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

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

Proof

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

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

Proof:

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

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

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

Then $(b_n)$ converges to the same limit as $(a_n)$ (should $(a_n)$ converge)