Finding limit of a (Laurent?) series












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I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
$1 + frac{1}{1 + 2} + frac{1}{1 + 2 + 3} + ... + frac{1}{1 + 2 + 3 + ... + n}$
The task is to find the limit.






sequences-and-series limits







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
    $1 + frac{1}{1 + 2} + frac{1}{1 + 2 + 3} + ... + frac{1}{1 + 2 + 3 + ... + n}$
    The task is to find the limit.






    sequences-and-series limits







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
      $1 + frac{1}{1 + 2} + frac{1}{1 + 2 + 3} + ... + frac{1}{1 + 2 + 3 + ... + n}$
      The task is to find the limit.






      sequences-and-series limits







      share|cite|improve this question











      $endgroup$




      I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
      $1 + frac{1}{1 + 2} + frac{1}{1 + 2 + 3} + ... + frac{1}{1 + 2 + 3 + ... + n}$
      The task is to find the limit.






      sequences-and-series limits




      sequences-and-series limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 3 hours ago









      Max

      556116




      556116










      asked 4 hours ago









      RoseRose

      166




      166






















          5 Answers
          5






          active

          oldest

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          4












          $begingroup$

          We have $$1+2+...+k=frac{k(k+1)}2$$
          So the sum you need to compute is
          $$begin{split}
          sum_{k=1}^n frac 2{k(k+1)} &= sum_{k=1}^n 2left ( frac 1 k - frac 1 { k+1} right )\
          &=2-frac 2 {n+1}\
          &=frac {2n} {n+1}
          end{split}$$
          Now you can take the limit.






          share|cite|improve this answer











          $endgroup$





















            5












            $begingroup$

            Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
            $$1+2+cdots+n=frac{n(n+1)}{2}$$
            Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
            $$
            frac{2}{n(n+1)}=2(frac{1}{n}-frac{1}{n+1})$$
            After that, take the limit.






            share|cite|improve this answer









            $endgroup$





















              3












              $begingroup$

              Hint: Use the fact that $$1+2+...+n=frac{n(n+1)}{2}$$



              The series then becomes $$2sumlimits_{n=1}^infty left(frac{1}{n} - frac{1}{n+1} right)$$ which is a telescoping series.






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                As $1+2+ldots +n = frac{n}{2}(n+1)$ you are looking for the sum of the series
                $S = sum_{n=1}^{infty} frac{1}{frac{k}{2}(k+1)} = 2 sum_{k=1}^{infty} frac{1}{k(k+1)}$.



                Nos separating you get that $S_n = 2sum_{k=1}^{n} frac{1}{k} - frac{1}{k+1} = 2(1 - frac{1}{n+1})$. Now taking limit when $n rightarrow infty$ you get that $S=2$






                share|cite|improve this answer









                $endgroup$





















                  0












                  $begingroup$

                  $small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$



                  $small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$



                  Telescopic sum



                  $1+sum_{k=2}^{infty} a_k =?$






                  share|cite|improve this answer











                  $endgroup$













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                    5 Answers
                    5






                    active

                    oldest

                    votes








                    5 Answers
                    5






                    active

                    oldest

                    votes









                    active

                    oldest

                    votes






                    active

                    oldest

                    votes









                    4












                    $begingroup$

                    We have $$1+2+...+k=frac{k(k+1)}2$$
                    So the sum you need to compute is
                    $$begin{split}
                    sum_{k=1}^n frac 2{k(k+1)} &= sum_{k=1}^n 2left ( frac 1 k - frac 1 { k+1} right )\
                    &=2-frac 2 {n+1}\
                    &=frac {2n} {n+1}
                    end{split}$$
                    Now you can take the limit.






                    share|cite|improve this answer











                    $endgroup$


















                      4












                      $begingroup$

                      We have $$1+2+...+k=frac{k(k+1)}2$$
                      So the sum you need to compute is
                      $$begin{split}
                      sum_{k=1}^n frac 2{k(k+1)} &= sum_{k=1}^n 2left ( frac 1 k - frac 1 { k+1} right )\
                      &=2-frac 2 {n+1}\
                      &=frac {2n} {n+1}
                      end{split}$$
                      Now you can take the limit.






                      share|cite|improve this answer











                      $endgroup$
















                        4












                        4








                        4





                        $begingroup$

                        We have $$1+2+...+k=frac{k(k+1)}2$$
                        So the sum you need to compute is
                        $$begin{split}
                        sum_{k=1}^n frac 2{k(k+1)} &= sum_{k=1}^n 2left ( frac 1 k - frac 1 { k+1} right )\
                        &=2-frac 2 {n+1}\
                        &=frac {2n} {n+1}
                        end{split}$$
                        Now you can take the limit.






                        share|cite|improve this answer











                        $endgroup$



                        We have $$1+2+...+k=frac{k(k+1)}2$$
                        So the sum you need to compute is
                        $$begin{split}
                        sum_{k=1}^n frac 2{k(k+1)} &= sum_{k=1}^n 2left ( frac 1 k - frac 1 { k+1} right )\
                        &=2-frac 2 {n+1}\
                        &=frac {2n} {n+1}
                        end{split}$$
                        Now you can take the limit.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited 7 mins ago









                        RQM

                        1213




                        1213










                        answered 4 hours ago









                        Stefan LafonStefan Lafon

                        1,99218




                        1,99218























                            5












                            $begingroup$

                            Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
                            $$1+2+cdots+n=frac{n(n+1)}{2}$$
                            Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
                            $$
                            frac{2}{n(n+1)}=2(frac{1}{n}-frac{1}{n+1})$$
                            After that, take the limit.






