description/proof of that series of \(1\)-increasing positive natural numbers to powers of minus natural number larger than \(1\) converges smaller than this
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of natural numbers set.
- The reader knows a definition of Lebesgue integral of measurable extended real function over measurable subset of measure space.
Target Context
- The reader will have a description and a proof of the proposition that any series of \(1\)-increasing positive natural numbers to the powers of minus any natural number larger than \(1\) converges smaller than this.
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:
\(l\): \(\in \mathbb{N}\), such that \(0 \lt l\)
\(n\): \(\in \mathbb{N}\), such that \(1 \lt n\)
\(s\): \(= \sum_{j \in \{l, l + 1, ...\}} j^{- n}\)
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Statements:
\(s \lt l^{- n} (n - 1 + l) / (n - 1) \le l^{- n + 1} n / (n - 1)\)
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2: Proof
Whole Strategy: Step 1: see that \((j + 1)^{- n} \lt \int^{j + 1}_j x^{- n} d x\); Step 2: see that \(s \lt l^{- n} + \int^{\infty}_l x^{- n} d x\); Step 3: see that \(l^{- n} + \int^{\infty}_l x^{- n} d x = l^{- n} (n - 1 + l) / (n - 1) \le l^{- n + 1} n / (n - 1)\); Step 4: conclude the proposition.
Step 1:
Let \(j \in \mathbb{N}\) be any such that \(1 \le j\).
Over \([j, j + 1] \subseteq \mathbb{R}\), \((j + 1)^{- n} \le x^{- n}\), because \(x^{- n}: [1, \infty) \to \mathbb{R}\) is decreasing while \((j + 1)^{- n} \le (j + 1)^{- n}\).
So, \(\int^{j + 1}_j (j + 1)^{- n} d x \le \int^{j + 1}_j x^{- n} d x\).
But the left hand side is \((j + 1)^{- n}\).
So, \((j + 1)^{- n} \lt \int^{j + 1}_j x^{- n} d x\).
Step 2:
\(s = \sum_{j \in \{l, l + 1, ...\}} j^{- n} = l^{- n} + \sum_{j \in \{l + 1, l + 2, ...\}} j^{- n} = l^{- n} + \sum_{j \in \{l, l + 1, ...\}} (j + 1)^{- n} \lt l^{- n} + \sum_{j \in \{l, l + 1, ...\}} \int^{j + 1}_j x^{- n} d x\), because as \(1 \le l\), \(1 \le j\), \(= l^{- n} + \int^{\infty}_l x^{- n} d x\).
Step 3:
\(l^{- n} + \int^{\infty}_l x^{- n} d x = l^{- n} + [1 / (- n + 1) x^{- n + 1}]^{\infty}_l = l^{- n} + (1 / (- n + 1) {\infty}^{- n + 1} - 1 / (- n + 1) l^{- n + 1}) = l^{- n} + (0 - 1 / (- n + 1) l^{- n + 1}) = l^{- n} - 1 / (- n + 1) l^{- n + 1} = l^{- n} (1 - 1 / (- n + 1) l) = l^{- n} ((- n + 1) / (- n + 1) - 1 / (- n + 1) l) = l^{- n} (- n + 1 - l) / (- n + 1) = l^{- n} (n - 1 + l) / (n - 1)\).
\(n - 1 + l \le n l\), because when \(l = 1\), \(n = n - 1 + 1 \le n 1 = n\), and as \(l\) increases by \(1\), the left hand side increases by \(1\) while the right hand side increases by \(n\).
So, \(l^{- n} (n - 1 + l) / (n - 1) \le l^{- n} n l / (n - 1) = l^{- n + 1} n / (n - 1)\).
Step 4:
So, \(s \lt l^{- n} (n - 1 + l) / (n - 1) \le l^{- n + 1} n / (n - 1)\).
