Find the limit of a multiplying term function when n tends to infinity.Help in Evaluate the following...

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Find the limit of a multiplying term function when n tends to infinity.


Help in Evaluate the following limit?Limit n tends to infinityThe Limit of $xleft(sqrt[x]{a}-1right)$ as $xtoinfty$.Using L'Hospital's Rule to evaluate limit to infinityEvaluate the following limit without L'Hopitallimit of function with one fractional termLimit of exp function in infinityTrigonometric Limit 0*infinityFind the limit of $e^x/2^x$ as $x$ approaches infinityWhat is the limit of series when n tends to infinity?













2












$begingroup$


How do I evaluate the limit,



$$lim_{n to infty} left(1-frac1{2^2}right)left(1-frac1{3^2}right)left(1-frac1{4^2}right)...left(1-frac1{n^2}right)$$



I tried to break the $n^{text{th}}$ term into $$frac{(n+1)(n-1)}{n.n}$$ and then tried to do something but couldn't get anywhere. Please help as to how to solve such questions in general. Thanks.






limits convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you see that the numerator in each term will cancel with the denominators of the terms on either side. So there ought to be some mass cancellation - write the first few terms in the form you have discovered o see what is going on.
    $endgroup$
    – Mark Bennet
    9 hours ago
















2












$begingroup$


How do I evaluate the limit,



$$lim_{n to infty} left(1-frac1{2^2}right)left(1-frac1{3^2}right)left(1-frac1{4^2}right)...left(1-frac1{n^2}right)$$



I tried to break the $n^{text{th}}$ term into $$frac{(n+1)(n-1)}{n.n}$$ and then tried to do something but couldn't get anywhere. Please help as to how to solve such questions in general. Thanks.






limits convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you see that the numerator in each term will cancel with the denominators of the terms on either side. So there ought to be some mass cancellation - write the first few terms in the form you have discovered o see what is going on.
    $endgroup$
    – Mark Bennet
    9 hours ago














2












2








2





$begingroup$


How do I evaluate the limit,



$$lim_{n to infty} left(1-frac1{2^2}right)left(1-frac1{3^2}right)left(1-frac1{4^2}right)...left(1-frac1{n^2}right)$$



I tried to break the $n^{text{th}}$ term into $$frac{(n+1)(n-1)}{n.n}$$ and then tried to do something but couldn't get anywhere. Please help as to how to solve such questions in general. Thanks.






limits convergence







share|cite|improve this question











$endgroup$




How do I evaluate the limit,



$$lim_{n to infty} left(1-frac1{2^2}right)left(1-frac1{3^2}right)left(1-frac1{4^2}right)...left(1-frac1{n^2}right)$$



I tried to break the $n^{text{th}}$ term into $$frac{(n+1)(n-1)}{n.n}$$ and then tried to do something but couldn't get anywhere. Please help as to how to solve such questions in general. Thanks.






limits convergence




limits convergence






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share|cite|improve this question













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share|cite|improve this question








edited 9 hours ago









cmk

1,640214




1,640214










asked 9 hours ago









Divyesh ShahDivyesh Shah

255




255












  • $begingroup$
    Can you see that the numerator in each term will cancel with the denominators of the terms on either side. So there ought to be some mass cancellation - write the first few terms in the form you have discovered o see what is going on.
    $endgroup$
    – Mark Bennet
    9 hours ago


















  • $begingroup$
    Can you see that the numerator in each term will cancel with the denominators of the terms on either side. So there ought to be some mass cancellation - write the first few terms in the form you have discovered o see what is going on.
    $endgroup$
    – Mark Bennet
    9 hours ago
















$begingroup$
Can you see that the numerator in each term will cancel with the denominators of the terms on either side. So there ought to be some mass cancellation - write the first few terms in the form you have discovered o see what is going on.
$endgroup$
– Mark Bennet
9 hours ago




