Sum 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

Let $\epsilon \in \R_{>0}$.

From Tail of Convergent Series tends to Zero, it follows that there exists $M \in \N$ such that:

$\ds \sum_{n \mathop = M + 1}^\infty \cmod {a_n} < \dfrac \epsilon 2$

and:

$\ds\sum_{n \mathop = M + 1}^\infty \cmod {b_n} < \dfrac \epsilon 2$

For all $m \ge M$, it follows that:

\(\ds \cmod {\sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n - \sum_{n \mathop = 1}^m \paren {a_n + b_n} }\) \(=\) \(\ds \cmod {\sum_{n \mathop = m + 1}^\infty a_n + \sum_{n \mathop = m + 1}^\infty b_n}\)
\(\ds \) \(\le\) \(\ds \sum_{n \mathop = m + 1}^\infty \cmod {a_n} + \sum_{n \mathop = m + 1}^\infty \cmod {b_n}\) by Triangle Inequality
\(\ds \) \(\le\) \(\ds \sum_{n \mathop = M + 1}^\infty \cmod {a_n} + \sum_{n \mathop = M + 1}^\infty \cmod {b_n}\)
\(\ds \) \(<\) \(\ds \epsilon\)


By definition of convergent series, it follows that:

\(\ds \sum_{n \mathop = 1}^\infty a_n + \sum_{n \mathop = 1}^\infty b_n\) \(=\) \(\ds \lim_{m \mathop \to \infty} \sum_{n \mathop = 1}^m \paren {a_n + b_n}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}\)


To show that $\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is absolutely convergent, note that:

\(\ds \sum_{n \mathop = 1}^\infty \cmod {a_n} + \sum_{n \mathop = 1}^\infty \cmod {b_n}\) \(=\) \(\ds \sum_{n \mathop = 1}^\infty \paren {\cmod {a_n} + \cmod {b_n} }\) as shown above
\(\ds \) \(\ge\) \(\ds \sum_{n \mathop = 1}^\infty \cmod {a_n + b_n}\) by Triangle Inequality

$\blacksquare$


Sources

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