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If the real and imaginary parts of a complex series converge absolutely, then the complex series converges absolutely.

Is this true? If we write our complex series $\sum_{k=0}^{\infty} b_k = \left(\sum_{k=0}^{\infty} u_k \right)+ i \left( \sum_{k=0}^{\infty} v_k \right)$ into real and complex parts and the individual terms converge absolutely, then we can conclude that they converge, so $\sum_{k=0}^{\infty} b_k$ is convergent, not necessarily absolutely. What would be a good counterexample for this?

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  • $\begingroup$ Observe that $\sum z\le \sum |z| \le \sum |a|+i\sum |b|$. If $a_n\le b_n,\forall n$ then $\lim a_n\le \lim b_n$. $\endgroup$ – Masacroso May 25 '16 at 1:33
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Yes, that is true.

$\sum_{n=x}^{y}(a+bi) = \sum_{n=x}^{y}a+\sum_{n=x}^{y}bi$, even if both a and b are functions of n and the series is infinite.

If the sums of both parts converge, then because of that identity, the complex series must also converge.

Hope that helped.

EDIT: It not only converges, but converges absolutely.

Proof (I'm kinda working backwards): $$|a+bi| = \sqrt{a^2+b^2}$$ For any real a and b, $$2|a||b|\ge0$$ Duh. Now add $|a|^2$ and $|b|^2$ to both sides. $$2|a||b|+|a|^2+|b|^2\ge|a|^2+|b|^2$$ Trust me. I'm getting somewhere with this. Now take the square root of both sides. $$|a|+|b|\ge\sqrt{a^2+b^2}$$ And therefore: $$|a|+|b|\ge|a+bi|$$

So, if the sums of $|a|$ and $|b|$ are not infinite, the sum of $|a+bi|$ cannot be infinite, and therefore must converge.

It took me a long time to think of this, so you better make good use of it :P.

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  • $\begingroup$ What can we say about absolute convergence though? $\endgroup$ – user167857 May 25 '16 at 1:12
  • $\begingroup$ Ooh... I'm not sure. If I can think of a counterexample I'll add it to my answer. $\endgroup$ – Polygon May 25 '16 at 1:35
  • $\begingroup$ @user167857 Yes, it does converge absolutely $\endgroup$ – Polygon May 25 '16 at 1:45

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