description/proof of that for convergent series on \(1\)-dimensional Euclidean metric space and real number, series with terms as corresponding terms multiplied by number converges with convergence multiplied by number
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of convergence of series on metric space.
Target Context
- The reader will have a description and a proof of the proposition that for any convergent series on the \(1\)-dimensional Euclidean metric space and any real number, the series with the terms as the corresponding terms multiplied by the number converges with the convergence multiplied by the number.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(J\): \(\subseteq \mathbb{N}\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(s\): \(: J \to \mathbb{R}\), such that \(\sum_{j \in J} s (j) = r \in \mathbb{R}\)
\(r'\): \(\in \mathbb{R}\)
\(r' s\): \(: J \to \mathbb{R}, j \mapsto r' s (j)\)
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Statements:
\(\sum_{j \in J} r' s (j) = r' r \in \mathbb{R}\)
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2: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite and with the case that \(r' = 0\), and suppose otherwise, thereafter; Step 2: for each \(\epsilon\), take an \(N\) such that for each \(N \lt n\), \(\vert r - \sum_{j \in \{J_1, ..., J_n\}} s (j) \vert \lt \epsilon / \vert r' \vert\); Step 3: see that for each \(N \lt n\), \(\vert r' r - \sum_{j \in \{J_1, ..., J_n\}} r' s (j) \vert \lt \epsilon\).
Step 1:
When \(J\) is finite, it holds, because they are some finite operations.
Let us suppose otherwise, hereafter.
Let us suppose that \(r' = 0\).
\(\sum_{j \in J} r' s (j) = \sum_{j \in J} 0 s (j) = \sum_{j \in J} 0 = 0\).
\(r' r = 0 r = 0\).
So, \(\sum_{j \in J} r' s (j) = r' r \in \mathbb{R}\).
Let us suppose that \(r' \neq 0\) hereafter.
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N \lt n\), \(\vert r - \sum_{j \in \{J_1, ..., J_n\}} s (j) \vert \lt \epsilon / \vert r' \vert\), by the definition of convergence.
Step 3:
For each \(n \in \mathbb{N}\) such that \(N \lt n\), \(\vert r' r - \sum_{j \in \{J_1, ..., J_n\}} r' s (j) \vert = \vert r' (r - \sum_{j \in \{J_1, ..., J_n\}} s (j)) \vert = \vert r' \vert \vert r - \sum_{j \in \{J_1, ..., J_n\}} s (j) \vert \lt \vert r' \vert \epsilon / \vert r' \vert = \epsilon\).
That means that \(\sum_{j \in J} r' s (j) = r' r \in \mathbb{R}\).
