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Series Converges.

  1. Aug 5, 2012 #1
    When I say a series [itex]\sum[/itex]a[itex]_{n}[/itex] converges, what exactly is it that I am saying?
    for example
    [itex]\sum^{∞}_{n=1}[/itex][itex]\frac{9n^{2}}{3n^{5}+5}[/itex] is convergent. what did I just say?
     
    Last edited: Aug 5, 2012
  2. jcsd
  3. Aug 5, 2012 #2

    HallsofIvy

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    Why would you say a series converges if you don't know what it means?

    If you have taken a course dealing with sequences and series, then you should have seen a definition of "convergence of a sequence": the series [itex]\sum_{n=1}^\infty a_n[/itex] converges if and only if the sequence of partial sums [itex]s_i= \sum_{n= 0}^i a_n[/itex] converges.

    (I hope you won't say that [itex]\sum_{n=1}^\infty \frac{9n^2}{3n^2+ 5}[/itex] is convergent. It obviously isn't.)
     
  4. Aug 5, 2012 #3
    Congrats on telling me exactly what the book told . so now if you don't mind tell it to me as if I was not a person studying Mathematics .
     
  5. Aug 5, 2012 #4

    mathman

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    In plain English a series is convergent if you keep adding terms of the series and it gets to a limit. For example 1 + 1/2 + 1/4 + 1/8 + ... gets closer and closer to 2 as you add more terms. On the other hand 1 + 1/2 + 1/3 + 1/4 + 1/5 + .... keeps getting bigger as you add more terms, so it is not convergent.
     
  6. Aug 5, 2012 #5

    Bacle2

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    A series converges to a value "s" if, by adding enough terms , you can get indefinitely

    close to the value s. This is made rigorous : if I want to be within, say, 1/100 of the

    value s, I must show that there is a term ,say "N", so that by adding N-or-more terms,

    the value of the expression: (a_1+a_2+......+a_N )-s

    of the sum will be within 1/100 of the value s. Take the series 1+1/2+1/4+.......

    Its limit is 2. After 1 term, you are within 1 unit of the limit. After adding two terms

    you are within 1/2 of the limit. Now, convergence means that I can guarantee that , no

    matter how close I want to get to 2, I just need to add enough terms, and my sum

    will be within this --or (almost) any other--distance from 2. We do not demand that the

    sum be exactly two, but that the sum be indefinitely close to it.
     
  7. Aug 5, 2012 #6
    Thank you guys very much I now understand.
     
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