View source for Operations on convergent sequences of real numbers

{{Stub page|grade=A*|msg=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}}
{{Requires work|grade=A*|msg=Cleanup, references, links to other pages. So forth}}
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==Statement==
Let {{M|(a_n)_{n\in\mathbb{N} }\subset\mathbb{R} }} and {{M|(b_n)_{n\in\mathbb{N} }\subset\mathbb{R} }} be [[real sequences]] (and are thus considered convergent with respect to the [[metric]] {{M|d(x,y):\eq\vert x-y\vert}}). Suppose that [[limit (sequence)|{{M|\lim_{n\rightarrow\infty}(a_n)\eq a}}]] and {{M|\lim_{n\rightarrow\infty}(b_n)\eq b}} then:
# {{MM|\lim_{n\rightarrow\infty}(a_n+b_n)\eq \lim_{n\rightarrow\infty}(a_n)+\lim_{n\rightarrow\infty}(b_n)\eq a+b}},
# {{MM|\lim_{n\rightarrow\infty}(a_nb_n)\eq \lim_{n\rightarrow\infty}(a_n)\cdot\lim_{n\rightarrow\infty}(b_n)\eq ab}}, and
# {{MM|\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}} {{XXX|what conditions}}
#* We can define this safely given: {{M|\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: {{link|eventually|sequence}} {{M|(a_n)_{n\in\mathbb{N} } }} becomes non-zero.
===[[Corollary|Corollaries]]===
Let {{M|c\in\mathbb{R} }} be any [[real number]]. Immediately we have the following results:
# {{M|\lim_{n\rightarrow\infty}(ca_n)\eq c\cdot\lim_{n\rightarrow\infty}(a_n)\eq ca}}
# {{M|\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
# {{M|\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}} {{XXX|what conditions}}
#* We can define this safely given: {{M|\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: {{link|eventually|sequence}} {{M|(b_n)_{n\in\mathbb{N} } }} becomes non-zero.
==Proof of claims==
===Claim 1===
{{Requires proof|grade=D|msg=Routine for first years, the gist of the proof is as follows:
* Let {{M|\epsilon>0}} be given. Then
** By hypothesis: {{M|\forall\epsilon'>0\exists N'\in\mathbb{N}\forall n'\in\mathbb{N}[n'>N'\implies d(a_{n'},a)<\epsilon']}}  and {{M|\forall\epsilon'>0\exists N'\in\mathbb{N}\forall n'\in\mathbb{N}[n'>N'\implies d(b_{n'},b)<\epsilon']}} 
*** Choose {{M|\epsilon':\eq\frac{1}{2}\epsilon}} in both cases, then you get two {{M|N'}}
** Define: {{M|N:\eq\text{max}(N_a',N_b')}} 
*** Let {{M|n\in\mathbb{N} }} be given
**** If {{M|n>N}} 
***** then we know  {{M|n>N_a'}} and {{M|n>N_b'}} so in either case we have:
****** {{M|d(a_n,a)<\frac{1}{2}\epsilon}} and {{M|d(b_n,b)<\frac{1}{2}\epsilon}} too
***** Noting that {{M|d(x,y):\eq \vert x-y\vert}} we see
****** {{M|a_n+b_n-a-b}} blah blah blah {{M|\le \vert a_n-a\vert + \vert b_n - b\vert < \frac{1}{2}\epsilon+\frac{1}{2}\epsilon \eq \epsilon}}
}}
===Claim 2===
{{Requires proof|grade=D|msg=Routine for first years. The key to the proof is:
* Noting that eventually {{M|a_n}} is within {{M|\pm\epsilon_a}} of {{M|a}}
* Then you just have to find a bound for {{M|b_n}} such that whenever {{M|x\in (b_n\pm\epsilon_b)}} and {{M|y\in(a_n\pm\epsilon_a)}} we have {{M|xy\in(ab\pm\epsilon)}}}}
===Claim 3===
===Corollaries===
====Corollary 1====
Let {{M|c\in\mathbb{R} }} be given. Then the sequence: {{M|(c_n)_{n\in\mathbb{N} } }} defined by: {{M|\forall n\in\mathbb{N}[c_n:\eq c]}} obviously {{link|converges|sequence}} to {{M|c}} (that is: {{M|\lim_{n\rightarrow\infty}(c_n)\eq c}})

Then note that, by claim 2:
* {{MM|1=\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 {{M|(-b_n)_{n\in\mathbb{N} } }} given by multiplying each term in {{M|(b_n)}} by {{M|-1}}. We see from corollary 1:
* {{MM|\lim_{n\rightarrow\infty}(-b_n)\eq -b}}.
Now we use the first claim, that {{M|\lim_{n\rightarrow\infty}(a_n+b_n)\eq a+b}} but with {{M|(a_n)}} and {{M|(-b_n)}} as the sequences. 

We see:
* {{MM|1=\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====
{{Requires proof|grade=D|msg=Routine for first years, the key to the proof is:
* Using claim 3, note that {{MM|\lim_{n\rightarrow\infty}\left(\frac{1}{b_n}\right)\eq\frac{1}{b} }}, then:
* Using claim 2, note that {{MM|\lim_{n\rightarrow\infty}\left(a_n\cdot\frac{1}{b_n}\right)\eq a\frac{1}{b}\eq \frac{a}{b} }}}}
==References==
<references/>
{{Requires references|grade=B|msg=Should be easy to find, see if that guy I emailed on 22/11/2016 got back to me}}
{{Theorem Of|Real Analysis|Analysis|Functional Analysis|Metric Space}}