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Use the result $1 + z + z^2...+z^n=\frac{z^{n+1}-1}{z-1}$ to sum the series to n terms

$1+\cos\theta+\cos2\theta+...$

also show that partial sums of series $\sum \cos (n\theta)$ is bounded when $0<\theta<\pi/2$

My attempt

so z can be written as $e^{i\theta}$ which means:

$1+ \cos \theta + \cos 2\theta ....+\cos n\theta + i(\sin \theta+\sin 2\theta+....+\sin n\theta)=\frac{z^{n+1}-1}{z-1}$

after this.. i dont know

Joe
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ketan
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1 Answers1

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Remember that $$ e^{it}=\cos t+i\sin t\;\;\;\;\forall t\in\Bbb C $$ and that $$ \sum_{j=0}^{n}z^j=\frac{1-z^{n+1}}{1-z}\;\;\forall z\in\Bbb C,\;\;|z|<1. $$ Thus $$ \sum_{j=0}^{n}\cos(j\theta)= \sum_{j=0}^{n}\Re{(e^{ij\theta})}= \Re\left(\sum_{j=0}^{n}(e^{ij\theta})\right)= \Re\left(\frac{1-e^{i\theta(n+1)}}{1-e^{i\theta}}\right) $$ The last term I wrote can be handled easily in order to be written explicitly and get the results you wanted.

Joe
  • 12,091
  • since we know the sum of series $\sum cos n\theta$ does this imply it is bounded irrespective of $\theta$? – ketan Aug 13 '15 at 10:28
  • For this purpose, write explicitly the last term I've written. Split $(1-e^{i\theta(n+1)})/(1-e^{i\theta})$ into its real and imaginary part. – Joe Aug 13 '15 at 10:58