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Thread: Sequences and series

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

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

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

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

    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 $\displaystyle x$? 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; Sep 27th 2009 at 09:53 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 $\displaystyle (x_{n})_{n}$ such that $\displaystyle limx_{n} = x$.

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

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

    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 $\displaystyle x$? 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 $\displaystyle x+100$.

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

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

    (b) This sequence converges to x.

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

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

    (c) It converges.

    Proof: $\displaystyle \sum_{n=K}^{\infty}a_n=\sum_{n=1}^{\infty}a_n -\sum_{n=1}^{K-1}a_n$. 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 $\displaystyle \sum_{n=K}^{\infty}a_n$ converges. $\displaystyle \square$
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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 $\displaystyle x+100$.

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

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

    (b) This sequence converges to x.

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

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

    (c) It converges.

    Proof: $\displaystyle \sum_{n=K}^{\infty}a_n=\sum_{n=1}^{\infty}a_n -\sum_{n=1}^{K-1}a_n$. 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 $\displaystyle \sum_{n=K}^{\infty}a_n$ converges. $\displaystyle \square$
    So, in general, if I have a new series $\displaystyle y_{n}=x_{n+c}$. I only have to show $\displaystyle |(x_{n}-c)-(x-c)|<\epsilon$ holds?

    For part (b) did you mean, $\displaystyle |z_{n}-x|<\epsilon$?
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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 $\displaystyle y_{n}=x_{n+c}$. I only have to show $\displaystyle |(x_{n}-c)-(x-c)|<\epsilon$ holds?

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

    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 $\displaystyle y_n=x_n+c$ (is that what you meant?), you have that $\displaystyle |(x_n+c)-(x+c)|=|x_n-x|<\epsilon$ so $\displaystyle y_n\to x+c$.

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

    It should be...

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

    Suppose $\displaystyle (x_{n})_{n}$ converges to $\displaystyle x$ and let $\displaystyle y_{n}=x_{n+100}$. Since $\displaystyle (x_{n})_{n}$ converges, there exists an $\displaystyle N\in\mathbb{N}$ such that $\displaystyle n\geq\\N$ implies $\displaystyle |x_{n}-x|<\epsilon$. Now, let $\displaystyle N$ be some natural number satisfying $\displaystyle N\geq\\n+100$ such that for all $\displaystyle n\geq\\N$ implies $\displaystyle |x_{n+100}-x|=|y_{n}-x|<\epsilon$. Therefore, $\displaystyle (y_{n})_{n}$ converges to x as well.
    Last edited by Danneedshelp; Sep 27th 2009 at 10:15 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 $\displaystyle \{x_n\}$ converges to $\displaystyle x$ if and only if every (infinite) subsequence of $\displaystyle \{x_n\}$ also converges to $\displaystyle x$.
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