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I have a problem

  1. Jan 8, 2005 #1
    [tex] x_{n}+px_{n-1} [/tex] converges [tex]\Leftrightarrow x_{n}[/tex] converges.
    For which p it is true?

  2. jcsd
  3. Jan 9, 2005 #2
    "Xn converges -> Xn + pXn-1 converges" is always true.
    "Xn + pXn-1 converges -> Xn converges" is true if |p| < 1
  4. Jan 9, 2005 #3


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    I'm guessing there are two series. The first being [tex]x_{n}[/tex] and the second being defined in terms of the first as: [tex]y_{n}=x_{n}+px_{n-1}[/tex]

    Assume [tex]x_{n}[/tex] converges. So there exists a finite L such that:
    [tex]\lim_{n \rightarrow \infty} x_{n} = L[/tex]

    To see whether or not [tex]y_{n}[/tex] converges, calculate the limit:
    [tex]\lim_{n \rightarrow \infty} x_{n} + px_{n-1}[/tex]

    [tex]= \lim_{n \rightarrow \infty} x_{n} + p \lim_{n \rightarrow \infty} x_{n-1}[/tex]

    [tex]= L + pL [/tex]

    L+pL exists and it is finite for all p, so [tex]y_{n}[/tex] converges.

    So we've shown
    [tex]x_{n}[/tex] converges [tex]\rightarrow x_{n} + px_{n-1}[/tex] converges for all p.

    Assume [tex]y_{n}[/tex] converges. So there is a finite M such that:
    [tex]\lim_{n \rightarrow \infty} x_{n} + px_{n-1} = M[/tex]

    we can rewrite the left side:
    [tex]\lim_{n \rightarrow \infty} x_{n} + p \lim_{n \rightarrow \infty} x_{n-1} = M[/tex]

    [tex]\lim_{n \rightarrow \infty} x_{n} + p \lim_{n \rightarrow \infty} x_{n} = M[/tex]

    [tex] (1+p) \lim_{n \rightarrow \infty} x_{n} = M[/tex]

    [tex] \lim_{n \rightarrow \infty} x_{n} = M/(1+p)[/tex]

    M/(1+p) exists and is finite for all p not equal to -1.

    [tex]x_{n} + px_{n-1} [/tex] converges [tex]\rightarrow x_{n} [/tex] converges for all p not equal -1
    Last edited: Jan 9, 2005
  5. Jan 9, 2005 #4
    That's only true if x_n converges, which is what you were trying to prove... ;)
  6. Jan 9, 2005 #5


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    Example where yn series converges and xn does not is as follows: xn =1 for even n and -1 for odd n, for p =1, all yn =0, while the xn series oscillates between 1 and 0 (assuming start with n=0).
  7. Jan 9, 2005 #6


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    Yes, you're right. All I did was show that if the limit exists then it is M/(p+1) and that the limit does not exist for p=-1. But I did not show where the limit does exist.

    Anyway, please ignore my solution... Only look if you want to know how NOT to do calculus!
    Last edited: Jan 9, 2005
  8. Jan 10, 2005 #7


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    It is easy to construct an example for any p, where |p|<=1.
    Let xn=(-1/p)n. Then the x series diverges or oscillates, while the y series is always 0.
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