Ordering of Series of Ordered Sequences/Proof 1

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:

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

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 $

$=$

$\ds \sum_{n \mathop = 0}^\infty \epsilon_n$

$\ds $

$=$

$\ds \epsilon$

$\ds $

$>$

$\ds 0$

Hence the result, by definition of inequality.

$\blacksquare$