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!

  • $\begingroup$ 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

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 $$B(x) = \frac{2\sin(x)}{5-4\cos(x)}.\tag 4$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.