$begingroup$
Can you see that the numerator in each term will cancel with the denominators of the terms on either side. So there ought to be some mass cancellation - write the first few terms in the form you have discovered o see what is going on.
$endgroup$
– Mark Bennet
9 hours ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

We have
$$ logleft(prod_{n =2}^N 1 - frac{1}{n^2} right) = sum_{n = 2}^N log(n-1) + log(n+1) - 2 log(n) = log(1) - log(2) - log(N) + log(N+1).$$
So we have that
$$prod_{n = 1}^N 1 - frac{1}{n^2} = frac{N+1}{2N} to frac{1}{2}$$






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    Hint: Prove by induction that $$prod_{i=2}^n 1-frac{1}{i^2}=frac{n+1}{2n}$$






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      Hint.



      $$
      frac{(2-1)(2+1)}{2cdot 2}cdots frac{(n-2)n}{(n-1)(n-1)}frac{(n-1)(n+1)}{nn}frac{n(n+2)}{(n+1)(n+1)}frac{(n+1)(n+3)}{(n+2)(n+2)} = frac 12frac{(n+3)}{(n+2)}
      $$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Did you try starting with small sequences and then seeing what cancels out each time?




        1. $frac{1*3}{2*2} = frac{3}{4}$

        2. $frac{3}{4} * frac{2*4}{3*3} = frac{2}{3}$

        3. $frac{2}{3} * frac{3*5}{4*4} = frac{5}{8}$

        4. $frac{5}{8} * frac{4*6}{5*5} = frac{3}{5}$






        share|cite|improve this answer









        $endgroup$














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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          We have
          $$ logleft(prod_{n =2}^N 1 - frac{1}{n^2} right) = sum_{n = 2}^N log(n-1) + log(n+1) - 2 log(n) = log(1) - log(2) - log(N) + log(N+1).$$
          So we have that
          $$prod_{n = 1}^N 1 - frac{1}{n^2} = frac{N+1}{2N} to frac{1}{2}$$






          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

            We have
            $$ logleft(prod_{n =2}^N 1 - frac{1}{n^2} right) = sum_{n = 2}^N log(n-1) + log(n+1) - 2 log(n) = log(1) - log(2) - log(N) + log(N+1).$$
            So we have that
            $$prod_{n = 1}^N 1 - frac{1}{n^2} = frac{N+1}{2N} to frac{1}{2}$$






            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              We have
              $$ logleft(prod_{n =2}^N 1 - frac{1}{n^2} right) = sum_{n = 2}^N log(n-1) + log(n+1) - 2 log(n) = log(1) - log(2) - log(N) + log(N+1).$$
              So we have that
              $$prod_{n = 1}^N 1 - frac{1}{n^2} = frac{N+1}{2N} to frac{1}{2}$$






              share|cite|improve this answer









              $endgroup$



              We have
              $$ logleft(prod_{n =2}^N 1 - frac{1}{n^2} right) = sum_{n = 2}^N log(n-1) + log(n+1) - 2 log(n) = log(1) - log(2) - log(N) + log(N+1).$$
              So we have that
              $$prod_{n = 1}^N 1 - frac{1}{n^2} = frac{N+1}{2N} to frac{1}{2}$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 9 hours ago









              Tychonoff3000Tychonoff3000

              1066




              1066























                  6












                  $begingroup$

                  Hint: Prove by induction that $$prod_{i=2}^n 1-frac{1}{i^2}=frac{n+1}{2n}$$






                  share|cite|improve this answer









                  $endgroup$


















                    6












                    $begingroup$

                    Hint: Prove by induction that $$prod_{i=2}^n 1-frac{1}{i^2}=frac{n+1}{2n}$$






                    share|cite|improve this answer









                    $endgroup$
















                      6












                      6








                      6





                      $begingroup$

                      Hint: Prove by induction that $$prod_{i=2}^n 1-frac{1}{i^2}=frac{n+1}{2n}$$






                      share|cite|improve this answer









                      $endgroup$



                      Hint: Prove by induction that $$prod_{i=2}^n 1-frac{1}{i^2}=frac{n+1}{2n}$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 9 hours ago









                      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                      82.1k42867




                      82.1k42867























                          3












                          $begingroup$

                          Hint.



