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Prove the complex $$\sum_{k=0}^\infty (k+k^2i)^{-1}$$ series converge absolutely

Solution: by triangle inequality, we have $$|k+k^2i|\ge k^2-k \ge {k^2\over 2}$$ if $$k \ge 2 $$ how to understand this??

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  • $\begingroup$ Looks good to me. Though I'm not sure about your index of summation starting from $0$. And maybe say something about the reverse triangle inequality? $\endgroup$ – GenericMathematician Dec 9 '18 at 23:26
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I am assuming that the sum begins with $k=1$.

What you did is correct and proves that the series converges absolutely. You can also use the fact that$$\frac1{k+k^2i}=\frac{k-k^2i}{k^2+k^k}=\frac1{k+k^3}-\frac1{1+k^2}i$$and that both series$$\sum_{k=1}^\infty\frac1{k+k^3}\text{ and }\sum_{k=1}^\infty\frac1{1+k^2}$$converge, by the comparaison test.

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  • $\begingroup$ how to understand that triangle inequality?? $\endgroup$ – Ziang Xing Dec 10 '18 at 0:09
  • $\begingroup$ If $z,w\in\mathbb C$, then $|z|=|z-w+w|\leqslant|z-w|+|w|$ and therefore $|z-w|\geqslant|z|-|w|$. $\endgroup$ – José Carlos Santos Dec 10 '18 at 0:12

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