Find the sum of the infinite series $sum_{n=3}^infty [3/(n(n+3))]$












0












$begingroup$


I just do not understand how to find the sum of this infinite series; I understand that through partial fraction composition you come out with a sum of that ends with $ frac{1}{n} - frac{1}{n+3} $ but I do not know how to apply this to find the sum of this series.



$$ sum_{i=3}^inftyfrac{3}{n(n+3)} $$






calculus sequences-and-series summation







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    0












    $begingroup$


    I just do not understand how to find the sum of this infinite series; I understand that through partial fraction composition you come out with a sum of that ends with $ frac{1}{n} - frac{1}{n+3} $ but I do not know how to apply this to find the sum of this series.



    $$ sum_{i=3}^inftyfrac{3}{n(n+3)} $$






    calculus sequences-and-series summation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I just do not understand how to find the sum of this infinite series; I understand that through partial fraction composition you come out with a sum of that ends with $ frac{1}{n} - frac{1}{n+3} $ but I do not know how to apply this to find the sum of this series.



      $$ sum_{i=3}^inftyfrac{3}{n(n+3)} $$






      calculus sequences-and-series summation







      share|cite|improve this question











      $endgroup$




      I just do not understand how to find the sum of this infinite series; I understand that through partial fraction composition you come out with a sum of that ends with $ frac{1}{n} - frac{1}{n+3} $ but I do not know how to apply this to find the sum of this series.



      $$ sum_{i=3}^inftyfrac{3}{n(n+3)} $$






      calculus sequences-and-series summation




      calculus sequences-and-series summation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 26 at 1:04









      Chris Custer

      14.2k3827




      14.2k3827










      asked Jan 26 at 0:38









      cat123cat123

      32




      32






















          4 Answers
          4






          active

          oldest

          votes


















          1












          $begingroup$

          The operating principle here is called telescoping. Think of an old fashioned shipboard (pirate) spyglass-style telescope that folds inward to become shorter.



          You already know how to decompose into partial fractions. Let's write down the first few terms of the series formed after that decomposition:



          $frac 13 - frac 16 + frac 14 - frac 17 + frac 15 - frac 18 + frac 16 - frac 19 + frac 17 - frac 1{10} + ...$



          Note how, starting from $frac 16$ onward every initial negative term has a positive counterpart later on. In essence, all negative terms will simply vanish by cancellation, as will all positive terms after $frac 15$. So your entire series "telescopes" to become the really short $frac 13 + frac 14 + frac 15 = frac{47}{60}$.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Hint: this is a telescopic series:
            $$
            sum_{n=3}^inftyleft[frac1n-frac{1}{n+3}right]=frac13+frac14+frac15,
            $$
            in which only three first positive terms survive, the other being cancelled by the preceeding negative ones.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The series sum starts from index $3$ not $1$.
              $endgroup$
              – Deepak
              Jan 26 at 0:48












            • $begingroup$
              @Deepak Thanks for noting this.
              $endgroup$
              – user
              Jan 26 at 0:50



















            0












            $begingroup$

            Hint:
            $$frac3{n(n+3}=frac 1n-frac 1{n+3}$$
            Write out the beginning of this sum, and you should find telescoping terms.






            share|cite|improve this answer









            $endgroup$





















              -1












              $begingroup$

              This is a special case
              of the following theorem:



              Given $k$ reals
              $(a_k)_{j=1}^k$,
              the sum
              $sum_{n=1}^{infty} sum_{j=1}^k dfrac{a_j}{(n-1)k+j}
              $
              converges
              if and only if
              $sum_{j=1}^k a_j= 0$.



