# Ordering of Series of Ordered Sequences

## 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.

$\blacksquare$