Ordering of Series of Ordered Sequences

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Theorem

Let $\sequence {a_n}$ and $\sequence {b_n}$ be two sequences.

Let $\ds \sum_{n \mathop = 1}^{\infty} a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be convergent series.

For each $n \in \N$, let $a_n < b_n$.


Then:

$\ds \sum_{n \mathop = 0}^\infty a_n < \sum_{n \mathop = 0}^\infty b_n$


Proof

Let $\sequence {\epsilon_n}$ be the sequence defined by:

$\forall n \in \N : b_n - a_n$


From Linear Combination of Convergent Series, $\ds \sum_{n \mathop = 0}^\infty \epsilon_n$ is convergent with sum $\epsilon > 0$.


Then:

\(\ds \sum_{n \mathop = 0}^\infty b_n - \sum_{n \mathop = 0}^\infty a_n\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {a_n + \epsilon_n} - \sum_{n \mathop = 0}^\infty a_n\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \epsilon_n\) Linear Combination of Convergent Series
\(\ds \) \(=\) \(\ds \epsilon\)
\(\ds \) \(>\) \(\ds 0\)

Hence the result, by definition of inequality.

$\blacksquare$