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Find the sum of the series $\sum_{n=0}^{\infty}\frac{\cos(nx)}{2^{n}}$ and $\sum_{n=0}^{\infty}\frac{\sin(nx)}{2^{n}}$.

Hint: Rewrite the trigonometric functions using complex exponentials. $$$$ This is what I have so far:

$\cos(nx) = \frac{e^{inx}+e^{-inx}}{2}$ and $\sin(nx) = \frac{e^{inx}-e^{-inx}}{2i}$

By substituting in the values into the given series, we get:

$\frac{1}{2}\sum_{n=0}^{\infty}\frac{e^{inx}+e^{-inx}}{2}$ + $\frac{1}{2}\sum_{n=0}^{\infty}\frac{e^{inx}-e^{-inx}}{2i}$.

I have no clue what to do from here, please help.

Edit: used latex codes \cos and \sin.

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  • $\begingroup$ Use the $C+iS$ method of summation, it will be easier. $\endgroup$ – Awe Kumar Jha Nov 7 '18 at 3:33
  • $\begingroup$ @AweKumarJha I'm confused by what you're suggesting $\endgroup$ – AmR Nov 7 '18 at 3:41
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First, $e^{inx} = \cos(nx)+i\sin(nx)$.

So, $\displaystyle\sum_{n=0}^\infty \left(\frac{\cos(nx)}{2^n}+i\frac{\sin(nx)}{2^n}\right) = \displaystyle\sum_{n=0}^\infty \frac{e^{inx}}{2^n} = \displaystyle\sum_{n=0}^\infty \left(\frac{e^{ix}}{2}\right)^n=\frac{2}{2-e^{ix}}$.

Convert the RHS to trig functions, then compare re and im parts.

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  • $\begingroup$ so is the way I did the substitution wrong? $\endgroup$ – AmR Nov 7 '18 at 3:40
  • $\begingroup$ @AmR somewhat wrong because the denominators should be $2^n$. Say you have that denominator, then it is not wrong, although you need to compute for both $\sum \frac{e^{inx}}{2^n}$ and $\sum \frac{e^{-inx}}{2^n}$, and do it twice; thus, more work. $\endgroup$ – Isko10986 Nov 7 '18 at 19:24

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