# Operations on convergent sequences of real numbers

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## Statement

Let [ilmath](a_n)_{n\in\mathbb{N} }\subset\mathbb{R} [/ilmath] and [ilmath](b_n)_{n\in\mathbb{N} }\subset\mathbb{R} [/ilmath] be real sequences (and are thus considered convergent with respect to the metric [ilmath]d(x,y):\eq\vert x-y\vert[/ilmath]). Suppose that [ilmath]\lim_{n\rightarrow\infty}(a_n)\eq a[/ilmath] and [ilmath]\lim_{n\rightarrow\infty}(b_n)\eq b[/ilmath] 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: [ilmath]\exists M\in\mathbb{N}\ \forall m\in\mathbb{N}[m>M\implies a_n\neq 0][/ilmath] I would have thought
• That is to say: eventually [ilmath](a_n)_{n\in\mathbb{N} } [/ilmath] becomes non-zero.

### Corollaries

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

1. [ilmath]\lim_{n\rightarrow\infty}(ca_n)\eq c\cdot\lim_{n\rightarrow\infty}(a_n)\eq ca[/ilmath]
2. [ilmath]\lim_{n\rightarrow\infty}(a_n-b_n)\eq \lim_{n\rightarrow\infty}(a_n)-\lim_{n\rightarrow\infty}(b_n)\eq a-b[/ilmath] and
3. [ilmath]\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} [/ilmath] - Caveat:under certain conditions
TODO: what conditions
• We can define this safely given: [ilmath]\exists M\in\mathbb{N}\ \forall m\in\mathbb{N}[m>M\implies b_n\neq 0][/ilmath] I would have thought
• That is to say: eventually [ilmath](b_n)_{n\in\mathbb{N} } [/ilmath] 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 [ilmath]\epsilon>0[/ilmath] be given. Then
• By hypothesis: [ilmath]\forall\epsilon'>0\exists N'\in\mathbb{N}\forall n'\in\mathbb{N}[n'>N'\implies d(a_{n'},a)<\epsilon'][/ilmath] and [ilmath]\forall\epsilon'>0\exists N'\in\mathbb{N}\forall n'\in\mathbb{N}[n'>N'\implies d(b_{n'},b)<\epsilon'][/ilmath]
• Choose [ilmath]\epsilon':\eq\frac{1}{2}\epsilon[/ilmath] in both cases, then you get two [ilmath]N'[/ilmath]
• Define: [ilmath]N:\eq\text{max}(N_a',N_b')[/ilmath]
• Let [ilmath]n\in\mathbb{N} [/ilmath] be given
• If [ilmath]n>N[/ilmath]
• then we know [ilmath]n>N_a'[/ilmath] and [ilmath]n>N_b'[/ilmath] so in either case we have:
• [ilmath]d(a_n,a)<\frac{1}{2}\epsilon[/ilmath] and [ilmath]d(b_n,b)<\frac{1}{2}\epsilon[/ilmath] too
• Noting that [ilmath]d(x,y):\eq \vert x-y\vert[/ilmath] we see
• [ilmath]a_n+b_n-a-b[/ilmath] blah blah blah [ilmath]\le \vert a_n-a\vert + \vert b_n - b\vert < \frac{1}{2}\epsilon+\frac{1}{2}\epsilon \eq \epsilon[/ilmath]

### Claim 2

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
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Routine for first years. The key to the proof is:
• Noting that eventually [ilmath]a_n[/ilmath] is within [ilmath]\pm\epsilon_a[/ilmath] of [ilmath]a[/ilmath]
• Then you just have to find a bound for [ilmath]b_n[/ilmath] such that whenever [ilmath]x\in (b_n\pm\epsilon_b)[/ilmath] and [ilmath]y\in(a_n\pm\epsilon_a)[/ilmath] we have [ilmath]xy\in(ab\pm\epsilon)[/ilmath]

### Corollaries

#### Corollary 1

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

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 [ilmath](-b_n)_{n\in\mathbb{N} } [/ilmath] given by multiplying each term in [ilmath](b_n)[/ilmath] by [ilmath]-1[/ilmath]. We see from corollary 1:

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

Now we use the first claim, that [ilmath]\lim_{n\rightarrow\infty}(a_n+b_n)\eq a+b[/ilmath] but with [ilmath](a_n)[/ilmath] and [ilmath](-b_n)[/ilmath] 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

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:
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 $\lim_{n\rightarrow\infty}\left(a_n\cdot\frac{1}{b_n}\right)\eq a\frac{1}{b}\eq \frac{a}{b}$