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In a question I am given the sequence $e^{i\theta}+\frac{1}{2}e^{2i\theta}+\frac{1}{4}e^{3i\theta}+\dots$

I can show that the sum to infinity is $S_\infty=\frac{2e^{i\theta}}{2-e^{i\theta}}=\frac{2\cos\theta+2i\sin\theta}{2-\cos\theta-i\sin\theta}$

From this it asks me to prove that $\cos\theta + \frac{1}{2}\cos2\theta + \frac{1}{4}\cos3\theta + \dots = \frac{4\cos\theta-2}{5-4\cos\theta}$

I know $\cos\theta + \frac{1}{2}\cos2\theta + \frac{1}{4}\cos3\theta + \dots = \frac{2\cos\theta+2i\sin\theta}{2-\cos\theta-i\sin\theta}-(i\sin\theta+\frac{1}{2}i\sin2\theta+\frac{1}{4}i\sin3\theta+\dots)$ Where do I go from there?

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    $\begingroup$ Just take the real part of $S_\infty$. $\endgroup$ – Lord Shark the Unknown Sep 26 '17 at 17:33
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Given Lord Shark the Unknnow's hint:

$$ S_{\infty}=\frac{2\cos(\theta)+2i\sin(\theta)}{2-\cos(\theta)-i\sin(\theta)}=\frac{2[\cos(\theta)+i\sin(\theta)][(2-\cos(\theta))+i\sin(\theta)]}{(2-\cos(\theta))^2 +\sin^2(\theta)}= \frac{2[2\cos(\theta)-(\cos^2(\theta)+\sin^2(\theta))+2i\sin(\theta)]}{5 -4\cos(\theta)}= \frac{4\cos(\theta)-2}{5-4\cos(\theta)}+i\frac{4\sin(\theta)}{5-4\cos(\theta)}$$

$$Re (S_{\infty}) =\frac{4\cos(\theta)-2}{5-4\cos(\theta)}$$

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Hint: Use $$\cos(n\theta)=\operatorname{Re}\left[e^{in\theta}\right]$$

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Th real part of $$\dfrac{2\cos t+2i\sin t}{2-\cos t-i\sin t}=\dfrac{(2\cos t+2i\sin t)(2-\cos t+i\sin t)}{(2-\cos t)^2+(\sin t)^2}$$

is $$=\dfrac{4\cos t-2(\cos^2t+\sin^2t)}{4+(\cos^2t+\sin^2t)-4\cos t}=?$$

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Hint: As you know $S_{\infty}$, You just need to equate real and complex parts. Calculate real part of $S_{\infty}$ by multiplying the numerator and denominator by $2-\cos \theta+i\sin \theta$.

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Well, we have:

$$\mathscr{S}_{\space\theta}:=\exp\left(\theta i\right)+\frac{1}{2}\cdot\exp\left(2\theta i\right)+\frac{1}{4}\cdot\exp\left(3\theta i\right)+\dots=$$ $$\frac{1}{1}\cdot\exp\left(1\theta i\right)+\frac{1}{2}\cdot\exp\left(2\theta i\right)+\frac{1}{4}\cdot\exp\left(3\theta i\right)+\dots=$$ $$\sum_{\text{n}=0}^\infty\frac{\exp\left(\left(\text{n}+1\right)\theta i\right)}{2^\text{n}}=\frac{2\exp\left(\theta i\right)}{2+\exp\left(\theta i\right)}\tag1$$

When:

$$\lim_{\text{n}\to\infty}\space\left|\frac{\left(\frac{\exp\left(\left(\left(\text{n}+1\right)+1\right)\theta i\right)}{2^{\text{n}+1}}\right)}{\left(\frac{\exp\left(\left(\text{n}+1\right)\theta i\right)}{2^\text{n}}\right)}\right|=\frac{1}{2}\cdot\frac{1}{\exp\left(\Im\left(\theta\right)\right)}<1\tag2$$

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