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Math Help - Analize convergence of an infinite series

  1. #1
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    Analize convergence of an infinite series

    This is the series:

    \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})}

    So this is what I did:

    \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})} = \sum_{n=1}^{\inf}{ \tfrac{1}{2n}} - \sum_{n=1}^{\inf} {\tfrac{1}{2(n+1)}}= <br />
\tfrac{1}{2} (\sum_{n=1}^{\inf}{ \tfrac{1}{n}} - \sum_{n=1}^{\inf} {\tfrac{1}{(n+1)})}
    So the first term diverges because it is harmonic and thus the hole series diverges.

    I have this marked as wrong, but I don't understand why. Can somebody explain it please?
    Last edited by ascii; July 20th 2008 at 02:36 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ascii View Post
    This is the series:

    \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})}

    So this is what I did:

    \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})} = \sum_{n=1}^{\inf}{ \tfrac{1}{2n}} - \sum_{n=1}^{\inf} {\tfrac{1}{2(n+1)}}= <br />
\tfrac{1}{2} (\sum_{n=1}^{\inf}{ \tfrac{1}{n}} - \sum_{n=1}^{\inf} {\tfrac{1}{(n+1)})}
    So the first term converges because it is harmonic and thus the hole series converges.

    I have this marked as wrong, but I don't understand why. Can somebody explain it please?
    Because if you actually write down the first couple terms of the series, you will see all but the first and the last terms cancel out

    Oh, and FYI \infty is given by \infty =)
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  3. #3
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    Oh thank you... so if there was a plus between the series they would diverge for sure.
    I'll try other way.

    Thanks again
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  4. #4
    MHF Contributor arbolis's Avatar
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    So the first term converges because it is harmonic
    If you're talking about the harmonic series, then it diverges! (you can see it by doing the integral test for example).
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    Quote Originally Posted by arbolis View Post
    If you're talking about the harmonic series, then it diverges! (you can see it by doing the integral test for example).
    Oopsy! I know that, my mistake typing. I meant diverges.
    If I had meant converges, then the question in the first place wouldn't have made any sense.
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  6. #6
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    Ok so I got this series:
    \tfrac{1}{2}\sum_{n=1}^{\infty}{ (\tfrac{1}{n(n+1)})}
    Which converges as the limit comparison with 1/n^2 confirms.
    But now I'm asked to express the result of the sum.
    I think this is a telescoping series, isn't it?
    If it is, then the sum is equal to the limit when N goes to infinity of the first term minus the N-th term?
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    Quote Originally Posted by ascii View Post
    Ok so I got this series:
    \tfrac{1}{2}\sum_{n=1}^{\infty}{ (\tfrac{1}{n(n+1)})}
    Which converges as the limit comparison with 1/n^2 confirms.
    But now I'm asked to express the result of the sum.
    I think this is a telescoping series, isn't it?
    If it is, then the sum is equal to the limit when N goes to infinity of the first term minus the N-th term?
    \sum_{n=1}^{\infty}\frac{1}{n} - \frac{1}{n+1} = 1 - \frac{1}{2}+\frac{1}{2} - \frac{1}{3}+\frac{1}{3} - .... = 1.
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  8. #8
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    So if I want to know the sum of a telescoping I have to express a few terms and see how they cancel?
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    The series \sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{2n}} - \frac{1}{{2n + 2}}} \right)} is very nice example of a collapsing sum.
    A partial sum of the series looks this:
    S_K  = \sum\limits_{n = 1}^K {\left( {\frac{1}{{2n}} - \frac{1}{{2n + 2}}} \right) = } \left( {\frac{1}{2} - \frac{1}{4}} \right) + \left( {\frac{1}{4} - \frac{1}{6}} \right) + \left( {\frac{1}{6} - \frac{1}{8}} \right) \cdots  + \left( {\frac{1}{{2K}} - \frac{1}{{2K + 2}}} \right) = \left( {\frac{1}{2} - \frac{1}{{2K + 2}}} \right)
    Now it simple to see that S_K  \to \frac{1}{2},
    Therefore the series converges,
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  10. #10
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    I quote myself and reask the quesiton
    Quote Originally Posted by ascii View Post
    the sum is equal to the limit when N goes to infinity of the first term minus the N-th term?
    Is it always like that?
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  11. #11
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    Quote Originally Posted by ascii View Post
    I quote myself and reask the quesiton Is it always like that?
    It is always dangerous to answer such a general question!
    However, it is always true that if the sequence of partial sums converges then the series converges and the sum of the series is the limit of the sequence of partial sums.
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  12. #12
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    Thank you.
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  13. #13
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    Hello,

    Be very careful !

    The sum of a divergent series and a convergent series is indeed divergent.

    But the sum of 2 divergent series is not necessarily divergent.
    ------------------
    Another way of doing it :

    \frac 12 \left(\sum_{n=1}^\infty \tfrac 1n-\sum_{n=1}^\infty \tfrac1{n+1}\right)

    We can transform the second one by changing the indice :
    \sum_{n=1}^\infty \tfrac1{n+1}=\sum_{n={\color{red}2}}^\infty \tfrac 1n


    So the sum is now :
    \frac 12 \left(\sum_{n=1}^\infty \tfrac 1n-\sum_{n=2}^\infty \tfrac 1n\right)=\frac 12 \left(1+\underbrace{\sum_{n=2}^\infty \tfrac 1n-\sum_{n=2}^\infty \tfrac 1n}_{=0}\right)=\frac 12
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  14. #14
    MHF Contributor kalagota's Avatar
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    can i also react?

    Quote Originally Posted by ascii View Post
    This is the series:

    \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})}

    So this is what I did:

    \sum_{n=1}^{\inf}{ (\tfrac{1}{2n} - \tfrac{1}{2(n+1)})} = \sum_{n=1}^{\inf}{ \tfrac{1}{2n}} - \sum_{n=1}^{\inf} {\tfrac{1}{2(n+1)}}= <br />
\tfrac{1}{2} (\sum_{n=1}^{\inf}{ \tfrac{1}{n}} - \sum_{n=1}^{\inf} {\tfrac{1}{(n+1)})}
    So the first term diverges because it is harmonic and thus the hole series diverges.

    I have this marked as wrong, but I don't understand why. Can somebody explain it please?
    you cannot do this: \sum (a_n + b_n) = \sum a_n + \sum b_n unless \sum a_n and \sum b_n both converge.
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  15. #15
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    Quote Originally Posted by kalagota View Post
    can i also react?



    you cannot do this: \sum (a_n + b_n) = \sum a_n + \sum b_n unless \sum a_n and \sum b_n both converge.
    Why ?

    \sum (a_n+b_n)=a_1+b_1+a_2+b_2+\dots

    \sum a_n+\sum b_n=a_1+a_2+\dots+b_1+b_2+\dots

    rearranging the terms won't change the convergence or divergence, so :

    \sum a_n+\sum b_n=a_1+b_1+a_2+b_2+\dots=\sum (a_n+b_n)

    The most important thing here, imo, is that the domain of n for the two sums is the same, isn't it ?
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