Operations on convergent sequences of real numbers

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Demote once cleaned up. Routine work for first years, documenting only

• make sure links on If a real series converges then its terms tend to zero agree

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Cleanup, references, links to other pages. So forth

Statement

Let $(a_n)_{n\in\mathbb{N} }\subset\mathbb{R}$ and $(b_n)_{n\in\mathbb{N} }\subset\mathbb{R}$ be real sequences (and are thus considered convergent with respect to the metric $d(x,y):\eq\vert x-y\vert$). Suppose that $\lim_{n\rightarrow\infty}(a_n)\eq a$ and $\lim_{n\rightarrow\infty}(b_n)\eq b$ then:

1. $$\lim_{n\rightarrow\infty}(a_n+b_n)\eq \lim_{n\rightarrow\infty}(a_n)+\lim_{n\rightarrow\infty}(b_n)\eq a+b$$ ,
2. $$\lim_{n\rightarrow\infty}(a_nb_n)\eq \lim_{n\rightarrow\infty}(a_n)\cdot\lim_{n\rightarrow\infty}(b_n)\eq ab$$ , and
3. $$\lim_{n\rightarrow\infty}\left(\frac{1}{a_n}\right)\eq \frac{1}{\lim_{n\rightarrow\infty}(a_n)}\eq\frac{1}{a}$$ - Caveat:under certain conditions
   TODO: what conditions
   • We can define this safely given: $\exists M\in\mathbb{N}\ \forall m\in\mathbb{N}[m>M\implies a_n\neq 0]$ I would have thought
     • That is to say: eventually $(a_n)_{n\in\mathbb{N} }$ becomes non-zero.

Corollaries

Let $c\in\mathbb{R}$ be any real number. Immediately we have the following results:

1. $\lim_{n\rightarrow\infty}(ca_n)\eq c\cdot\lim_{n\rightarrow\infty}(a_n)\eq ca$
2. $\lim_{n\rightarrow\infty}(a_n-b_n)\eq \lim_{n\rightarrow\infty}(a_n)-\lim_{n\rightarrow\infty}(b_n)\eq a-b$ and
3. $\lim_{n\rightarrow\infty}(a_n/b_n)\eq \frac{\lim_{n\rightarrow\infty}(a_n)}{\lim_{n\rightarrow\infty}(b_n)}\eq \frac{a}{b}$ - Caveat:under certain conditions
   TODO: what conditions
   • We can define this safely given: $\exists M\in\mathbb{N}\ \forall m\in\mathbb{N}[m>M\implies b_n\neq 0]$ I would have thought
     • That is to say: eventually $(b_n)_{n\in\mathbb{N} }$ becomes non-zero.

Proof of claims

Claim 1

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Routine for first years, the gist of the proof is as follows:

• Let $\epsilon>0$ be given. Then
  • By hypothesis: $\forall\epsilon'>0\exists N'\in\mathbb{N}\forall n'\in\mathbb{N}[n'>N'\implies d(a_{n'},a)<\epsilon']$ and $\forall\epsilon'>0\exists N'\in\mathbb{N}\forall n'\in\mathbb{N}[n'>N'\implies d(b_{n'},b)<\epsilon']$
    • Choose $\epsilon':\eq\frac{1}{2}\epsilon$ in both cases, then you get two $N'$
  • Define: $N:\eq\text{max}(N_a',N_b')$
    • Let $n\in\mathbb{N}$ be given
      • If $n>N$
        • then we know $n>N_a'$ and $n>N_b'$ so in either case we have:
          • $d(a_n,a)<\frac{1}{2}\epsilon$ and $d(b_n,b)<\frac{1}{2}\epsilon$ too
        • Noting that $d(x,y):\eq \vert x-y\vert$ we see
          • $a_n+b_n-a-b$ blah blah blah $\le \vert a_n-a\vert + \vert b_n - b\vert < \frac{1}{2}\epsilon+\frac{1}{2}\epsilon \eq \epsilon$

Claim 2

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Routine for first years. The key to the proof is:

• Noting that eventually $a_n$ is within $\pm\epsilon_a$ of $a$
• Then you just have to find a bound for $b_n$ such that whenever $x\in (b_n\pm\epsilon_b)$ and $y\in(a_n\pm\epsilon_a)$ we have $xy\in(ab\pm\epsilon)$

Corollaries

Corollary 1

Let $c\in\mathbb{R}$ be given. Then the sequence: $(c_n)_{n\in\mathbb{N} }$ defined by: $\forall n\in\mathbb{N}[c_n:\eq c]$ obviously converges to $c$ (that is: $\lim_{n\rightarrow\infty}(c_n)\eq c$)

Then note that, by claim 2:

• $$\lim_{n\rightarrow\infty}(c_na_n)=\lim_{n\rightarrow\infty}(c_n)\cdot\lim_{n\rightarrow\infty}(a_n)\eq c\cdot\lim_{n\rightarrow\infty}(a_n)=ca$$

Corollary 2

Consider the sequence $(-b_n)_{n\in\mathbb{N} }$ given by multiplying each term in $(b_n)$ by $-1$. We see from corollary 1:

• $$\lim_{n\rightarrow\infty}(-b_n)\eq -b$$ .

Now we use the first claim, that $\lim_{n\rightarrow\infty}(a_n+b_n)\eq a+b$ but with $(a_n)$ and $(-b_n)$ as the sequences.

We see:

• $$\lim_{n\rightarrow\infty}(a_n+-b_n)=\lim_{n\rightarrow\infty}(a_n)+\lim_{n\rightarrow\infty}(-b_n)\eq a+(-b)\eq a-b$$

Corollary 3

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Routine for first years, the key to the proof is:

• Using claim 3, note that $$\lim_{n\rightarrow\infty}\left(\frac{1}{b_n}\right)\eq\frac{1}{b}$$ , then:
• Using claim 2, note that $$\lim_{n\rightarrow\infty}\left(a_n\cdot\frac{1}{b_n}\right)\eq a\frac{1}{b}\eq \frac{a}{b}$$

References

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Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
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Should be easy to find, see if that guy I emailed on 22/11/2016 got back to me