Compute the series


Hint: the answer is in fact 0


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Recall that $\cos(x)=(e^{ix}+e^{-ix})/2$ and $\sin(x)=(e^{ix}-e^{-ix})/2i$, so your sum can be rewritten as

$$\sum_{j=1}^k\biggl[\frac{e^{j\pi i/k}+e^{-j\pi i/k}}2\biggr]^n\,\biggl[\frac{e^{nj\pi i/k}-e^{-nj\pi i/k}}{2i}\biggr]\,.$$

Applying binomial theorem on the first factor (of each summand) your sum becomes

$$\begin{align*} &\,\frac1{2^{n+1}i}\sum_{j=1}^k\sum_{r=0}^n\binom nr\bigl(e^{j\pi i/k}\bigr)^r\bigl(e^{-j\pi i/k}\bigr)^{n-r}\,\bigl[e^{nj\pi i/k}-e^{-nj\pi i/k}\bigr]\\[2mm] &\,\frac1{2^{n+1}i}\sum_{r=0}^n\binom nr\sum_{j=1}^k\bigl[e^{2rj\pi i/k}-e^{2j(r-n)\pi i/k}\,\bigr]\,. \end{align*}$$

Now $\sum_{j=1}^ke^{2rj\pi i/k}$ is the sum of a geometric progression, with sum equals to

$$e^{2r\pi i/k}\frac{e^{2rk\pi i/k}-1}{e^{2r\pi i/k}-1}=0\,,$$

because $e^{2r\pi i}=1$; similarly,

$$\sum_{j=1}^ke^{2j(r-n)\pi i/k}=e^{2(r-n)\pi i/k}\frac{e^{2(r-n)k\pi i/k}-1}{e^{2(r-n)\pi i/k}-1}=0\,,$$

and so all the inner summands above are equal to $0$.

  • 1
    $\begingroup$ You lost a factor of $i$ in the denominator of the $\sin$ expansion. Of course it doesn't matter here since the result is $0$. Nice work. $\endgroup$ – Daniel Fischer Aug 25 '13 at 21:45
  • $\begingroup$ @DanielFischer Thanks, correcting now. $\endgroup$ – Matemáticos Chibchas Aug 25 '13 at 21:50

Hint: by symmetry $\sum (\zeta+\zeta^{-1})^n\zeta^n=\sum (\zeta+\zeta^{-1})^n\zeta^{-n}$ as $\zeta$ is summed over the $k$th roots of $1$.


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