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Infinite series proof

  1. Sep 29, 2011 #1
    Prove that [itex]\sum_{i=1}^{\infty }x_{i}[/itex] = [itex]\sum_{i=1}^{\infty }(x_{2i} + x_{2i+1})[/itex] if [itex]\sum_{i=1}^{\infty }x_{i}[/itex] converges and if for any [itex]\varepsilon > 0[/itex] there is some m such that [itex]|x_{k}| < \varepsilon[/itex] for all [itex]k\geq m[/itex].

    I'm a little confused by this because for 1/(4^i) for i from 1 to infinity, this doesn't hold. The sum of 1/(4^2i) for i from 1 to infinity is 1/15 and the sum of 1/(4^(2i+1)) for i from 1 to infinity is 1/60. Together they have a sum of 1/12. But the sum of 1/(4^i) for i from 1 to infinity is 1/3. Doesn't this only hold when the index starts at i=0?
     
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  3. Sep 29, 2011 #2

    SammyS

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    There is a small problem with the indices. The sum on the right is missing x1.
     
  4. Sep 29, 2011 #3
    Yes! That's exactly what I was thinking. So there must be something wrong with the question, right?
     
  5. Sep 29, 2011 #4

    SammyS

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    It would work if the sums started at i = 0.

    It would also work if the index 2i+1 were changed to 2i-1 in the sum on the right side, or if the sum on the left side had i start at 2, etc.

    Of course the equality can only hold if both series converge.
     
  6. Sep 29, 2011 #5
    True.

    I'm not sure if I need to outline the different scenarios for which the equality holds, or just state that the equality as given in the problem is not true. I guess I'll have to ask my prof.

    Thanks for the resposnes though.
     
  7. Sep 29, 2011 #6

    SammyS

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    I suspect the both sums should start with i=0, & that's the version your prof. will want you work with.
     
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