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Math Help - Sum of series

  1. #1
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    Question Sum of series

    What is the sum of the following series:

    SUMMATION(n=1 to infiniti) -1/(n^2 + 9n + 20)

    I tried using the forumula:
    sum = a/(1-r)

    But i cant get an exact value for r so i am not sure

    Thanks
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  2. #2
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    Quote Originally Posted by taurus View Post
    What is the sum of the following series:

    SUMMATION(n=1 to infiniti) -1/(n^2 + 9n + 20)

    I tried using the forumula:
    sum = a/(1-r)

    But i cant get an exact value for r so i am not sure

    Thanks
    \frac{a}{1-r} is for a geometric series: \sum_{n=1}^{\infty} ar^n

    To sum the series: \sum_{n=1}^{\infty} \frac{-1}{(n+4)(n+5)}, try working out the first few terms...

    \frac{-1}{30} + \frac{-1}{42} + \frac{-1}{56} + \frac{-1}{72} + \frac{-1}{90} + \frac{-1}{110} + \ldots

    Not sure what the value is, but it seems that it lies in (-1,0)
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  3. #3
    Moo
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    Hello

    Quote Originally Posted by taurus View Post
    What is the sum of the following series:

    SUMMATION(n=1 to infiniti) -1/(n^2 + 9n + 20)

    I tried using the forumula:
    sum = a/(1-r)

    But i cant get an exact value for r so i am not sure

    Thanks
    n^2+9n+20=(n+4)(n+5)

    \frac{-1}{(n+4)(n+5)}=\frac{n-n+4-5}{(n+4)(n+5)}=\frac{(n+4)-(n+5)}{(n+4)(n+5)}

    =\frac{(n+4)}{(n+4)(n+5)}-\frac{(n+5)}{(n+4)(n+5)}=\frac{1}{n+5}-\frac{1}{n+4}


    So \sum_{n=1}^\infty \frac{-1}{n^2+9n+20}=\sum_{n=1}^\infty \left(-\frac{1}{n+4}+\frac{1}{n+5}\right) =\left(-\frac 15+{\color{red}\frac 16}\right)+\left({\color{red}-\frac 16}+{\color{blue}\frac 17}\right)+\left({\color{blue}-\frac 17}+\frac 18\right)+\dots

    What can you conclude ? (this is called a telescoping sum)




    @ colby : partial fractions is the key
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  4. #4
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    Question

    I still dont get it.

    What happened to your first line with n-n etc?

    Also i still dont get how to get the sum then?
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  5. #5
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    Quote Originally Posted by Moo View Post
    @ colby : partial fractions is the key
    I had to rush out of the room before I finished the solution. Thanks for finishing it for me.

    Taurus, Moo did some algebraic manipulation by having -1 = n - n - 5 + 4
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  6. #6
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    Quote Originally Posted by taurus View Post
    I still dont get it.

    What happened to your first line with n-n etc?

    Also i still dont get how to get the sum then?
    I grouped the terms..



    For the sum, you can notice that each term is annulated by the one which follows, apart from the first term, -1/5.

    Telescoping series - Wikipedia, the free encyclopedia
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  7. #7
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    Unhappy

    i cant conclude anything, all i see is a series of numbers with 1/7 repeated?
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  8. #8
    Moo
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    Quote Originally Posted by taurus View Post
    i cant conclude anything, all i see is a series of numbers with 1/7 repeated?
    ..

    It's -1/5+1/6-1/6+1/7-1/8+1/8-1/9+1/9-..............

    So you can see that fractions annulate themselves, apart from -1/5.

    Therefore, there remains only -1/5 at the end
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  9. #9
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    but that doesnt give me the sum.
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  10. #10
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    Therefore, there remains only -1/5 at the end
    From Moo. This is the result of your infinite sum.
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  11. #11
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    Question

    But when i enter that as an answer it says its wrong it says:
    Your answer is not correct.
    Hint: break up the summand as partial fractions and then write out the series to the first few terms.
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  12. #12
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    Quote Originally Posted by taurus View Post
    What is the sum of the following series:

    SUMMATION(n=1 to infiniti) -1/(n^2 + 9n + 20)

    I tried using the forumula:
    sum = a/(1-r)

    But i cant get an exact value for r so i am not sure

    Thanks

    \frac{-1}{(n+4)(n+5)}=\frac{-1+5-5+n-n}{(n+4)(n+5)}=\frac{(n+4)-(n+5)}{(n+4)(n+5)}=


     \frac{(n+4)}{(n+4)(n+5)}+\frac{-(n+5)}{(n+4)(n+5)} =\frac{1}{(n+5)}-\frac{1}{(n+4)}

     \sum_{n=1}^{\infty} \left( \frac{1}{(n+5)} - \frac{1}{(n+4)} \right)= \sum_{n=1}^{\infty} \frac{1}{n+5} - \sum_{n=1}^{\infty} \frac{1}{n+4}

    lets focus on the 2nd sum for a second

    \sum_{n=1}^{\infty} \frac{1}{n+4}=\frac{1}{5}+\sum_{n=2}^{\infty} \frac{1}{n+4}

    lets reindex the series k = n-1 theb n=k+1

     <br />
\frac{1}{5}+\sum_{n=2}^{\infty} \frac{1}{n+4}=\frac{1}{5}+\sum_{k=1}^{\infty} \frac{1}{k+5}=\frac{1}{5}+\sum_{n=1}^{\infty} \frac{1}{n+5}<br />

    for the last step just reindex again with n=k

    Now back to where we were before

    \sum_{n=1}^{\infty} \frac{1}{n+5} - \sum_{n=1}^{\infty} \frac{1}{n+4}=\sum_{n=1}^{\infty} \frac{1}{n+5} -\left(\frac{1}{5}+ \sum_{n=1}^{\infty} \frac{1}{n+5} \right)

    -\frac{1}{5}+\sum_{n=1}^{\infty}\left( \frac{1}{n+5}-\frac{1}{n-5}\right)=-\frac{1}{5}+\sum_{n=1}^{\infty}0=-\frac{1}{5}
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  13. #13
    Moo
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    Quote Originally Posted by taurus View Post
    But when i enter that as an answer it says its wrong it says:
    Your answer is not correct.
    Hint: break up the summand as partial fractions and then write out the series to the first few terms.
    This is what I did : break the summand...Then I wrote out the first few terms of the series...

    I see no mistake ~

    Was that \sum \frac{1}{n^2+9n+20} or \sum \frac{{\color{red}-}1}{n^2+9n+20}
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  14. #14
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    Question

    oh sorry, this si totally my fault, typo
    correct question is:

    SUMMATION(n=1 to infiniti) -2/(n^2 + 9n + 20)
    ?

    thanks
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  15. #15
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    Quote Originally Posted by taurus View Post
    oh sorry, this si totally my fault, typo
    correct question is:

    SUMMATION(n=1 to infiniti) -2/(n^2 + 9n + 20)
    ?

    thanks
    \sum \frac{-2}{n^2+9n+20}=2 \cdot \sum \frac{-1}{n^2+9n+20}=2 \cdot \left(-\frac 15\right)

    ...
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