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Math Help - Convergence of the series 1 / n(n+2)

  1. #1
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    Convergence of the series 1 / n(n+2)

    Hi,

    I'm just about to hit my head on a table or a wall...

    I have to determine if the series (sum symbol) (n=1 to infinity) 1 / n(n+2) is convergent or not, and if it is, what is its limit. In the book, the answer is "3/4".

    So I began writing its partial sum :
    sn = 1/3 + 1/8 + 1/15 + 1/24 + ... + 1 / n(n+2) + ...

    I noticed that every term in sn is multiplied by : n(n+2) / (n+1)(n+2+1) = n(n+2) / (n+1)(n+3)

    So I thought it was a geometric series with
    a = 1 / a(n+2)
    r = n(n+2) / (n+1)(n+3)

    Since r will always be < 1 for every n >= 1, I tried to resolve a / (1 - r), but I never got 3/4.

    Thinking I had to set a equal to "1/3", I did not get 3/4 either.

    Then, I told myself it was a combinaison of two series :
    1. 1 / n2
    2. 1 / 2n

    But the second one does not have a limit since it is 1 / 2 multiplied by the harmonic series (1 / n) which does not converge.

    What am I doing wrong ? What did I not see ?

    I just started this calculus class in university and my college maths are soooooooo far away (11 years).

    Thanks for your help
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Convergence of the series 1 / n(n+2)

    Last edited by MaxJasper; October 12th 2012 at 10:59 AM.
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  3. #3
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    Re: Convergence of the series 1 / n(n+2)

    Quote Originally Posted by NZAU1984 View Post
    Hi,

    I'm just about to hit my head on a table or a wall...

    I have to determine if the series (sum symbol) (n=1 to infinity) 1 / n(n+2) is convergent or not, and if it is, what is its limit. In the book, the answer is "3/4".

    So I began writing its partial sum :
    sn = 1/3 + 1/8 + 1/15 + 1/24 + ... + 1 / n(n+2) + ...

    I noticed that every term in sn is multiplied by : n(n+2) / (n+1)(n+2+1) = n(n+2) / (n+1)(n+3)

    So I thought it was a geometric series with
    a = 1 / a(n+2)
    r = n(n+2) / (n+1)(n+3)

    Since r will always be < 1 for every n >= 1, I tried to resolve a / (1 - r), but I never got 3/4.

    Thinking I had to set a equal to "1/3", I did not get 3/4 either.

    Then, I told myself it was a combinaison of two series :
    1. 1 / n2
    2. 1 / 2n

    But the second one does not have a limit since it is 1 / 2 multiplied by the harmonic series (1 / n) which does not converge.

    What am I doing wrong ? What did I not see ?

    I just started this calculus class in university and my college maths are soooooooo far away (11 years).

    Thanks for your help
    This is a telescopic series!

    We can rewrite the summand as

    \frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n} + \frac{-1}{n+2} \right)

    The above can be found by partial fraction decomposition (or many other ways)

    \sum_{n=1}^{\infty}\frac{1}{n(n+2)}=\frac{1}{2} \sum_{n=1}^{\infty}\left( \frac{1}{n}+\frac{-1}{n+2} \right)

    Notice if you write out a few terms you will see that pattern...

    \frac{1}{2} \left[ \left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ ...\right)-\left( \frac{1}{3}+\frac{1}{4}+ ...\right) \right]

    \frac{1}{2} (\left( 1+\frac{1}{2} \right)=\frac{3}{4}
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    Re: Convergence of the series 1 / n(n+2)

    I just forgot about simplification, I really don't see that stuff anymore (I used to be so good in maths).

    Thanks !

    Just a little question,

    At the very end of sn, there would remain \frac{-1}{n+2} which tends towards 0 as n tends to zero, right ? So that's why we only have left 1 \cdot (\frac{1}{1}+\frac{1}{2}) ?
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    Re: Convergence of the series 1 / n(n+2)

    Quote Originally Posted by NZAU1984 View Post
    Just a little question,
    At the very end of sn, there would remain \frac{-1}{n+2} which tends towards 0 as n tends to zero, right ? So that's why we only have left 1 \cdot (\frac{1}{1}+\frac{1}{2}) ?
    For n\ge 3, each partial sum collapses to S_n  = \frac{1}{2}\left[ {1 + \frac{1}{2} - \frac{1}{{n + 1}} - \frac{1}{{n + 2}}} \right]

    The sequence of partial sums converges \left( {S_n } \right) \to \frac{1}{2}\left[ {1 + \frac{1}{2}} \right] = \frac{3}{4}
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    Re: Convergence of the series 1 / n(n+2)

    Quote Originally Posted by NZAU1984 View Post
    Hi,

    I'm just about to hit my head on a table or a wall...

    I have to determine if the series (sum symbol) (n=1 to infinity) 1 / n(n+2) is convergent or not, and if it is, what is its limit. In the book, the answer is "3/4".

    So I began writing its partial sum :
    sn = 1/3 + 1/8 + 1/15 + 1/24 + ... + 1 / n(n+2) + ...

    I noticed that every term in sn is multiplied by : n(n+2) / (n+1)(n+2+1) = n(n+2) / (n+1)(n+3)

    So I thought it was a geometric series with
    a = 1 / a(n+2)
    r = n(n+2) / (n+1)(n+3)
    That's not a geometric series. A Geometric series is of the form \sum ar^n with a and r constants, not depending on n.

    Since r will always be < 1 for every n >= 1, I tried to resolve a / (1 - r), but I never got 3/4.

    Thinking I had to set a equal to "1/3", I did not get 3/4 either.

    Then, I told myself it was a combinaison of two series :
    1. 1 / n2
    2. 1 / 2n

    But the second one does not have a limit since it is 1 / 2 multiplied by the harmonic series (1 / n) which does not converge.

    What am I doing wrong ? What did I not see ?

    I just started this calculus class in university and my college maths are soooooooo far away (11 years).

    Thanks for your help
    Follow Math Help Forum on Facebook and Google+

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    Re: Convergence of the series 1 / n(n+2)

    Quote Originally Posted by TheEmptySet View Post

    \frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n} + \frac{-1}{n+2} \right)
    This is the trick to solving this problem and others like it.

    - Hollywood
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: Convergence of the series 1 / n(n+2)

    There is a general result:if q is a positive integer and \varphi (n) a convergent sequence, S=\sum_{n=1}^{\infty}(\varphi (n+q)-\varphi(n))=q\lim_{n\to \infty}\varphi(n)-(\varphi(1)+\ldots +\varphi(q)). In our case, \varphi(n)=\frac{-1/2}{n} and q=2, so S=-\varphi(1)-\varphi(2)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}.
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