Ordering of Series of Ordered Sequences

Theorem
Let $\left\langle{ a_n }\right\rangle$ and $\left\langle{ b_n }\right\rangle$ be two sequences.

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

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

Then:
 * $\displaystyle \sum_{n \mathop = 0}^{\infty} a_n < \displaystyle \sum_{n \mathop = 0}^{\infty} b_n$

Proof
Let $\left\langle{ \epsilon_n }\right\rangle$ be the sequence defined by:
 * $\forall n \in \N : b_n - a_n$

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

Then:

Hence the result, by definition of $\displaystyle \sum_{n \mathop = 0}^{\infty} a_n < \displaystyle \sum_{n \mathop = 0}^{\infty} b_n$.