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I can't seem to solve this one. I've tried Dirichlet's test but to no avail. Any help is greatly appreciated.

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If you want, you can actually calculate what it converges to by appealing to complex numbers. The sum is the imaginary part of $$ \sum \frac{e^{nix}}{2^n} = \sum \left(\frac{e^{ix}}{2}\right)^n = \frac{2}{2-e^{ix}}, $$ so your sum converges to $\frac{2 \sin x}{5 - 4\cos x}$.

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HINT: To show absolute convergence use that $\sin$ is bounded by 1 (I assume that you want $x$ to take real values).

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    right, i forgot that. sorry for being such a beginner. thank you!2010-10-09