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Infinite series

  1. May 12, 2005 #1
    if we know that an infinite series is convergent from an integer T, to infinity, then the series is convergent from 1 to infinity. conversely, if a series is convergent from 1 to infinity then it is convergent from T to infinity (i.e. starting point of the series does not affect convergence/divergence) This seems obvious but can anyone help me prove it please.
     
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  3. May 12, 2005 #2

    Galileo

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    This is obvious from the definition of a convergent series. So check the definition again.
     
  4. May 12, 2005 #3

    Zurtex

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    I pulled my lecturer up on this recently, consider the series:

    [tex]\sum_{x=10}^{\infty} \frac{1}{(x-7)^2}[/tex]

    It converges and if you are interested to:

    [tex]\frac{1}{12} \left(2 \pi^2 - 15\right)[/tex]

    However:

    [tex]\sum_{x=1}^{\infty} \frac{1}{(x-7)^2}[/tex]

    Clearly does not converge, so be careful how you word it. Anyway, it's not too difficult to prove, just think of it like:

    [tex]a_1 + a_2 + \ldots + a_{t-1} + \sum_{n=t}^{\infty} a_n[/tex]
     
  5. May 12, 2005 #4

    mathman

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    Is the series adamg asking about a power series? If so, the answer to his question is yes. However in the more general case as Zurtex showed, it is not true.
     
  6. May 13, 2005 #5

    Galileo

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    I'd rather say his example

    [tex]\sum_{x=1}^{\infty} \frac{1}{(x-7)^2}[/tex]

    is not a series, since the 7'th term is not defined.
     
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