Difference of Absolutely Convergent Series
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Theorem
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent.
Then the series $\ds \sum_{n \mathop = 1}^\infty \paren {a_n - b_n}$ is absolutely convergent, and:
- $\ds \sum_{n \mathop = 1}^\infty \paren {a_n - b_n} = \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n$
Proof
The series $\ds \sum_{n \mathop = 1}^\infty \paren {-b_n}$ is absolutely convergent, as $\cmod {-b_n} = \cmod {b_n}$ for all $n \in \N$.
Then:
\(\ds \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty a_n + \paren {-1} \sum_{n \mathop = 1}^\infty b_n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty \paren {-1} b_n\) | Manipulation of Absolutely Convergent Series: Scale Factor | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \paren {a_n - b_n}\) | Sum of Absolutely Convergent Series |
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4.4$. Power Series