              This problem is the case
              $k=4, a_1=1, a_2=0, a_3 = 0, a_4 = -1$.
              Therefore the sum converges.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                The problem appears to be to find the value of the sum, not whether it converges.
                $endgroup$
                – 6005
                Jan 26 at 2:49












              • $begingroup$
                Yep. I posted too fast. I think I'll make my answer into a question with a proof. Thanks.
                $endgroup$
                – marty cohen
                Jan 26 at 5:19











              Your Answer





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              4 Answers
              4






              active

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              4 Answers
              4






              active

              oldest

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              active

              oldest

              votes






              active

              oldest

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              1












              $begingroup$

              The operating principle here is called telescoping. Think of an old fashioned shipboard (pirate) spyglass-style telescope that folds inward to become shorter.



              You already know how to decompose into partial fractions. Let's write down the first few terms of the series formed after that decomposition:



              $frac 13 - frac 16 + frac 14 - frac 17 + frac 15 - frac 18 + frac 16 - frac 19 + frac 17 - frac 1{10} + ...$



              Note how, starting from $frac 16$ onward every initial negative term has a positive counterpart later on. In essence, all negative terms will simply vanish by cancellation, as will all positive terms after $frac 15$. So your entire series "telescopes" to become the really short $frac 13 + frac 14 + frac 15 = frac{47}{60}$.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                The operating principle here is called telescoping. Think of an old fashioned shipboard (pirate) spyglass-style telescope that folds inward to become shorter.



                You already know how to decompose into partial fractions. Let's write down the first few terms of the series formed after that decomposition:



                $frac 13 - frac 16 + frac 14 - frac 17 + frac 15 - frac 18 + frac 16 - frac 19 + frac 17 - frac 1{10} + ...$



                Note how, starting from $frac 16$ onward every initial negative term has a positive counterpart later on. In essence, all negative terms will simply vanish by cancellation, as will all positive terms after $frac 15$. So your entire series "telescopes" to become the really short $frac 13 + frac 14 + frac 15 = frac{47}{60}$.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The operating principle here is called telescoping. Think of an old fashioned shipboard (pirate) spyglass-style telescope that folds inward to become shorter.



                  You already know how to decompose into partial fractions. Let's write down the first few terms of the series formed after that decomposition:



                  $frac 13 - frac 16 + frac 14 - frac 17 + frac 15 - frac 18 + frac 16 - frac 19 + frac 17 - frac 1{10} + ...$



                  Note how, starting from $frac 16$ onward every initial negative term has a positive counterpart later on. In essence, all negative terms will simply vanish by cancellation, as will all positive terms after $frac 15$. So your entire series "telescopes" to become the really short $frac 13 + frac 14 + frac 15 = frac{47}{60}$.






                  share|cite|improve this answer











                  $endgroup$



                  The operating principle here is called telescoping. Think of an old fashioned shipboard (pirate) spyglass-style telescope that folds inward to become shorter.



                  You already know how to decompose into partial fractions. Let's write down the first few terms of the series formed after that decomposition:



                  $frac 13 - frac 16 + frac 14 - frac 17 + frac 15 - frac 18 + frac 16 - frac 19 + frac 17 - frac 1{10} + ...$



                  Note how, starting from $frac 16$ onward every initial negative term has a positive counterpart later on. In essence, all negative terms will simply vanish by cancellation, as will all positive terms after $frac 15$. So your entire series "telescopes" to become the really short $frac 13 + frac 14 + frac 15 = frac{47}{60}$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 26 at 0:53

























                  answered Jan 26 at 0:47









                  DeepakDeepak

                  17.5k11539




                  17.5k11539























                      1












                      $begingroup$

                      Hint: this is a telescopic series:
                      $$
                      sum_{n=3}^inftyleft[frac1n-frac{1}{n+3}right]=frac13+frac14+frac15,
                      $$
                      in which only three first positive terms survive, the other being cancelled by the preceeding negative ones.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        The series sum starts from index $3$ not $1$.
                        $endgroup$
                        – Deepak
                        Jan 26 at 0:48












                      • $begingroup$
                        @Deepak Thanks for noting this.
                        $endgroup$
                        – user
                        Jan 26 at 0:50
















