrm{sin}}^{k+1}\left(a\right)/\left(k+1\right)}{{\mathrm{sin}}^{k}\left(a\right)/k}|=|\mathrm{sin}\left(a\right)|×\frac{k+1}{k}⟶|\mathrm{sin}\left(a\right)|$
Because $|\mathrm{sin}\left(a\right)|<1$, then by the ratio test, the series converges.
In conclusion, the series converges iff $\mathrm{sin}\left(a\right)\ne 1$, i.e.