Find the exact value of the infinite series












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Find the exact value of the infinite series given by $$S = frac{1}{(3^2+1)} + frac{1}{(4^2+2)} + frac{1}{(5^2+3)} + ... $$



I found the notation of how it would be written with sigma notation:
$$sum_{x=1}^∞ frac{1}{(x+1)(x+4)}$$



I don't know how to get an exact value for the sum after that.



Help would be appreciated, thanks!






sequences-and-series summation







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    Please use MathJax YOu'll get more help if your questions are easy to read.
    $endgroup$
    – saulspatz
    Jan 28 at 18:53
















0












$begingroup$


Find the exact value of the infinite series given by $$S = frac{1}{(3^2+1)} + frac{1}{(4^2+2)} + frac{1}{(5^2+3)} + ... $$



I found the notation of how it would be written with sigma notation:
$$sum_{x=1}^∞ frac{1}{(x+1)(x+4)}$$



I don't know how to get an exact value for the sum after that.



Help would be appreciated, thanks!






sequences-and-series summation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please use MathJax YOu'll get more help if your questions are easy to read.
    $endgroup$
    – saulspatz
    Jan 28 at 18:53














0












0








0





$begingroup$


Find the exact value of the infinite series given by $$S = frac{1}{(3^2+1)} + frac{1}{(4^2+2)} + frac{1}{(5^2+3)} + ... $$



I found the notation of how it would be written with sigma notation:
$$sum_{x=1}^∞ frac{1}{(x+1)(x+4)}$$



I don't know how to get an exact value for the sum after that.



Help would be appreciated, thanks!






sequences-and-series summation







share|cite|improve this question











$endgroup$




Find the exact value of the infinite series given by $$S = frac{1}{(3^2+1)} + frac{1}{(4^2+2)} + frac{1}{(5^2+3)} + ... $$



I found the notation of how it would be written with sigma notation:
$$sum_{x=1}^∞ frac{1}{(x+1)(x+4)}$$



I don't know how to get an exact value for the sum after that.



Help would be appreciated, thanks!






sequences-and-series summation




sequences-and-series summation






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share|cite|improve this question













share|cite|improve this question




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edited Jan 28 at 20:39







ShadyAF

















asked Jan 28 at 18:48









ShadyAFShadyAF

448




448








  • 1




    $begingroup$
    Please use MathJax YOu'll get more help if your questions are easy to read.
    $endgroup$
    – saulspatz
    Jan 28 at 18:53














  • 1




    $begingroup$
    Please use MathJax YOu'll get more help if your questions are easy to read.
    $endgroup$
    – saulspatz
    Jan 28 at 18:53








1




1




$begingroup$
Please use MathJax YOu'll get more help if your questions are easy to read.
$endgroup$
– saulspatz
Jan 28 at 18:53




$begingroup$
Please use MathJax YOu'll get more help if your questions are easy to read.
$endgroup$
– saulspatz
Jan 28 at 18:53










2 Answers
2






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2












$begingroup$

Hint:
$$frac{1}{(x+1)(x+4)}=frac{1}{3}left(frac{1}{x+1}-frac{1}{x+4}right).$$
Can you see how to compute the telescopic sum?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, but I'm not too sure on what telescopic sums are.
    $endgroup$
    – ShadyAF
    Jan 28 at 20:40










  • $begingroup$
    @ShadyAF to see that, just write write down $frac{1}{n+1}-frac{1}{n+4}$, for the few first values of $n$, one row for each $n$, and sum
    $endgroup$
    – G Cab
    Jan 28 at 22:56










  • $begingroup$
    Hey, thanks! I got it!
    $endgroup$
    – ShadyAF
    Jan 29 at 2:07



















0












$begingroup$

You can write your sum as
$$
sum_{n=1}^{+infty}frac{1}{n+left(n+2right)^2}
$$
and
$$
frac{1}{n+left(n+2right)^2}=frac{1}{n^2+5n+4}
$$
Then you need to write
$$
frac{1}{n^2+5n+4}=frac{a}{n-alpha_1}+frac{b}{n-alpha_2}
$$
with $alpha_1, alpha_2$ being the two roots of $n^2+5n+4$.
Can you take it from there ?






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Hint:
    $$frac{1}{(x+1)(x+4)}=frac{1}{3}left(frac{1}{x+1}-frac{1}{x+4}right).$$
    Can you see how to compute the telescopic sum?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks, but I'm not too sure on what telescopic sums are.
      $endgroup$
      – ShadyAF
      Jan 28 at 20:40










    • $begingroup$
      @ShadyAF to see that, just write write down $frac{1}{n+1}-frac{1}{n+4}$, for the few first values of $n$, one row for each $n$, and sum
      $endgroup$
      – G Cab
      Jan 28 at 22:56










    • $begingroup$
      Hey, thanks! I got it!
      $endgroup$
      – ShadyAF
      Jan 29 at 2:07
















    2












    $begingroup$

    Hint:
    $$frac{1}{(x+1)(x+4)}=frac{1}{3}left(frac{1}{x+1}-frac{1}{x+4}right).$$
    Can you see how to compute the telescopic sum?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks, but I'm not too sure on what telescopic sums are.
      $endgroup$
      – ShadyAF
      Jan 28 at 20:40