                      1












                      $begingroup$

                      Hint: this is a telescopic series:
                      $$
                      sum_{n=3}^inftyleft[frac1n-frac{1}{n+3}right]=frac13+frac14+frac15,
                      $$
                      in which only three first positive terms survive, the other being cancelled by the preceeding negative ones.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        The series sum starts from index $3$ not $1$.
                        $endgroup$
                        – Deepak
                        Jan 26 at 0:48












                      • $begingroup$
                        @Deepak Thanks for noting this.
                        $endgroup$
                        – user
                        Jan 26 at 0:50














                      1












                      1








                      1





                      $begingroup$

                      Hint: this is a telescopic series:
                      $$
                      sum_{n=3}^inftyleft[frac1n-frac{1}{n+3}right]=frac13+frac14+frac15,
                      $$
                      in which only three first positive terms survive, the other being cancelled by the preceeding negative ones.






                      share|cite|improve this answer











                      $endgroup$



                      Hint: this is a telescopic series:
                      $$
                      sum_{n=3}^inftyleft[frac1n-frac{1}{n+3}right]=frac13+frac14+frac15,
                      $$
                      in which only three first positive terms survive, the other being cancelled by the preceeding negative ones.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 26 at 1:03

























                      answered Jan 26 at 0:47









                      useruser

                      5,58911030




                      5,58911030












                      • $begingroup$
                        The series sum starts from index $3$ not $1$.
                        $endgroup$
                        – Deepak
                        Jan 26 at 0:48












                      • $begingroup$
                        @Deepak Thanks for noting this.
                        $endgroup$
                        – user
                        Jan 26 at 0:50


















                      • $begingroup$
                        The series sum starts from index $3$ not $1$.
                        $endgroup$
                        – Deepak
                        Jan 26 at 0:48












                      • $begingroup$
                        @Deepak Thanks for noting this.
                        $endgroup$
                        – user
                        Jan 26 at 0:50
















                      $begingroup$
                      The series sum starts from index $3$ not $1$.
                      $endgroup$
                      – Deepak
                      Jan 26 at 0:48






                      $begingroup$
                      The series sum starts from index $3$ not $1$.
                      $endgroup$
                      – Deepak
                      Jan 26 at 0:48














                      $begingroup$
                      @Deepak Thanks for noting this.
                      $endgroup$
                      – user
                      Jan 26 at 0:50




                      $begingroup$
                      @Deepak Thanks for noting this.
                      $endgroup$
                      – user
                      Jan 26 at 0:50











                      0












                      $begingroup$

                      Hint:
                      $$frac3{n(n+3}=frac 1n-frac 1{n+3}$$
                      Write out the beginning of this sum, and you should find telescoping terms.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Hint:
                        $$frac3{n(n+3}=frac 1n-frac 1{n+3}$$
                        Write out the beginning of this sum, and you should find telescoping terms.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Hint:
                          $$frac3{n(n+3}=frac 1n-frac 1{n+3}$$
                          Write out the beginning of this sum, and you should find telescoping terms.






                          share|cite|improve this answer









                          $endgroup$



                          Hint:
                          $$frac3{n(n+3}=frac 1n-frac 1{n+3}$$
                          Write out the beginning of this sum, and you should find telescoping terms.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 26 at 0:49









                          BernardBernard

                          123k741117




                          123k741117























                              -1












                              $begingroup$

                              This is a special case
                              of the following theorem:



                              Given $k$ reals
                              $(a_k)_{j=1}^k$,
                              the sum
                              $sum_{n=1}^{infty} sum_{j=1}^k dfrac{a_j}{(n-1)k+j}
                              $
                              converges
                              if and only if
                              $sum_{j=1}^k a_j= 0$.



                              This problem is the case
                              $k=4, a_1=1, a_2=0, a_3 = 0, a_4 = -1$.
                              Therefore the sum converges.






