UTLOC:1

Unnamed Theorem Lemma or Corollary: 1

Grade: A*

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This needs a "still unnamed" classification, so one has a category of unnamed theorems, as this page will persist even if it is given a name

Statement

Let [ilmath](a_n)_{n\in\mathbb{N} } \subset\mathbb{R} [/ilmath] be a real sequence. Suppose [ilmath]\lim_{n\rightarrow\infty}(a_n)\eq\ell[/ilmath]. Then:

• [ilmath](b_n)_{n\in\mathbb{N} }\subset\mathbb{R} [/ilmath] defined by: [ilmath]b_n:\eq\left\{\begin{array}{lr}a_{n-1}&\text{if }n\ge 2\\c\in\mathbb{R}&\text{if }n\eq 1\end{array}\right.[/ilmath] also converges to [ilmath]\ell[/ilmath].
  • Here [ilmath]c\in\mathbb{R} [/ilmath] is any value. I used [ilmath]0\in\mathbb{R} [/ilmath] on paper as I was working with series (summation) and this could be used to show [ilmath]\lim_{n\rightarrow\infty}(\sum^n_{i\eq 1}a_n)\eq\lim_{n\rightarrow\infty}(\sum^n_{i\eq 1}b_n)[/ilmath] as [ilmath]\sum_{i=1}^n b_n\eq 0+\sum_{i=2}^n b_n=0+\sum_{i=1}^{n-1}a_n[/ilmath]
    • Which is to show: [ilmath]\lim_{n\rightarrow\infty}(S_n)\eq\lim_{n\rightarrow\infty}(S_{n-1})\eq\ell[/ilmath] - a vital part in If a real series converges then the limit of the terms of the series is zero

Proof

Suppose that [ilmath]\lim_{n\rightarrow\infty}(a_n)\eq\ell[/ilmath], and [ilmath](b_n)_{n\in\mathbb{N} } [/ilmath] is defined as above, for an arbitrary constant [ilmath]c\in\mathbb{R} [/ilmath]. We wish to show that:

• [ilmath]\lim_{n\rightarrow\infty}(b_n)\eq\ell[/ilmath] which can be stated symbolically: [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(b_n,\ell)<\epsilon][/ilmath]

Proof:

• Let [ilmath]\epsilon>0[/ilmath] be given.
  • Choose [ilmath]N:\eq N'+1[/ilmath] where [ilmath]N'\in\mathbb{N} [/ilmath] exists by hypothesis and means: [ilmath]\forall n'\in\mathbb{N}[n'>N'\implies d(a_n,\ell)<\epsilon][/ilmath]
    • Let [ilmath]n\in\mathbb{N} [/ilmath] be given
      • Suppose [ilmath]n\le N[/ilmath] - by the nature of logical implication we do not care whether or not [ilmath]d(b_n,\ell)<\epsilon[/ilmath] the implication is true regardless.
      • Suppose [ilmath]n > N[/ilmath]
        • This means [ilmath]n>N'+1[/ilmath] by definition of [ilmath]N[/ilmath], so [ilmath]n-1>N'[/ilmath]
          • This means [ilmath]d(a_{n-1},\ell)<\epsilon[/ilmath] (by hypothesis and what [ilmath]N'[/ilmath] is defined to be)
        • Caution:The reader must be sure [ilmath]a_{n-1} [/ilmath] exists. I.E that [ilmath]n-1\ge 1[/ilmath] or [ilmath]n\ge 2[/ilmath]. As [ilmath]N'\in\mathbb{N} [/ilmath] we see [ilmath]N'\ge 1[/ilmath], then [ilmath]N:\eq N'+1[/ilmath] so [ilmath]N'+1\ge 2[/ilmath] giving [ilmath]N\ge 2[/ilmath]. Then we have [ilmath]n>N\ge 2[/ilmath] giving [ilmath]n> 2[/ilmath] or [ilmath]n\ge 3[/ilmath]. As [ilmath]n[/ilmath] is at least 3, we can be sure [ilmath]a_{n-1} [/ilmath] exists.
          • Warning:This varies slightly depending on your definition of convergent sequence, if we had had: [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n\ge N\implies d(x_n,x)<\epsilon][/ilmath] - notice the [ilmath]n\ge N[/ilmath] rather than our [ilmath]n>N[/ilmath] we would have to consider this differently
          • Ultimately all of this can be avoided by saying [ilmath]N:\eq\max(N'+1,2)[/ilmath]
          • We also probably already have this from the definition of [ilmath]N'[/ilmath] - I didn't check. My sanity check is in the "caution" above.
        • We now know [ilmath]n\ge 3[/ilmath] and [ilmath]d(a_{n-1},\ell)<\epsilon[/ilmath]. Note that for such an [ilmath]n[/ilmath] we have:
          • [ilmath]b_n:\eq a_{n-1} [/ilmath]
        • Thus we see [ilmath]d(b_n,\ell)<\epsilon[/ilmath] as required.

Possible generalisation

I almost created the page:

• If a sequence converges then all sequences that are eventually equal to it converge also

Where "eventually equal"

The statement I have written (but scrapped, in favour of this page) is included below.

Statement

Let [ilmath](a_n)_{n\in\mathbb{N} } [/ilmath] be a convergent sequence in a metric space [ilmath](X,d)[/ilmath]. Then any sequence eventually equal to it, [ilmath](b_n)_{n\in\mathbb{N} } [/ilmath] converges also. Furthermore [ilmath](b_n)[/ilmath] converges to the same limit as [ilmath](a_n)[/ilmath]

That is to say [ilmath](b_n)_{n\in\mathbb{N} } [/ilmath] is a sequence such that:

• [ilmath]\exists M\in\mathbb{N}\forall m\in\mathbb{N}[a_m\eq b_{m+M}][/ilmath]

Then [ilmath](b_n)[/ilmath] converges to the same limit as [ilmath](a_n)[/ilmath] (should [ilmath](a_n)[/ilmath] converge)