    • $begingroup$
      @ShadyAF to see that, just write write down $frac{1}{n+1}-frac{1}{n+4}$, for the few first values of $n$, one row for each $n$, and sum
      $endgroup$
      – G Cab
      Jan 28 at 22:56










    • $begingroup$
      Hey, thanks! I got it!
      $endgroup$
      – ShadyAF
      Jan 29 at 2:07














    2












    2








    2





    $begingroup$

    Hint:
    $$frac{1}{(x+1)(x+4)}=frac{1}{3}left(frac{1}{x+1}-frac{1}{x+4}right).$$
    Can you see how to compute the telescopic sum?






    share|cite|improve this answer









    $endgroup$



    Hint:
    $$frac{1}{(x+1)(x+4)}=frac{1}{3}left(frac{1}{x+1}-frac{1}{x+4}right).$$
    Can you see how to compute the telescopic sum?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 28 at 18:53









    Eclipse SunEclipse Sun

    7,9851438




    7,9851438












    • $begingroup$
      Thanks, but I'm not too sure on what telescopic sums are.
      $endgroup$
      – ShadyAF
      Jan 28 at 20:40










    • $begingroup$
      @ShadyAF to see that, just write write down $frac{1}{n+1}-frac{1}{n+4}$, for the few first values of $n$, one row for each $n$, and sum
      $endgroup$
      – G Cab
      Jan 28 at 22:56










    • $begingroup$
      Hey, thanks! I got it!
      $endgroup$
      – ShadyAF
      Jan 29 at 2:07


















    • $begingroup$
      Thanks, but I'm not too sure on what telescopic sums are.
      $endgroup$
      – ShadyAF
      Jan 28 at 20:40










    • $begingroup$
      @ShadyAF to see that, just write write down $frac{1}{n+1}-frac{1}{n+4}$, for the few first values of $n$, one row for each $n$, and sum
      $endgroup$
      – G Cab
      Jan 28 at 22:56










    • $begingroup$
      Hey, thanks! I got it!
      $endgroup$
      – ShadyAF
      Jan 29 at 2:07
















    $begingroup$
    Thanks, but I'm not too sure on what telescopic sums are.
    $endgroup$
    – ShadyAF
    Jan 28 at 20:40




    $begingroup$
    Thanks, but I'm not too sure on what telescopic sums are.
    $endgroup$
    – ShadyAF
    Jan 28 at 20:40












    $begingroup$
    @ShadyAF to see that, just write write down $frac{1}{n+1}-frac{1}{n+4}$, for the few first values of $n$, one row for each $n$, and sum
    $endgroup$
    – G Cab
    Jan 28 at 22:56




    $begingroup$
    @ShadyAF to see that, just write write down $frac{1}{n+1}-frac{1}{n+4}$, for the few first values of $n$, one row for each $n$, and sum
    $endgroup$
    – G Cab
    Jan 28 at 22:56












    $begingroup$
    Hey, thanks! I got it!
    $endgroup$
    – ShadyAF
    Jan 29 at 2:07




    $begingroup$
    Hey, thanks! I got it!
    $endgroup$
    – ShadyAF
    Jan 29 at 2:07











    0












    $begingroup$

    You can write your sum as
    $$
    sum_{n=1}^{+infty}frac{1}{n+left(n+2right)^2}
    $$
    and
    $$
    frac{1}{n+left(n+2right)^2}=frac{1}{n^2+5n+4}
    $$
    Then you need to write
    $$
    frac{1}{n^2+5n+4}=frac{a}{n-alpha_1}+frac{b}{n-alpha_2}
    $$
    with $alpha_1, alpha_2$ being the two roots of $n^2+5n+4$.
    Can you take it from there ?






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You can write your sum as
      $$
      sum_{n=1}^{+infty}frac{1}{n+left(n+2right)^2}
      $$
      and
      $$
      frac{1}{n+left(n+2right)^2}=frac{1}{n^2+5n+4}
      $$
      Then you need to write
      $$
      frac{1}{n^2+5n+4}=frac{a}{n-alpha_1}+frac{b}{n-alpha_2}
      $$
      with $alpha_1, alpha_2$ being the two roots of $n^2+5n+4$.
      Can you take it from there ?






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You can write your sum as
        $$
        sum_{n=1}^{+infty}frac{1}{n+left(n+2right)^2}
        $$
        and
        $$
        frac{1}{n+left(n+2right)^2}=frac{1}{n^2+5n+4}
        $$
        Then you need to write
        $$
        frac{1}{n^2+5n+4}=frac{a}{n-alpha_1}+frac{b}{n-alpha_2}
        $$
        with $alpha_1, alpha_2$ being the two roots of $n^2+5n+4$.
        Can you take it from there ?






        share|cite|improve this answer









        $endgroup$



        You can write your sum as
        $$
        sum_{n=1}^{+infty}frac{1}{n+left(n+2right)^2}
        $$
        and
        $$
        frac{1}{n+left(n+2right)^2}=frac{1}{n^2+5n+4}
        $$
        Then you need to write
        $$
        frac{1}{n^2+5n+4}=frac{a}{n-alpha_1}+frac{b}{n-alpha_2}
        $$
        with $alpha_1, alpha_2$ being the two roots of $n^2+5n+4$.
        Can you take it from there ?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 28 at 18:52









        AtmosAtmos

        4,830420




        4,830420






























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