How can I find the solution to the series $\sum_{n=1} ^\infty n^2(\cos(nx) + i \sin(nx))$
complex/trigonometric series
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complex-analysis
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2Fabian is right, it converges for complex $x$ if \Im(x)>0. In that case, you can get the limit by writing the terms as $n^2\mathrm{e}^{\mathrm{i}nx}$, writing that as the second derivative of $-\mathrm{e}^{\mathrm{i}nx}$ with respect to $x$, summing the geometric series and then taking the second derivative. (To make it a bit easier, you can add the $n=0$ term, since it's $0$ anyway.) – 2011-04-05
1 Answers
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As joriki pointed out, the series diverges for real $x$. I will in the following assume that $\text{Im} x >0$.
We have $ \begin{align} \sum_{n=1}^\infty n^2[\cos(nx) + i \sin(nx)] &= \sum_{n=0} ^\infty n^2 e^{i n x} = -\partial_x^2 \sum_{n=0} ^\infty e^{i n x}\\ &=\partial_x^2 \frac{1} {e^{ix}-1}. \end{align}$
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0@joriki: you are right (but still does not look very simple to me ;-)) I will implement your suggestion... – 2011-04-05