Absolutely Convergent Series is Convergent/Complex Numbers
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Theorem
Let $\ds \sum_{n \mathop = 1}^\infty z_n$ be an absolutely convergent series in $\C$.
Then $\ds \sum_{n \mathop = 1}^\infty z_n$ is convergent.
Proof
Let $z_n = u_n + i v_n$.
We have that:
\(\ds \cmod {z_n}\) | \(=\) | \(\ds \sqrt { {u_n}^2 + {v_n}^2}\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds \sqrt { {u_n}^2}\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds \size {u_n}\) |
and similarly:
- $\cmod {z_n} > \size {v_n}$
From the Comparison Test, the series $\ds \sum_{n \mathop = 1}^\infty u_n$ and $\ds \sum_{n \mathop = 1}^\infty v_n$ are absolutely convergent.
From Absolutely Convergent Real Series is Convergent, $\ds \sum_{n \mathop = 1}^\infty u_n$ and $\ds \sum_{n \mathop = 1}^\infty v_n$ are convergent.
By Convergence of Series of Complex Numbers by Real and Imaginary Part, it follows that $\ds \sum_{n \mathop = 1}^\infty z_n$ is convergent.
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4.3$. Series: Theorem