Finding the n-th partial sum of a series Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How to find the partial sum of a given series?Partial sum formula of a polynomial series?Find the sum to n terms of the series $frac11.2.3+frac32.3.4+frac53.4.5+frac74.5.6+cdots$..Sum $sum _ n=1 ^ infty frac 2n left( n+1 right) ! $Summing up series which is similar to Taylor expansionFinding the sum of the infinite seriesSome of series involving factorial in the denominatorFinding a partial sumFind the sum to n terms as well as the sum to infinity of the series:Partial sum of divergent series

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Finding the n-th partial sum of a series



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How to find the partial sum of a given series?Partial sum formula of a polynomial series?Find the sum to n terms of the series $frac11.2.3+frac32.3.4+frac53.4.5+frac74.5.6+cdots$..Sum $sum _ n=1 ^ infty }{ frac 2n left( n+1 right) ! $Summing up series which is similar to Taylor expansionFinding the sum of the infinite seriesSome of series involving factorial in the denominatorFinding a partial sumFind the sum to n terms as well as the sum to infinity of the series:Partial sum of divergent series










0












$begingroup$


I am trying to find the partial sum of the series defined by:



$$frac91.2.3+frac92.3.4+frac93.4.5+....+frac9n(n+1)(n+2)+...$$



I know the answer is $$frac9n(n+3)4(n+1)(n+2)$$t



I tried the telescopic series but didn't work.
Any suggestions?










share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    Induction is always a possibility (not particularly neat one but will get you there)
    $endgroup$
    – Radost
    Apr 1 at 19:05















0












$begingroup$


I am trying to find the partial sum of the series defined by:



$$frac91.2.3+frac92.3.4+frac93.4.5+....+frac9n(n+1)(n+2)+...$$



I know the answer is $$frac9n(n+3)4(n+1)(n+2)$$t



I tried the telescopic series but didn't work.
Any suggestions?










share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    Induction is always a possibility (not particularly neat one but will get you there)
    $endgroup$
    – Radost
    Apr 1 at 19:05













0












0








0





$begingroup$


I am trying to find the partial sum of the series defined by:



$$frac91.2.3+frac92.3.4+frac93.4.5+....+frac9n(n+1)(n+2)+...$$



I know the answer is $$frac9n(n+3)4(n+1)(n+2)$$t



I tried the telescopic series but didn't work.
Any suggestions?










share|cite|improve this question









$endgroup$




I am trying to find the partial sum of the series defined by:



$$frac91.2.3+frac92.3.4+frac93.4.5+....+frac9n(n+1)(n+2)+...$$



I know the answer is $$frac9n(n+3)4(n+1)(n+2)$$t



I tried the telescopic series but didn't work.
Any suggestions?







sequences-and-series convergence summation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 1 at 19:04









Rabih AssafRabih Assaf

524




524







  • 2




    $begingroup$
    Induction is always a possibility (not particularly neat one but will get you there)
    $endgroup$
    – Radost
    Apr 1 at 19:05












  • 2




    $begingroup$
    Induction is always a possibility (not particularly neat one but will get you there)
    $endgroup$
    – Radost
    Apr 1 at 19:05







2




2




$begingroup$
Induction is always a possibility (not particularly neat one but will get you there)
$endgroup$
– Radost
Apr 1 at 19:05




$begingroup$
Induction is always a possibility (not particularly neat one but will get you there)
$endgroup$
– Radost
Apr 1 at 19:05










1 Answer
1






active

oldest

votes


















0












$begingroup$

Use partial fractions to get
$$a_n=frac92 left (frac1r-frac1r+1 right )-frac92 left (frac1r+1-frac1r+2 right )$$
Now use telescoping.






share|cite|improve this answer









$endgroup$













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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Use partial fractions to get
    $$a_n=frac92 left (frac1r-frac1r+1 right )-frac92 left (frac1r+1-frac1r+2 right )$$
    Now use telescoping.






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      Use partial fractions to get
      $$a_n=frac92 left (frac1r-frac1r+1 right )-frac92 left (frac1r+1-frac1r+2 right )$$
      Now use telescoping.






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        Use partial fractions to get
        $$a_n=frac92 left (frac1r-frac1r+1 right )-frac92 left (frac1r+1-frac1r+2 right )$$
        Now use telescoping.






        share|cite|improve this answer









        $endgroup$



        Use partial fractions to get
        $$a_n=frac92 left (frac1r-frac1r+1 right )-frac92 left (frac1r+1-frac1r+2 right )$$
        Now use telescoping.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 1 at 19:15









        Anurag AAnurag A

        26.4k12351




        26.4k12351



























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