Ordering of Series of Ordered Sequences/Proof 2

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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

\(\ds \sum_{n \mathop = 0}^\infty b_n - \sum_{n \mathop = 0}^\infty a_n\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \paren {b_n - a_n}\) 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 $b_n - a_n > 0$
\(\ds \) \(>\) \(\ds 0\)

$\blacksquare$