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yo what the fuck
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literally trivial, just use the definition of convergence
also put this brainlet tier question into the stupid question thread
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>>8749457
yeah i get why its true but idk how to apply the definition to the question as its a general integer sequence and not a specific one
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If it's convergent then grab any epsilon<1 and you can find an N such that for all indices creater than that N, you get that the values must be be les than 1 unity apart. If you have only integers, theb clearly no pair of distinct integers satisfy this so they must be the same for all the indices.
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>>8749482
okay cool that sorta makes sense i wrote "fix epsilon<1. |a_n - L|<1 and no two distinct integers satisfy this" but idk how to logically and rigourously link this with the idea that there must exist an N such that a_i=a_j for all i,j>N
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>>8749526
Have you seen cauchy sequences?
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>>8749531
no im doing this new high-level math course that my uni just created and it feels like they're writing the curriculum as we do the course, i've done fields, the rigorous definition of limit, triangle inequality, squeeze theorem, a definition of e using the bernoulli inequality and some other basic proofs about the product and sum of converging limits. i'll read up on it right now though
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>>8749539
btw this is a course intended to be taken in the first semester so i meant "high level maths course" in a relative sense,
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>>8749539
>>8749542
Okey, you should look them but you dont actually need them. let [math]0<\epsilon<1[/math] from the definitión of convergence [math]\exists N\in \mathbb{N}[/math] such that [math]i>N \rightarrow |x_{i}-x_{j}|<\frac{\epsilon}{2}[/math] Then let [math]i,j\in \mathbb{N}, i,j>N[/math] then [math]|x_{i}-x_{j}|=|(x_{i}-c) + (c-x_{j})|\leq |(x_{i}-c)| + |(c-x_{j})|<\frac{\epsilon}{2} + \frac{\epsilon}{2}=\epsilon[/math] You can fill in the rest
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>>8749567
we havent been shown the cauchy sequence so i think it'll be suspicious if i use it, but is this still rigourous?
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>>8749596
but i think you do by how it says a_n is "a convergence sequence of integers" right?
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>>8749599
Not really. It obvious, but I suppose the point of your class is rigor. You are actually proving that, but ypur hypothesis only says that the sequence is convergent. From the definition, nothing readily tells you the limit must be then an integer.
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>>8749599
only as a consequence of this problem
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>>8749603
are you sure? im no expert but it sounds like "a sequence of integers" refers to a sequence that contains only integers so mustn't the limit also be an integer? we're not supposed to use cauchy as we've never been shown it so i dont know how else it'd be done
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>>8749608
If your class is focusing a lot in your proofs never go for what something "sounds" like. Is there anything in the definition of convergent sequence that tells you about the the nature of the limit? You are actually talking about a corollary of this proof. Also, if ypu prove something you can use it, that's why you if you prove the first half of cauchy criterion you can use it. Just don't calñ it that.
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>>8749478
Just give up dude, you're a brainlet.
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>>8749608
Take for example 1/n. This is a sequence taking values only in an open interval (0,1), but as you can see this doesn't imply that the limit also lies there.
in our case it is of course true that any convergent sequence taking values in integers has an integer for a limit. that's because the integers are what is called a "closed set" - closed sets are precisely those which contain all limits of its convergent sequences. in my example I actually proved that (0.1) is not a closed set - the sequence 1/n lives there, but its limit lives somewhere else. check wikipedia for "metric spaces", "closed set", "cauchy sequence"
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is this sufficiently rigourous then?
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>>8749876
No.
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>>8749912
how about this then?
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Proof by contradiction:
Let [math](a_n)_{n=1}^\infty [/math] be a convergent sequence of integers with the limit [math] L [/math] and let [math] \varepsilon = 0.5 [/math].
Since [math](a_n)_{n=1}^\infty [/math] is a convergent sequence there is an [math]N \in \mathbb{N} [/math] such that
[math] |a_n - L| < \varepsilon = 0.5 [/math] for all [math] n > N [/math].
Suppose there exist [math]i,j > N [/math] with [math]a_i \neq a_j [/math] then
[eqn] 1 \underset{\text{Since }a_i\text{ and }a_j\text{ are distinct integers.}}{\leq} |a_i - a_j| \underset{\text{Triangle inequality}}{\leq} |a_i - L| + |L - a_j| < 0.5 + 0.5 = 1[/eqn]
Contradiction thus we must have [math]a_i = a_j [/math]. for all [math]i, j > N [/math].
q.e.d.
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>>8749926
I dont think this is logically coherent because the |a_n - L| definition should hold be arbitrarily small values of epsilon whereas you've set epsilon to 0.5
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>>8749932
We take a convergent sequence so we already know that for all epsilon>0 there is such an N. Since it works all epsilon>0 it must also work for specific one like 0.5.
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>>8749922
if you havent done cauchy sequences, you have to prove that convergent seq are cauchy seq.
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First: for any integers [math]m. n[/math] their difference is either 0 or [math]|m-n| \ge 1[/math]. Since the sequence converges to some [math]a \in \mathbb{Z}[/math], for every [math]\varepsilon > 0[/math], there is an integer [math]N>0[/math] such that [math]|a_n - a|<\varepsilon[/math], for all [math]n>N[/math]. Choosing [math]\varepsilon = \frac{1}{2}[/math], one obtains an [math]N[/math] such that [math]|a_n - a|<\frac{1}{2} \Leftrightarrow |a_n-a|=0 \Leftrightarrow a_n=a[/math] for all [math]n > N[/math].
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>>8750851
>m. n
m, n
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>>8749455
Literally pick epsilon be 0.5
As it is convergent there exist a satisfactory delta so that the terms and the value are less than 0.5 apart
But the value is an integer, and the terms are integers. And there no integers that are less than 1 apart, so you can see that such integers are 0 apart and thus are equal.
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>>8750851
very c-cute anime girl a-anon
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>>8750864
T-thanks. You might not believe it, but it's really me in the pic.
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>>8750857
You don't know the limit is an integer fuckface.
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>>8750872
It has to be if the sequence converges. If it's not, then there is always a positive distance between the closest integer and the limit, and one can choose epsilon to be half of that, and thus contradict the assumption that the limit is not an integer.
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>>8750878
Yea, but you are using the result. OP cannot assume that without already using the result he has.
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>>8750885
That's not the result. The result uses that the limit is an integer, this proves the limit is an integer. This is just the definition of convergence combined with the fact that the distance between an integer and a non-integer is positive. What is used in the proof of OP's claim is that the limit is an integer, and then the sequence eventually becomes constant.
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lel
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>>8750872
You can easily prove it. Suppose it is not an integer. That means it is between integers. Let the limit be c. Then pick an epsilon neighbourhood of c such that the neighbourhood is completely between the two integers around c.
This epsilon neighbourhood should have infinitely many sequence terms, but as every term is an integer, none are.
Easy contradiction.
Is this thread "I got a D in real analysis" general?
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>>8751096
OP was using it without proof.
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>>8749596
brainlet here. why don't you know L is an integer if the series given was a series of integers?
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