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I am trying to determine whether the two series as follows converge and then evaluate their sums: $$A(x):=\sum_{n=0}^{\infty}2^{-n}cos(nx)$$ $$B(x):=\sum_{n=0}^{\infty}2^{-n}sin(nx)$$ I am thinking of using Dirichlet' theorem to prove that they converge:

a)The partial sums $A_{n}$ form a bounded sequence.

b) $b_{0} \geq b_{1} \geq b_{2} \geq...$

c) $\lim_{n \rightarrow \infty}b_{n} = 0;$

Then $\sum b_{n}a_{n}$ converges.

I think I can prove convergence but I cannot find a way to evaluate its sums. Thanks in advance for any tips!

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  • $\begingroup$ The convergence can be obtained using the comparison criterion. (The absolute value of cos(nx) or sin(nx) is bounded by 1 and $\sum_n 2^{-n}$ is a geometric series. $\endgroup$ – Thibaut Dumont Nov 9 '17 at 23:04
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Using $e^{i\theta}=\cos(\theta)+i\sin(\theta)$, we get $$A(x)+iB(x) =\sum_{n=0}^{\infty}\frac{\cos(nx)+i\sin(nx)}{2^n} =\sum_{n=0}^{\infty}\left(\frac{e^{ix}}{2}\right)^n = \frac{2}{2-e^{ix}}. \tag 1$$ But $$\frac{2}{2-e^{ix}} = \frac{4-2\cos(x)}{5-4\cos(x)}+i \frac{2\sin(x)}{5-4\cos(x)}. \tag 2$$ Therefore, $$A(x) = \frac{4-2\cos(x)}{5-4\cos(x)},\tag 3$$ and $$B(x) = \frac{2\sin(x)}{5-4\cos(x)}.\tag 4$$

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  • $\begingroup$ Hello, could you please give me another hint to evaluate that sum? $\endgroup$ – Joseph Nov 9 '17 at 21:27
  • $\begingroup$ if I expand the term $e^{ix}$ and then multiple by a conjugate them, after cleaning up using algebra, I have the real part as $\frac{4-2cosx}{5-4cosx}$ and the imaginary part as $\frac{2sinx}{5-4cosx}$ which perfectly correspond to A(x) and B(x) on the left. Am I doing alright here? Thanks so much! $\endgroup$ – Joseph Nov 9 '17 at 21:45
  • $\begingroup$ @Hugh Check the updated answer! $\endgroup$ – Math Lover Nov 9 '17 at 21:46
  • $\begingroup$ yay! I got it. Thanks so much, Math Lover $\endgroup$ – Joseph Nov 9 '17 at 21:47

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