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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2If by "solution" you mean "limit", then you can't -- this series diverges for all $x$. To see this, note that the terms are $n^2\mathrm{e}^{\mathrm{i}nx}$, so the absolute value of the terms diverges, instead of going to $0$ as it must if the series is to converge. – 2011-04-05
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0@joriki: I will diverge for real $x$ but might well converge for complex $x$. – 2011-04-05
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0@Fabian: Well I hope you converge again :-) But seriously: You're right, although that would be rather an unusual way to write it down if $x$ were meant to be complex. – 2011-04-05
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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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0As I pointed out above, you can simplify it somewhat by including the $n=0$ term, which is $0$; then you don't have to take out a factor of $\mathrm{e}^{\mathrm{i}x}$. – 2011-04-05
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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