                              share|cite|improve this answer









                              $endgroup$









                              • 1




                                $begingroup$
                                The problem appears to be to find the value of the sum, not whether it converges.
                                $endgroup$
                                – 6005
                                Jan 26 at 2:49












                              • $begingroup$
                                Yep. I posted too fast. I think I'll make my answer into a question with a proof. Thanks.
                                $endgroup$
                                – marty cohen
                                Jan 26 at 5:19
















                              -1












                              $begingroup$

                              This is a special case
                              of the following theorem:



                              Given $k$ reals
                              $(a_k)_{j=1}^k$,
                              the sum
                              $sum_{n=1}^{infty} sum_{j=1}^k dfrac{a_j}{(n-1)k+j}
                              $
                              converges
                              if and only if
                              $sum_{j=1}^k a_j= 0$.



                              This problem is the case
                              $k=4, a_1=1, a_2=0, a_3 = 0, a_4 = -1$.
                              Therefore the sum converges.






                              share|cite|improve this answer









                              $endgroup$









                              • 1




                                $begingroup$
                                The problem appears to be to find the value of the sum, not whether it converges.
                                $endgroup$
                                – 6005
                                Jan 26 at 2:49












                              • $begingroup$
                                Yep. I posted too fast. I think I'll make my answer into a question with a proof. Thanks.
                                $endgroup$
                                – marty cohen
                                Jan 26 at 5:19














                              -1












                              -1








                              -1





                              $begingroup$

                              This is a special case
                              of the following theorem:



                              Given $k$ reals
                              $(a_k)_{j=1}^k$,
                              the sum
                              $sum_{n=1}^{infty} sum_{j=1}^k dfrac{a_j}{(n-1)k+j}
                              $
                              converges
                              if and only if
                              $sum_{j=1}^k a_j= 0$.



                              This problem is the case
                              $k=4, a_1=1, a_2=0, a_3 = 0, a_4 = -1$.
                              Therefore the sum converges.






                              share|cite|improve this answer









                              $endgroup$



                              This is a special case
                              of the following theorem:



                              Given $k$ reals
                              $(a_k)_{j=1}^k$,
                              the sum
                              $sum_{n=1}^{infty} sum_{j=1}^k dfrac{a_j}{(n-1)k+j}
                              $
                              converges
                              if and only if
                              $sum_{j=1}^k a_j= 0$.



                              This problem is the case
                              $k=4, a_1=1, a_2=0, a_3 = 0, a_4 = -1$.
                              Therefore the sum converges.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 26 at 2:44









                              marty cohenmarty cohen

                              74.5k549129




                              74.5k549129








                              • 1




                                $begingroup$
                                The problem appears to be to find the value of the sum, not whether it converges.
                                $endgroup$
                                – 6005
                                Jan 26 at 2:49












                              • $begingroup$
                                Yep. I posted too fast. I think I'll make my answer into a question with a proof. Thanks.
                                $endgroup$
                                – marty cohen
                                Jan 26 at 5:19














                              • 1




                                $begingroup$
                                The problem appears to be to find the value of the sum, not whether it converges.
                                $endgroup$
                                – 6005
                                Jan 26 at 2:49












                              • $begingroup$
                                Yep. I posted too fast. I think I'll make my answer into a question with a proof. Thanks.
                                $endgroup$
                                – marty cohen
                                Jan 26 at 5:19








                              1




                              1




                              $begingroup$
                              The problem appears to be to find the value of the sum, not whether it converges.
                              $endgroup$
                              – 6005
                              Jan 26 at 2:49






                              $begingroup$
                              The problem appears to be to find the value of the sum, not whether it converges.
                              $endgroup$
                              – 6005
                              Jan 26 at 2:49














                              $begingroup$
                              Yep. I posted too fast. I think I'll make my answer into a question with a proof. Thanks.
                              $endgroup$
                              – marty cohen
                              Jan 26 at 5:19




                              $begingroup$
                              Yep. I posted too fast. I think I'll make my answer into a question with a proof. Thanks.
                              $endgroup$
                              – marty cohen
                              Jan 26 at 5:19


















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