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Convergence of sequences proof

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  • #1
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Given a sequence ## <x_n> ##, let ## <x_{n+1}> ## denote the sequence whose nth term for each ## n \in \mathbb{N} ## is ## x_{n+1} ##. Show that if ## <x_n> ## converges then ## < x_{n+1} ## converges and they have the same limit.

my attempt thus far

given ## \epsilon > 0 ## ##\exists N \in \mathbb{R} ## s.t. n > N implies ## |x_n - l | < \epsilon ## where l is the limit of ## x_n##.

now we need to prove that for given ## \epsilon > 0 ## ##\exists N_1 \in \mathbb{R} ## s.t. ##n > N_1 ## implies ## |x_{n+1} - m | < \epsilon ## where m is the limit of ## x_{n+1} ## and m = l

heres what I have thus far

consider ## |x_n - l |## :

## |x_n - l | = |x_n - l + x_{n+1} - x_{n+1} | = |(x_n - x_{n+1}) + (x_{n+1} - l ) | \leq |x_n - x_{n+1} | + |x_{n+1} - l| \leq |x - x_{n+1} | + |x_{n+1} - l | \leq |x_{n+1} - x | + |x_{n+1} -l | ##

any ideas how to proceed? or if I'm even doing it correctly
 

Answers and Replies

  • #2
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Not a bad try. But you want to show [itex] |x_{n+1}-l|\le \epsilon [/itex] so you should start your inequality with that instead of [itex] |x_n-l| [/itex]. Also when you break up the absolute value into two pieces, use what you know about the size of [itex] |x_n-l|[/itex] and [itex]|x_n-x_{n+1}| [/itex]. I'm not sure why you introduced "x", what is it?
 
  • #3
HallsofIvy
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If [itex]\{x_n\}[/itex] converges to L, then, given any [itex]\epsilon>0[/itex], there exist N such that if n> N then [itex]|x_n- L|< \epsilon[/itex]. Here, your second sequence, lets call it [itex]\{a_m\}[/itex] has the property that [itex]a_m= x_{n+1}[/itex]. Do you see that, for the same [itex]\epsilon[/itex], we can our new "N" to be the old N plus 1?

(This is a different method than ArcanaNoir's. Both will work.)
 
  • #4
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Not a bad try. But you want to show [itex] |x_{n+1}-l|\le \epsilon [/itex] so you should start your inequality with that instead of [itex] |x_n-l| [/itex]. Also when you break up the absolute value into two pieces, use what you know about the size of [itex] |x_n-l|[/itex] and [itex]|x_n-x_{n+1}| [/itex]. I'm not sure why you introduced "x", what is it?
The x is the first term of the sequence

if I do x_n - x_(n+1) = x_1, x_2, x_3 ... x_n - (x_2, x_3, x_4... x_(n+1))

how do I evaluate [itex]|x_n-x_{n+1}| [/itex]? I have no idea of it's size
 
  • #5
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If [itex]\{x_n\}[/itex] converges to L, then, given any [itex]\epsilon>0[/itex], there exist N such that if n> N then [itex]|x_n- L|< \epsilon[/itex]. Here, your second sequence, lets call it [itex]\{a_m\}[/itex] has the property that [itex]a_m= x_{n+1}[/itex]. Do you see that, for the same [itex]\epsilon[/itex], we can our new "N" to be the old N plus 1?

(This is a different method than ArcanaNoir's. Both will work.)
No, I don't see that - I mean it kind of makes sense, but I'm not all that convinced.

Thank you for your help
 
  • #6
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Have you learned about Cauchy sequences?

At any rate, you know [itex] |x_n-l|<\epsilon [/itex] for all n>N. Well if n>N, isn't n+1 > N?
 
  • #7
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Have you learned about Cauchy sequences?

At any rate, you know [itex] |x_n-l|<\epsilon [/itex] for all n>N. Well if n>N, isn't n+1 > N?
I have not learned about Cauchy sequences, all we have done is the definition of convergence and this is a question on my problem sheet.


I agree with the highlighted part, but I do not see how it helps.

thanks,
 
  • #8
Dick
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Maybe I just don't get what you don't get, but if

At any rate, you know [itex] |x_n-l|<\epsilon [/itex] for all n>N. Well if n>N, isn't n+1 > N?
doesn't that mean [itex] |x_{n+1}-l|<\epsilon [/itex]? It's really simpler than what you think.
 
Last edited:
  • #9
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Maybe I just don't get what you don't get, but if



doesn't that mean [itex] |x_{n+1}-l|<\epsilon [/itex]? It's really simpler than what you think.
Yeah that makes sense actually,

but for a different proof:

I got from here ## |x_{n+1} - l| \leq |x_{n+1} - x_n| + |x_n - l | < |x_{n+1} - x_n| + \epsilon ## how do I evaluate ## |x_{n+1} - x_n|##?
 
  • #10
Dick
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Yeah that makes sense actually,

but for a different proof:

I got from here ## |x_{n+1} - l| \leq |x_{n+1} - x_n| + |x_n - l | < |x_{n+1} - x_n| + \epsilon ## how do I evaluate ## |x_{n+1} - x_n|##?
That's really not a very good way to go. Because about the only thing you can say about ## |x_{n+1} - x_n|## is that since ##|x_n-l|<\epsilon## AND ## |x_{n+1} - l| < \epsilon## then ## |x_{n+1} - x_n| < 2\epsilon##, but you need the bound on ## |x_{n+1} - l|## to say that.
 
  • #11
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That's really not a very good way to go. Because about the only thing you can say about ## |x_{n+1} - x_n|## is that since ##|x_n-l|<\epsilon## AND ## |x_{n+1} - l| < \epsilon## then ## |x_{n+1} - x_n| < 2\epsilon##, but you need the bound on ## |x_{n+1} - l|## to say that.
hm I see, but I just cannot seem to construct a proof using that if x_n is bounded then there n>N for some N then n+1 > N also
 
  • #12
Dick
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hm I see, but I just cannot seem to construct a proof using that if x_n is bounded then there n>N for some N then n+1 > N also
I don't understand why you don't see it. If n>N then n+1>N. It's automatic. The same N that works for the sequence ##a_n## works for the sequence ##b_n## where ##b_n=a_{n+1}##.
 
  • #13
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I don't understand why you don't see it. If n>N then n+1>N. It's automatic. The same N that works for the sequence ##a_n## works for the sequence ##b_n## where ##b_n=a_{n+1}##.
So that's all I'd have to state for the proof?
 
  • #14
Dick
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So that's all I'd have to state for the proof?
Yes, a proof could be that simple. Write it in a way that you understand it.
 
  • #15
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Thank you very much Dick, and everyone else who helped :)!
 

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