                            share|cite|improve this answer









                            $endgroup$


















                              5












                              $begingroup$

                              Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
                              $$1+2+cdots+n=frac{n(n+1)}{2}$$
                              Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
                              $$
                              frac{2}{n(n+1)}=2(frac{1}{n}-frac{1}{n+1})$$
                              After that, take the limit.






                              share|cite|improve this answer









                              $endgroup$
















                                5












                                5








                                5





                                $begingroup$

                                Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
                                $$1+2+cdots+n=frac{n(n+1)}{2}$$
                                Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
                                $$
                                frac{2}{n(n+1)}=2(frac{1}{n}-frac{1}{n+1})$$
                                After that, take the limit.






                                share|cite|improve this answer









                                $endgroup$



                                Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
                                $$1+2+cdots+n=frac{n(n+1)}{2}$$
                                Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
                                $$
                                frac{2}{n(n+1)}=2(frac{1}{n}-frac{1}{n+1})$$
                                After that, take the limit.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 4 hours ago









                                weilam06weilam06

                                14511




                                14511























                                    3












                                    $begingroup$

                                    Hint: Use the fact that $$1+2+...+n=frac{n(n+1)}{2}$$



                                    The series then becomes $$2sumlimits_{n=1}^infty left(frac{1}{n} - frac{1}{n+1} right)$$ which is a telescoping series.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      3












                                      $begingroup$

                                      Hint: Use the fact that $$1+2+...+n=frac{n(n+1)}{2}$$



                                      The series then becomes $$2sumlimits_{n=1}^infty left(frac{1}{n} - frac{1}{n+1} right)$$ which is a telescoping series.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        3












                                        3








                                        3





                                        $begingroup$

                                        Hint: Use the fact that $$1+2+...+n=frac{n(n+1)}{2}$$



                                        The series then becomes $$2sumlimits_{n=1}^infty left(frac{1}{n} - frac{1}{n+1} right)$$ which is a telescoping series.






                                        share|cite|improve this answer









                                        $endgroup$



                                        Hint: Use the fact that $$1+2+...+n=frac{n(n+1)}{2}$$



                                        The series then becomes $$2sumlimits_{n=1}^infty left(frac{1}{n} - frac{1}{n+1} right)$$ which is a telescoping series.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 4 hours ago









                                        Haris GusicHaris Gusic

                                        17910




                                        17910























                                            0












                                            $begingroup$

                                            As $1+2+ldots +n = frac{n}{2}(n+1)$ you are looking for the sum of the series
                                            $S = sum_{n=1}^{infty} frac{1}{frac{k}{2}(k+1)} = 2 sum_{k=1}^{infty} frac{1}{k(k+1)}$.



                                            Nos separating you get that $S_n = 2sum_{k=1}^{n} frac{1}{k} - frac{1}{k+1} = 2(1 - frac{1}{n+1})$. Now taking limit when $n rightarrow infty$ you get that $S=2$






                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$

                                              As $1+2+ldots +n = frac{n}{2}(n+1)$ you are looking for the sum of the series
                                              $S = sum_{n=1}^{infty} frac{1}{frac{k}{2}(k+1)} = 2 sum_{k=1}^{infty} frac{1}{k(k+1)}$.



                                              Nos separating you get that $S_n = 2sum_{k=1}^{n} frac{1}{k} - frac{1}{k+1} = 2(1 - frac{1}{n+1})$. Now taking limit when $n rightarrow infty$ you get that $S=2$






                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                As $1+2+ldots +n = frac{n}{2}(n+1)$ you are looking for the sum of the series
                                                $S = sum_{n=1}^{infty} frac{1}{frac{k}{2}(k+1)} = 2 sum_{k=1}^{infty} frac{1}{k(k+1)}$.



                                                Nos separating you get that $S_n = 2sum_{k=1}^{n} frac{1}{k} - frac{1}{k+1} = 2(1 - frac{1}{n+1})$. Now taking limit when $n rightarrow infty$ you get that $S=2$






                                                share|cite|improve this answer









                                                $endgroup$



                                                As $1+2+ldots +n = frac{n}{2}(n+1)$ you are looking for the sum of the series
                                                $S = sum_{n=1}^{infty} frac{1}{frac{k}{2}(k+1)} = 2 sum_{k=1}^{infty} frac{1}{k(k+1)}$.



                                                Nos separating you get that $S_n = 2sum_{k=1}^{n} frac{1}{k} - frac{1}{k+1} = 2(1 - frac{1}{n+1})$. Now taking limit when $n rightarrow infty$ you get that $S=2$







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered 4 hours ago









                                                JoseSquareJoseSquare

                                                56812




                                                56812























                                                    0












                                                    $begingroup$

                                                    $small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$



                                                    $small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$



                                                    Telescopic sum



                                                    $1+sum_{k=2}^{infty} a_k =?$






                                                    share|cite|improve this answer











                                                    $endgroup$


















                                                      0












                                                      $begingroup$

                                                      $small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$



                                                      $small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$



                                                      Telescopic sum



                                                      $1+sum_{k=2}^{infty} a_k =?$






                                                      share|cite|improve this answer











                                                      $endgroup$
















                                                        0












                                                        0








                                                        0





                                                        $begingroup$

                                                        $small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$



                                                        $small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$



                                                        Telescopic sum



                                                        $1+sum_{k=2}^{infty} a_k =?$






                                                        share|cite|improve this answer











                                                        $endgroup$



                                                        $small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$



                                                        $small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$



                                                        Telescopic sum



                                                        $1+sum_{k=2}^{infty} a_k =?$







                                                        share|cite|improve this answer














                                                        share|cite|improve this answer



                                                        share|cite|improve this answer








                                                        edited 2 hours ago

























                                                        answered 4 hours ago









                                                        Peter SzilasPeter Szilas

                                                        11.3k2822




                                                        11.3k2822






























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