Sum of Summations equals Summation of Sum/Infinite Sequence/Proof 1
Jump to navigation
Jump to search
Theorem
Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers $\Z$.
Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.
Let the fiber of truth of $R$ be infinite.
Let $\ds \sum_{\map R i} b_i$ and $\ds \sum_{\map R i} c_i$ be convergent.
Then:
- $\ds \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$
Proof
Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
\(\ds \sum_{\map R i} \paren {b_i + c_i}\) | \(=\) | \(\ds \sum_{\map R i} \paren {a_{i 1} + a_{i 2} }\) | by definition | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{\map R i} \paren {\sum_{1 \mathop \le j \mathop \le 2} a_{i j} }\) | Definition of Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{1 \mathop \le j \mathop \le 2} \paren {\sum_{\map R i} a_{i j} }\) | Exchange of Order of Summation: Finite and Infinite Series | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{\map R i} a_{i 1} + \sum_{\map R i} a_{i 2}\) | Definition of Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{\map R i} b_i + \sum_{\map R i} c_i\) | by definition |
$\blacksquare$