description/proof of that for sequence on \(1\)-dimensional Euclidean metric space, if sequence converges, sequence of arithmetic means of leading elements converges with convergence
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of convergence of sequence on metric space.
Target Context
- The reader will have a description and a proof of the proposition that for any sequence on the \(1\)-dimensional Euclidean metric space, if the sequence converges, the sequence of the arithmetic means of the leading elements converges with the convergence.
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:
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(s\): \(: \mathbb{N} \to \mathbb{R}\)
\(s'\): \(: \mathbb{N} \to \mathbb{R}, n \mapsto \sum_{j \in \{0, ..., n\}} s (j) / (n + 1)\)
\(r\): \(\in \mathbb{R}\)
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Statements:
\(lim s = r\)
\(\implies\)
\(lim s' = r\)
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2: Note
In fact, for any \(m \in \mathbb{N}\), \(s'': \mathbb{N} \to \mathbb{R}, n \mapsto \sum_{j \in \{0, ..., n\}} s (j) / (n + m)\) converges to \(r\), because \(s'' (n) = \sum_{j \in \{0, ..., n\}} s (j) / (n + m) = \sum_{j \in \{0, ..., n\}} s (j) / (n + 1) (n + 1) / (n + m) = s' (n) (n + 1) / (n + m)\), and \((n + 1) / (n + m)\) converges to \(1\), so, \(s''\) converges to \(r 1 = r\).
The reverse of this proposition does not necessarily hold: as a counterexample, let \(s (j) = 1\) for each even \(j\) and \(s (j) = -1\) for each odd \(j\), then, \(s' (j) = 1 / (j + 1)\) for each even \(j\) and \(s' (j) = 0\) for each odd \(j\), so, \(s' (j) \lt 1 / (2 (j + 1))\) for each \(j\), so, \(s'\) converges to \(0\), but \(s\) does not converge.
3: Proof
Whole Strategy: Step 1: take an \(N_1\) such that for each \(N_1 \lt j\), \(\vert s (j) - r \vert \lt \epsilon / 2\) and an \(N_2\) such that \(N_1 \lt N_2\) and for each \(N_2 \lt l\), \(1 / (l + 1) \sum_{j \in \{0, ..., N_1\}} \vert s (j) - r \vert \lt \epsilon / 2\), and see that for each \(N_2 \lt l\), \(\vert s' (l) - r \vert \lt \epsilon\).
Step 1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N_1 \in \mathbb{N}\) such that for each \(j \in \mathbb{N}\) such that \(N_1 \lt j\), \(\vert s (j) - r \vert \lt \epsilon / 2\), because \(s\) converges to \(r\).
There is an \(N_2 \in \mathbb{N}\) such that \(N_1 \lt N_2\) and for each \(l \in \mathbb{N}\) such that \(N_2 \lt l\), \(1 / (l + 1) \sum_{j \in \{0, ..., N_1\}} \vert s (j) - r \vert \lt \epsilon / 2\), because as \(\sum_{j \in \{0, ..., N_1\}} \vert s (j) - r \vert\) is fixed, making \(l\) larger makes it any smaller.
For each \(N_2 \lt l\), \(\vert s' (l) - r \vert = \vert \sum_{j \in \{0, ..., l\}} s (j) / (l + 1) - r \vert = \vert \sum_{j \in \{0, ..., l\}} (s (j) / (l + 1) - r / (l + 1)) \vert = 1 / (l + 1) \vert \sum_{j \in \{0, ..., l\}} (s (j) - r) \vert \le 1 / (l + 1) (\sum_{j \in \{0, ..., N_1\}} \vert s (j) - r \vert + \sum_{j \in \{N_1 + 1, ..., l\}} \vert s (j) - r \vert) = 1 / (l + 1) \sum_{j \in \{0, ..., N_1\}} \vert s (j) - r \vert + 1 / (l + 1) \sum_{j \in \{N_1 + 1, ..., l\}} \vert s (j) - r \vert \lt \epsilon / 2 + 1 / (l + 1) \sum_{j \in \{N_1 + 1, ..., l\}} \epsilon / 2 = \epsilon / 2 + 1 / (l + 1) (l - N_1) \epsilon / 2 \lt \epsilon / 2 + 1 / (l + 1) (l + 1) \epsilon / 2 = \epsilon\).
So, \(lim s' = r\).
