Absolutely Convergent Series is Convergent/Complex Numbers

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Theorem

Let $\displaystyle \sum_{n \mathop = 1}^\infty z_n$ be an absolutely convergent series in $\C$.


Then $\displaystyle \sum_{n \mathop = 1}^\infty z_n$ is convergent.


Proof

Let $z_n = u_n + i v_n$.

We have that:

\(\displaystyle \cmod {z_n}\) \(=\) \(\displaystyle \sqrt { {u_n}^2 + {v_n}^2}\)
\(\displaystyle \) \(>\) \(\displaystyle \sqrt { {u_n}^2}\)
\(\displaystyle \) \(>\) \(\displaystyle \size {u_n}\)

and similarly:

$\cmod {z_n} > \size {v_n}$

From the Comparison Test, the series $\displaystyle \sum_{n \mathop = 1}^\infty u_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty v_n$ are absolutely convergent.

From Absolutely Convergent Series is Convergent: Real Numbers, $\displaystyle \sum_{n \mathop = 1}^\infty u_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty v_n$ are convergent.

By Convergence of Series of Complex Numbers by Real and Imaginary Part, it follows that $\displaystyle \sum_{n \mathop = 1}^\infty z_n$ is convergent.

$\blacksquare$


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