Difference of Absolutely Convergent Series

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Theorem

Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent.


Then the series $\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n - b_n}$ is absolutely convergent, and:

$\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n - b_n} = \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n$


Proof

The series $\displaystyle \sum_{n \mathop = 1}^\infty \paren {-b_n}$ is absolutely convergent, as $\cmod {-b_n} = \cmod {b_n}$ for all $n \in \N$.

Then:

\(\displaystyle \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty a_n + \paren {-1} \sum_{n \mathop = 1}^\infty b_n\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty \paren {-1} b_n\) Manipulation of Absolutely Convergent Series: Scale Factor
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n - b_n}\) Sum of Absolutely Convergent Series

$\blacksquare$


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