Ordering of Series of Ordered Sequences
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Theorem
Let $\sequence {a_n}$ and $\sequence {b_n}$ be two real 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 1
Let $\sequence {\epsilon_n}$ be the real 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.
$\blacksquare$
Proof 2
\(\ds \sum_{n \mathop = 0}^\infty b_n - \sum_{n \mathop = 0}^\infty a_n\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \paren {b_n - a_n}\) | Linear Combination of Convergent Series | |||||||||||
\(\ds \) | \(=\) | \(\ds b_0 - a_0 + \sum_{n \mathop = 1}^\infty \paren {b_n - a_n}\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds b_0 - a_0\) | as $b_n - a_n > 0$ | |||||||||||
\(\ds \) | \(>\) | \(\ds 0\) |
$\blacksquare$