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Math Help - Convergence problem

  1. #1
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    Convergence problem

    Test the convergence of the series
    \frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+..  .
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  2. #2
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    \sum_{n=1}^{\infty}\frac{2n-1}{n(n+1)(n+2)}

    You can use any number of tests to see if it converges or not(ratio test), but if it converges what does it converges to?.
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  3. #3
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    which test to use

    Which test should i use to evaluate this ..
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  4. #4
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    Does the fact that \frac{2n-1}{n(n+1)(n+2)}< \frac{2}{n^2} help at all?
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  5. #5
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    Quote Originally Posted by galactus View Post
    \sum_{n=1}^{\infty}\frac{2n-1}{n(n+1)(n+2)}

    You can use any number of tests to see if it converges or not(ratio test), but if it converges what does it converges to?.
    Now that's a good question!
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  6. #6
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    Split that into two series (both of them converge because of direct comparison test) and then:

    \frac{1}{(n+1)(n+2)}=\frac{n!}{(n+2)!}=\frac{\Gamm  a (n+1)\Gamma (2)}{\Gamma (n+3)}=\beta (n+1,2).

    Hence, turn that into a beta integral and switch the integral by the series. Use geometric series and solve the remaining integral. Do the same with the other.

    In general for n\ge2, \sum\limits_{k=0}^{\infty }{\frac{1}{(k+1)(k+2)\cdots (k+n)}}=\frac{1}{(n-1)(n-1)!}, which can be proved in the same fashion when using Beta function.
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    Split that into two series (both of them converge because of direct comparison test) and then:

    \frac{1}{(n+1)(n+2)}=\frac{n!}{(n+2)!}=\frac{\Gamm  a (n+1)\Gamma (2)}{\Gamma (n+3)}=\beta (n+1,2).

    Hence, turn that into a beta integral and switch the integral by the series. Use geometric series and solve the remaining integral. Do the same with the other.

    In general for n\ge2, \sum\limits_{k=0}^{\infty }{\frac{1}{(k+1)(k+2)\cdots (k+n)}}=\frac{1}{(n-1)(n-1)!}, which can be proved in the same fashion when using Beta function.
    There's also a very nice pattern if you decompose the series, i.e.

    \frac{1}{n(n+1) }= \frac{1}{n}-\frac{1}{n+1}

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

    \frac{1}{n(n+1) (n+2)(n+3)}= \frac{1}{6}\left(\frac{1}{n}-\frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)

    \frac{1}{n(n+1) (n+2)(n+3)(n+4)}= \frac{1}{4!}\left(\frac{1}{n}-\frac{4}{n+1} + \frac{6}{n+2} - \frac{4}{n+3} + \frac{1}{n+4}\right)

    Then one can group the terms as to make it a telescopic series. Just my 2 cents.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    This is similar to Krizalid's post except I am not sure this is what he had in mind

    Consider

    \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n~~|x|<1

    I dont feel like justifying interchanging the integral and summation, but just notice that your sum

    \begin{aligned}\sum_{n=1}^{\infty}\frac{2n-1}{n(n+1)(n+2)}&=\sum_{n=1}^{\infty}\frac{2}{(n+1)  (n+2)}-\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}\\<br />
&=2\int_0^1\left\{\int\frac{dx}{1-x}\right\}-\int_0^1\left\{\iint\frac{dx}{x(1-x)}\right\}\\<br />
&=\frac{3}{4}\end{aligned}
    Last edited by Mathstud28; December 29th 2008 at 01:45 PM.
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    This is similar to Krizalid's post except I am not sure this is what he had in mind

    Consider

    \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n~~|x|<1

    I dont feel like justifying interchanging the integral and summation, but just notice that your sum

    \begin{aligned}\sum_{n=1}^{\infty}\frac{2n-1}{n(n+1)(n+2)}&=\sum_{n=1}^{\infty}\frac{2}{(n+1)  (n+2)}-\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}\\<br />
&=2\int_0^1\left\{\int\frac{x}{1-x}dx\right\}-\int_0^1\left\{\iint\frac{dx}{1-x}\right\}\\<br />
&=\frac{3}{4}\end{aligned}
    But the series doesn't converge at x = 1
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by danny arrigo View Post
    But the series doesn't converge at x = 1
    It is obvious that any integral or derivative of a power series evaluated at its center (in this case 0) is 0. So instead let us consider

    \begin{aligned}\iiint\frac{1}{x(1-x)}\bigg|_{x=1}&=\iiint\sum_{n=1}^{\infty}x^{n-1}\bigg|_{x=1}\\<br />
&=\sum_{n=1}^{\infty}\iiint x^{n-1} dx|_{x=1}\\<br />
&=\sum_{n=0}^{\infty}\frac{x^{n+2}}{n(n+1)(n+2)}\b  igg|_{x=1}\\<br />
&=\sum_{n=0}^{\infty}\frac{1}{n(n+1)(n+2)}\end{ali  gned}
    Is that what you mean?
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  11. #11
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    Quote Originally Posted by Mathstud28 View Post
    It is obvious that any integral or derivative of a power series evaluated at its center (in this case 0) is 0. So instead let us consider

    \begin{aligned}\iiint\frac{1}{x(1-x)}\bigg|_{x=1}&=\iiint\sum_{n=1}^{\infty}x^{n-1}\bigg|_{x=1}\\<br />
&=\sum_{n=1}^{\infty}\iiint x^{n-1} dx|_{x=1}\\<br />
&=\sum_{n=0}^{\infty}\frac{x^{n+2}}{n(n+1)(n+2)}\b  igg|_{x=1}\\<br />
&=\sum_{n=0}^{\infty}\frac{1}{n(n+1)(n+2)}\end{ali  gned}
    Is that what you mean?
    What I mean is that you're using
    \frac{1}{1-x} = \sum_{n=0}^\infty x^n but it doesn't converge at x=1
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by danny arrigo View Post
    What I mean is that you're using
    \frac{1}{1-x} = \sum_{n=0}^\infty x^n but it doesn't converge at x=1
    So? The endpoints of a radius of convergence are subject to change when integrating.
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  13. #13
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    Quote Originally Posted by Mathstud28 View Post
    So? The endpoints of a radius of convergence are subject to change when integrating.
    Okay - after two integrations it does.
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  14. #14
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    What is the right answer

    Can any body tell me the right answer
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  15. #15
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by varunnayudu View Post
    Can any body tell me the right answer
    Do you mean the actual value? Mathcad gives \frac{3}{4}
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