Absolutely Convergent Series is Convergent/Complex Numbers

From ProofWiki
Jump to navigation Jump to search

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