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I have to write $\sum_{k=0}^{\infty} \cos(k \pi / 6)$ in form: $a+bi$.

I think I should consider $z=\cos(k \pi / 6)+i\sin(k \pi / 6)$ and also use the fact that $\sum _{k=0}^{\infty}x^n=\frac{1}{1-x}$ if $|x|<1$

But i don't know if thats correct and how can I transform this.

luka5z
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  • See http://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression. But $$|e^{ia}|=1\not<1$$ So, we can not apply http://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series – lab bhattacharjee Oct 19 '14 at 17:44
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    Sure you are not asked about $$\sum_kr^k\cos(k\pi/6)$$ instead? – Did Oct 19 '14 at 18:03

2 Answers2

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$$\cos\frac{k\pi}6=\begin{cases}1&\;,\;\;k=0,12,24,\ldots=0\pmod{12}\\ \frac{\sqrt3}2&\;,\;\;k=1,11,13,23,\ldots=1,11\pmod{12}\\ \frac12&\;,\;\;k=2,10,14,22,\ldots=2,10\pmod{12}\\etc.\end{cases}$$

You can fill up the lines missing in "etc."...and thus the series cannot possibly converge (why?)

Timbuc
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You want divergent summation of real numbers written a complex number. Perhaps the goal of the exercise is to check whether this sum is convergent?

If this is the case, some hints:

  • complete your summation with $i\sum_{k=0}^\infty \sin(k\pi/6) $, you want the real part of this new summation.

  • Remember that $cis(kx) = cis(x)^k$

  • Observe that the summation can't converge because $cis(x)^n$ diverges

villasv
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