minwaardekn

2022-07-01

Finding convergence/abs convergence/divergence of series with trig function
$\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(4n\right)}{{4}^{n}}$

svirajueh

Expert

We can note that
$\sum _{n\ge 1}\frac{\mathrm{sin}\left(4n\right)}{{4}^{n}}=\text{Im}\left(\sum _{n\ge 1}{\left(\frac{{e}^{4i}}{4}\right)}^{n}\right)$
and since
$|\frac{{e}^{4i}}{4}|=\frac{1}{4}$
$|\frac{{e}^{4i}}{4}|=\frac{1}{4}$
we can calculate the sum
$\sum _{n\ge 1}{\left(\frac{{e}^{4i}}{4}\right)}^{n}=\frac{{e}^{4i}}{4-{e}^{4i}}$
and so
$\text{Im}\left(\sum _{n\ge 1}{\left(\frac{{e}^{4i}}{4}\right)}^{n}\right)=\text{Im}\left(\frac{{e}^{4i}}{4-{e}^{4i}}\right)=\frac{4\mathrm{sin}\left(4\right)}{17-8\mathrm{cos}\left(4\right)}\approx -0.13619.$

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