Limit of $a_n=(1-frac13)^2cdot(1-frac16)^2ldots(1-frac{1}{frac{(n)(n+1)}{2}})^2 ; ;forall n geq 2$












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$$a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2ldotsleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
; ;forall n geq 2$$



I have no idea how to solve this, however I will give it a try using Cauchy's second theorem... the following is probably wrong since all terms will depend on $'n'$ after taking $nth$ root



$$lim_{nrightarrowinfty} a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
= lim_{nrightarrowinfty}left[left(1-frac13right)^{2n}cdotleft(1-frac16right)^{2n}cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}right]^{frac1n} = lim_{nrightarrowinfty}left(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}=1 $$










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    0












    $begingroup$


    $$a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2ldotsleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
    ; ;forall n geq 2$$



    I have no idea how to solve this, however I will give it a try using Cauchy's second theorem... the following is probably wrong since all terms will depend on $'n'$ after taking $nth$ root



    $$lim_{nrightarrowinfty} a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
    = lim_{nrightarrowinfty}left[left(1-frac13right)^{2n}cdotleft(1-frac16right)^{2n}cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}right]^{frac1n} = lim_{nrightarrowinfty}left(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}=1 $$










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      0












      0








      0





      $begingroup$


      $$a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2ldotsleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
      ; ;forall n geq 2$$



      I have no idea how to solve this, however I will give it a try using Cauchy's second theorem... the following is probably wrong since all terms will depend on $'n'$ after taking $nth$ root



      $$lim_{nrightarrowinfty} a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
      = lim_{nrightarrowinfty}left[left(1-frac13right)^{2n}cdotleft(1-frac16right)^{2n}cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}right]^{frac1n} = lim_{nrightarrowinfty}left(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}=1 $$










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      $endgroup$




      $$a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2ldotsleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
      ; ;forall n geq 2$$



      I have no idea how to solve this, however I will give it a try using Cauchy's second theorem... the following is probably wrong since all terms will depend on $'n'$ after taking $nth$ root



      $$lim_{nrightarrowinfty} a_n = left(1-frac13right)^2cdotleft(1-frac16right)^2cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^2
      = lim_{nrightarrowinfty}left[left(1-frac13right)^{2n}cdotleft(1-frac16right)^{2n}cdotldotscdotleft(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}right]^{frac1n} = lim_{nrightarrowinfty}left(1-frac{1}{frac{n(n+1)}{2}}right)^{2n}=1 $$







      sequences-and-series limits limits-without-lhopital cauchy-sequences






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      edited Jan 17 at 21:33









      user1337

      46110




      46110










      asked Jan 17 at 21:12









      AbhayAbhay

      3569




      3569






















          2 Answers
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          3












          $begingroup$

          Hint: $$
          1-frac{2}{n(n+1)}=frac{(n-1)(n+2)}{n(n+1)}.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
            $endgroup$
            – Abhay
            Jan 17 at 21:20










          • $begingroup$
            Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
            $endgroup$
            – Nick Peterson
            Jan 17 at 21:23












          • $begingroup$
            @learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
            $endgroup$
            – Henry
            Jan 17 at 21:24



















          1












          $begingroup$

          One can generalise $a_n = frac{(n+3)^2}{9cdot (n+1)^2}$ by considering the cancellation of each additional term during multiplication. By taking the limit as $n to infty$ we see that the resulting value is $frac{1}{9}$.






          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









            3












            $begingroup$

            Hint: $$
            1-frac{2}{n(n+1)}=frac{(n-1)(n+2)}{n(n+1)}.
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
              $endgroup$
              – Abhay
              Jan 17 at 21:20










            • $begingroup$
              Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
              $endgroup$
              – Nick Peterson
              Jan 17 at 21:23












            • $begingroup$
              @learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
              $endgroup$
              – Henry
              Jan 17 at 21:24
















            3












            $begingroup$

            Hint: $$
            1-frac{2}{n(n+1)}=frac{(n-1)(n+2)}{n(n+1)}.
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
              $endgroup$
              – Abhay
              Jan 17 at 21:20










            • $begingroup$
              Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
              $endgroup$
              – Nick Peterson
              Jan 17 at 21:23












            • $begingroup$
              @learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
              $endgroup$
              – Henry
              Jan 17 at 21:24














            3












            3








            3





            $begingroup$

            Hint: $$
            1-frac{2}{n(n+1)}=frac{(n-1)(n+2)}{n(n+1)}.
            $$






            share|cite|improve this answer









            $endgroup$



            Hint: $$
            1-frac{2}{n(n+1)}=frac{(n-1)(n+2)}{n(n+1)}.
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 17 at 21:18









            SongSong

            15.1k1636




            15.1k1636












            • $begingroup$
              True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
              $endgroup$
              – Abhay
              Jan 17 at 21:20










            • $begingroup$
              Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
              $endgroup$
              – Nick Peterson
              Jan 17 at 21:23












            • $begingroup$
              @learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
              $endgroup$
              – Henry
              Jan 17 at 21:24


















            • $begingroup$
              True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
              $endgroup$
              – Abhay
              Jan 17 at 21:20










            • $begingroup$
              Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
              $endgroup$
              – Nick Peterson
              Jan 17 at 21:23












            • $begingroup$
              @learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
              $endgroup$
              – Henry
              Jan 17 at 21:24
















            $begingroup$
            True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
            $endgroup$
            – Abhay
            Jan 17 at 21:20




            $begingroup$
            True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go
            $endgroup$
            – Abhay
            Jan 17 at 21:20












            $begingroup$
            Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
            $endgroup$
            – Nick Peterson
            Jan 17 at 21:23






            $begingroup$
            Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around.
            $endgroup$
            – Nick Peterson
            Jan 17 at 21:23














            $begingroup$
            @learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
            $endgroup$
            – Henry
            Jan 17 at 21:24




            $begingroup$
            @learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples)
            $endgroup$
            – Henry
            Jan 17 at 21:24











            1












            $begingroup$

            One can generalise $a_n = frac{(n+3)^2}{9cdot (n+1)^2}$ by considering the cancellation of each additional term during multiplication. By taking the limit as $n to infty$ we see that the resulting value is $frac{1}{9}$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              One can generalise $a_n = frac{(n+3)^2}{9cdot (n+1)^2}$ by considering the cancellation of each additional term during multiplication. By taking the limit as $n to infty$ we see that the resulting value is $frac{1}{9}$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                One can generalise $a_n = frac{(n+3)^2}{9cdot (n+1)^2}$ by considering the cancellation of each additional term during multiplication. By taking the limit as $n to infty$ we see that the resulting value is $frac{1}{9}$.






                share|cite|improve this answer









                $endgroup$



                One can generalise $a_n = frac{(n+3)^2}{9cdot (n+1)^2}$ by considering the cancellation of each additional term during multiplication. By taking the limit as $n to infty$ we see that the resulting value is $frac{1}{9}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 17 at 21:33









                Peter ForemanPeter Foreman

                2,399214




                2,399214






























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