                          $$
                          frac{(2-1)(2+1)}{2cdot 2}cdots frac{(n-2)n}{(n-1)(n-1)}frac{(n-1)(n+1)}{nn}frac{n(n+2)}{(n+1)(n+1)}frac{(n+1)(n+3)}{(n+2)(n+2)} = frac 12frac{(n+3)}{(n+2)}
                          $$






                          share|cite|improve this answer









                          $endgroup$


















                            3












                            $begingroup$

                            Hint.



                            $$
                            frac{(2-1)(2+1)}{2cdot 2}cdots frac{(n-2)n}{(n-1)(n-1)}frac{(n-1)(n+1)}{nn}frac{n(n+2)}{(n+1)(n+1)}frac{(n+1)(n+3)}{(n+2)(n+2)} = frac 12frac{(n+3)}{(n+2)}
                            $$






                            share|cite|improve this answer









                            $endgroup$
















                              3












                              3








                              3





                              $begingroup$

                              Hint.



                              $$
                              frac{(2-1)(2+1)}{2cdot 2}cdots frac{(n-2)n}{(n-1)(n-1)}frac{(n-1)(n+1)}{nn}frac{n(n+2)}{(n+1)(n+1)}frac{(n+1)(n+3)}{(n+2)(n+2)} = frac 12frac{(n+3)}{(n+2)}
                              $$






                              share|cite|improve this answer









                              $endgroup$



                              Hint.



                              $$
                              frac{(2-1)(2+1)}{2cdot 2}cdots frac{(n-2)n}{(n-1)(n-1)}frac{(n-1)(n+1)}{nn}frac{n(n+2)}{(n+1)(n+1)}frac{(n+1)(n+3)}{(n+2)(n+2)} = frac 12frac{(n+3)}{(n+2)}
                              $$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 9 hours ago









                              CesareoCesareo

                              11.1k3519




                              11.1k3519























                                  0












                                  $begingroup$

                                  Did you try starting with small sequences and then seeing what cancels out each time?




                                  1. $frac{1*3}{2*2} = frac{3}{4}$

                                  2. $frac{3}{4} * frac{2*4}{3*3} = frac{2}{3}$

                                  3. $frac{2}{3} * frac{3*5}{4*4} = frac{5}{8}$

                                  4. $frac{5}{8} * frac{4*6}{5*5} = frac{3}{5}$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    Did you try starting with small sequences and then seeing what cancels out each time?




                                    1. $frac{1*3}{2*2} = frac{3}{4}$

                                    2. $frac{3}{4} * frac{2*4}{3*3} = frac{2}{3}$

                                    3. $frac{2}{3} * frac{3*5}{4*4} = frac{5}{8}$

                                    4. $frac{5}{8} * frac{4*6}{5*5} = frac{3}{5}$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      Did you try starting with small sequences and then seeing what cancels out each time?




                                      1. $frac{1*3}{2*2} = frac{3}{4}$

                                      2. $frac{3}{4} * frac{2*4}{3*3} = frac{2}{3}$

                                      3. $frac{2}{3} * frac{3*5}{4*4} = frac{5}{8}$

                                      4. $frac{5}{8} * frac{4*6}{5*5} = frac{3}{5}$






                                      share|cite|improve this answer









                                      $endgroup$



                                      Did you try starting with small sequences and then seeing what cancels out each time?




                                      1. $frac{1*3}{2*2} = frac{3}{4}$

                                      2. $frac{3}{4} * frac{2*4}{3*3} = frac{2}{3}$

                                      3. $frac{2}{3} * frac{3*5}{4*4} = frac{5}{8}$

                                      4. $frac{5}{8} * frac{4*6}{5*5} = frac{3}{5}$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 9 hours ago









                                      Simpson17866Simpson17866

                                      3672513




                                      3672513






























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