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Thread: Convergence of Infinite Series

  1. #1
    Member mohammadfawaz's Avatar
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    Convergence of Infinite Series

    Hello,

    Suppose $\displaystyle 0<\alpha<\frac{\pi}{2}$ and $\displaystyle \{z_n\}$ is a sequence of complex numbers satisfying $\displaystyle -\alpha<arg(z_n)<\alpha$. Prove that the two infinite series $\displaystyle \sum_{n=1}^\infty {z_n}$ and $\displaystyle \sum_{n=1}^\infty |{z_n}|$ are both convergent or both divergent.

    First, I'm trying to show that if the first converges, then the second also converges. Setting $\displaystyle z_n=x_n+iy_n$, I know that all $\displaystyle x_n$ are positive and hence $\displaystyle \sum_{n=1}^\infty |x_n|$ converges. But for $\displaystyle y_n$, this is still unclear for me.

    Thank you

    Mohammad
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    For all $\displaystyle n$ we have $\displaystyle x_n\neq 0$ (why?). Then,

    $\displaystyle \left |\dfrac{y_n}{x_n}\right |=\left | \tan (\arg z_n)\right |\leq \left |{\tan \alpha}\right |=r<+\infty$

    So, $\displaystyle |y_n|\leq r|x_n|$ for all $\displaystyle n$ and $\displaystyle \sum_{n=1}^{+\infty}r|x_n|$ is convergent which implies $\displaystyle \sum_{n=1}^{+\infty}|y_n|$ is convergent.
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