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Math Help - need help with series limit

  1. #1
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    Post need help with series limit

    i know it's not very complicated, i just need help understanding this one


    lim(1/1*2+1/2*3+...+1/(n*(n+1)))= ?
    n->inf

    i already know the answer is 1, and i can see how the fractions are adding up to something that is <1 but i don't know how to get to that conclusion

    sorry i still don't know how to use latex, it will be different on my next posts

    thanks!
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  2. #2
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    Re: need help with series limit

    I'm thinking I'd resort to a partial fraction decomposition of the nth term.
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  3. #3
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    Re: need help with series limit

    can you elaborate on that?..
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  4. #4
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    Re: need help with series limit

    Did you do it?

    \frac{1}{n\cdot(n+1)}\;=\;\frac{?}{n}+\frac{??}{n+  1}
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  5. #5
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    Re: need help with series limit

    no because i don't understand what your'e getting at

    i know that the solution for what your'e asking is

    1/n - 1/(n+1)

    but what next?
    Last edited by idom87; September 20th 2011 at 08:22 AM.
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  6. #6
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    Re: need help with series limit

    Quote Originally Posted by idom87 View Post
    no because i don't understand what your'e getting at
    Sometimes this is just necessary:
    \sum\limits_{n - 1}^N {\frac{1}{{n(n + 1)}}}  = \sum\limits_{n - 1}^N {\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right] = 1 - \frac{1}{{N + 1}}}
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    Re: need help with series limit

    Quote Originally Posted by Plato View Post
    Sometimes this is just necessary:
    \sum\limits_{n - 1}^N {\frac{1}{{n(n + 1)}}}  = \sum\limits_{n - 1}^N {\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right] = 1 - \frac{1}{{N + 1}}}
    so the limit of 1 - \frac{1}{{N + 1}}
    is 1, ok

    but i still can't wrap my head around this part
    \sum\limits_{n - 1}^N {\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right] = 1 - \frac{1}{{N + 1}}
    can you explain why?
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  8. #8
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    Re: need help with series limit

    Quote Originally Posted by idom87 View Post
    so the limit of 1 - \frac{1}{{N + 1}}
    is 1, ok

    but i still can't wrap my head around this part
    \sum\limits_{n - 1}^N {\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right] = 1 - \frac{1}{{N + 1}}
    can you explain why?
    Let N=4 actually write the four terms.
    See how all but the first and the last 'collapse' .
    These are called collapsing sums .
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  9. #9
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    Re: need help with series limit

    thanks i got that after i wrote down your example

    what about if i have a more complex nth term?..
    like this one

    lim (1/(1*4) + 1/(4*7) + ... + 1/((3n-2)(3n+1)))
    n->inf

    ?
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  10. #10
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    Re: need help with series limit

    Same, split in partial fractions:
    \frac{1}{(3n-2)\cdot (3n+1)}=\frac{A}{3n-2}+\frac{B}{3n+1}=...
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  11. #11
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    Re: need help with series limit

    how do i find A and B?

    all i know is that

    A(3n+1)+B(3n-2)=1
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    Re: need help with series limit

    Quote Originally Posted by idom87 View Post
    how do i find A and B?
    A(3n+1)+B(3n-2)=1
    Have a look at this webpage.
    Be sure to click the show steps button.
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  13. #13
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    Re: need help with series limit

    damn i feel stupid now
    and rusty
    awsome page
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  14. #14
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    Re: need help with series limit

    Quote Originally Posted by idom87 View Post
    damn i feel stupid now
    and rusty
    awsome page
    You should really explore that website.
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