Im trying to figure out if this sum $$\sum_{n=1}^\infty \frac{e^{ik}}{n^{2}} $$ is absolutely convergent. There is a step in my solution that Im not 100% sure of. $$\sum_{n=1}^\infty \left| \frac{e^{i*n}}{n^{2}}\right| =\sum_{n=1}^\infty \left| \frac{cos{(n)}+isin{(n)}}{n^{2}}\right| \le \sum_{n=1}^\infty \frac{\sqrt{(cos{(n))^{2}}+(sin{(n)}})^{2}}{n^{2}} = \sum_{n=1}^\infty \frac{1}{n^{2}} $$ Hence it is absolutely convergent also. But Im not sure if the inequality that I used can be used. Can one say the absolute value of a complex number is less than the distance?
Abs.convergence of a complex-valued sum
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analysis
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0Actually, you can write $=$ instead of $\le$. – 2017-01-29
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0Okay, so it is valid to write = ? And does my solution look ok? – 2017-01-29
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0Yes, it is right. – 2017-01-29
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1For any real number a, we have $|e^{ia}|$ = 1, because $\sin^2 a + \cos^2 a = 1$. – 2017-01-29
2 Answers
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Since the series is in the form of $$\sum_{n=1}^\infty \frac{a_n}{n^s},$$ where $s=2$ and $a_n=e^{in}$, it converges absolutely, since $a_n$ is bounded and Re$(s)=2>1$. This comes from a more general result - https://en.wikipedia.org/wiki/Dirichlet_series
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It's right except that it is not "$\sqrt{(cos{(n))^{2}}+(sin{(n)}})^{2}$" but "$\sqrt{(\cos{(n))^{2}}+(\sin{(n)})^{2}}$" (the line of the square-root. In fact, $|z|$ is defined either as the Euclidean distance of $z$ from the origin in the complex plane, or as $\sqrt{zz^*}$ (where $z^*$ is the conjugate of $z$). In any case, $|x+yi| = \sqrt{x^2+y^2}$ for any reals $x,y$.