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Math Help - Sequences and series

  1. #1
    Senior Member Danneedshelp's Avatar
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    Sequences and series

    Consider a convergent sequence (x_{n})_{n} such that limx_{n} = x.

    (a) Let (y_{n})_{n} be built such that y_{n} = x_{n+100} , for all n\in\mathbb{N}. Is (y_{n})_{n} convergent? If yes, what is the
    limit?
    (b) Let (z_{n})_{n} be now the sequence \{0, \pi\\, e, -\sqrt{2}, ln(2), x_{1}, x_{2} , x_{3} , . . . , x-{n} , . . . \}. In other words, we added 5 extra terms at the beginning of (x_{n})_{n} and we called this new sequence (z_{n})_{n} . Is (z_{n})_{n}convergent?
    If yes, what is the limit?
    (c) Consider now a convergent series \sum_{<br />
n=1}a_{n} . Fix some natural number K > 1. Is \sum_{n=K}a_{n} convergent?

    (a) Assume (x_{n})_{n} converges to x and let y_{n} = x_{n+10}0. Since (x_{n})_{n} is convergent, (x_{n})_{n} has convergent subsequences. Moreover, these subsequence congerge to the same limit as (x_{n})_{n}, x. Cleary, y_{n} = x_{n+100} is a subsequence of (x_{n})_{n}. Thus, (y_{n})_{n} converges to x.

    Is it enough just to use a theorem to prove this one?

    (b) Can I just note that the first 5 terms are constants and pull them out of the limit and conclude the sequence still converges to x? I am not sure how to formalize this. Do I need to use the definition of convergence or the cauchy definition to prove this?

    (c) Not totaly sure how to approach this one. I would think that is iss convergent, because, even though K is some number latter on in the sequence, there will still be a rank for which the terms in the sequence are stuck in some epsilon neighborhood.

    Can anyone help me with some formal proofs?

    Thank you
    Last edited by Danneedshelp; September 27th 2009 at 10:53 PM.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    Consider a convergent sequence (x_{n})_{n} such that limx_{n} = x.

    (a) Let (y_{n})_{n} be built such that y_{n} = x_{n}+100 , for all n\in\mathbb{N}. Is (y_{n})_{n}convergent? If yes, what is the limit?
    (b) Let (z_{n})_{n} be now the sequence \{0, \pi\\, e, -\sqrt{2}, ln(2), x_{1}, x_{2} , x_{3} , . . . , x-{n} , . . . \}. In other words, we added 5 extra terms at the beginning of (x_{n})_{n} and we called this new sequence (z_{n})_{n} . Is (z_{n})_{n}convergent?
    If yes, what is the limit?
    (c) Consider now a convergent series \sum_{<br />
n=1}a_{n} . Fix some natural number K > 1. Is \sum_{n=K}a_{n} convergent?

    (a) Assume converges to x and let y_{n} = x_{n}+100. Since (x_{n})_{n} is convergent, (x_{n})_{n} has convergent subsequences. Moreover, these subsequence congerge to the same limit as (x_{n})_{n}, x. Cleary, y_{n} = x_{n}+100 is a subsequence of (x_{n})_{n}. Thus, (y_{n})_{n} converges to x.

    Is it enough just to use a theorem to prove this one?

    (b) Can I just note that the first 5 terms are constants and pull them out of the limit and conclude the sequence still converges to x? I am not sure how to formalize this. Do I need to use the definition of convergence or the cauchy definition to prove this?

    (c) Not totaly sure how to approach this one. I would think that is iss convergent, because, even though K is some number latter on in the sequence, there will still be a rank for which the terms in the sequence are stuck in some epsilon neighborhood.

    Can anyone help me with some formal proofs?

    Thank you
    (a) This sequence converges to x+100.

    Proof: Since \{x_n\}\to x, then \forall~\epsilon>0, \exists~N such that n>N implies that |x_n-x|<\epsilon. Thus, n>N also implies that |(x_n+100)-(x+100)|=|x_n-x|<\epsilon. So \{y_n\}\to x+100. \square

    ---------------

    (b) This sequence converges to x.

    Proof: Since \{x_n\}\to x, then \forall~\epsilon>0, \exists~N such that n>N implies that |x_n-x|<\epsilon. Thus, n>N+5 implies that |z_n-z|<\epsilon. So \{z_n\}\to x. \square

    ---------------

    (c) It converges.

    Proof: \sum_{n=K}^{\infty}a_n=\sum_{n=1}^{\infty}a_n -\sum_{n=1}^{K-1}a_n. Looking at the right side of the equation, we have a finite sum subtracted from a convergent series, so it is obvious that the right side is finite, and therefore \sum_{n=K}^{\infty}a_n converges. \square
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by redsoxfan325 View Post
    (a) This sequence converges to x+100.

    Proof: Since \{x_n\}\to x, then \forall~\epsilon>0, \exists~N such that n>N implies that |x_n-x|<\epsilon. Thus, n>N also implies that |(x_n+100)-(x+100)|=|x_n-x|<\epsilon. So \{y_n\}\to x+100. \square

    ---------------

    (b) This sequence converges to x.

    Proof: Since \{x_n\}\to x, then \forall~\epsilon>0, \exists~N such that n>N implies that |x_n-x|<\epsilon. Thus, n>N+5 implies that |z_n-z|<\epsilon. So \{z_n\}\to x. \square

    ---------------

    (c) It converges.

    Proof: \sum_{n=K}^{\infty}a_n=\sum_{n=1}^{\infty}a_n -\sum_{n=1}^{K-1}a_n. Looking at the right side of the equation, we have a finite sum subtracted from a convergent series, so it is obvious that the right side is finite, and therefore \sum_{n=K}^{\infty}a_n converges. \square
    So, in general, if I have a new series y_{n}=x_{n+c}. I only have to show |(x_{n}-c)-(x-c)|<\epsilon holds?

    For part (b) did you mean, |z_{n}-x|<\epsilon?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Danneedshelp View Post
    So, in general, if I have a new series y_{n}=x_{n+c}. I only have to show |(x_{n}-c)-(x-c)|<\epsilon holds?

    For part (b) did you mean, |z_{n}-x|<\epsilon?
    Yes, for a series y_n=x_n+c (is that what you meant?), you have that |(x_n+c)-(x+c)|=|x_n-x|<\epsilon so y_n\to x+c.

    And for part (b), yes, that's what I meant.
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  5. #5
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by redsoxfan325 View Post
    Yes, for a series y_n=x_n+c (is that what you meant?), you have that |(x_n+c)-(x+c)|=|x_n-x|<\epsilon so y_n\to x+c.

    And for part (b), yes, that's what I meant.
    Oh boy, I wrote the first problem down wrong!

    It should be...

    (a) Let (y_{n})_{n} be built such that y_{n} = x_{n+100} , for all n\in\mathbb{N}. Is (y_{n})_{n} convergent? If yes, what is the
    limit?

    Suppose (x_{n})_{n} converges to x and let y_{n}=x_{n+100}. Since (x_{n})_{n} converges, there exists an N\in\mathbb{N} such that n\geq\\N implies |x_{n}-x|<\epsilon. Now, let N be some natural number satisfying N\geq\\n+100 such that for all n\geq\\N implies |x_{n+100}-x|=|y_{n}-x|<\epsilon. Therefore, (y_{n})_{n} converges to x as well.
    Last edited by Danneedshelp; September 27th 2009 at 11:15 PM.
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  6. #6
    Super Member redsoxfan325's Avatar
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    Yes. There is a theorem that states that a sequence \{x_n\} converges to x if and only if every (infinite) subsequence of \{x_n\} also converges to x.
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