Evaluate the limit of $prod_{k=2}^n(1-frac2{k(k+1)})^2$












0












$begingroup$


Evaluate $lim_{n to infty} x_{n}$, where $x_{n} = (1-(1/3))^{2}(1-(1/6))^{2}(1-(1/10))^{2} dots (1-(2/(n(n+1)))^{2}, n≥2$



I couldn't find any pattern of terms to find the limit.
I applied $log$ both sides to get -
$log x_{n} = lim_{n to infty} 2sum_{k=1}^{n} log (1-(2/(k(k+1))$



Then I am stuck.



I also tried this -



$x_{n} = [(2/3)(5/6)(9/10) dots (frac{n(n+1)-2}{n(n+1)})]^{2}$
But, I am finding nothing.



How to solve this problem?






sequences-and-series







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Note that $$n(n+1)-2=(n-1)(n+2)$$ hence $$left(1-frac2{n(n+1)}right)^2=frac{(n-1)^2(n+2)^2}{n^2(n+1)^2}$$ In view of the above, I am surprised that you "couldn't find any pattern" in the partial products... Maybe a new look at these could help?
    $endgroup$
    – Did
    Jan 2 at 12:56










  • $begingroup$
    In an old notebook i have found this here $$frac{1}{9}left(frac{n+2}{n}right)^2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 13:03












  • $begingroup$
    we have proved it via induction.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 13:04






  • 1




    $begingroup$
    Who needs "old notebooks" here when the cancellations are so visible?
    $endgroup$
    – Did
    Jan 2 at 13:10










  • $begingroup$
    @Did I got the pattern $x_{n} = (frac{1(4)}{2(3)} frac{2(5)}{3(4)} frac{3(6)}{4(5)} dots )^{2}$ by applying limit $n to infty$, I got $1/9$
    $endgroup$
    – Mathsaddict
    Jan 2 at 13:21


















0












$begingroup$


Evaluate $lim_{n to infty} x_{n}$, where $x_{n} = (1-(1/3))^{2}(1-(1/6))^{2}(1-(1/10))^{2} dots (1-(2/(n(n+1)))^{2}, n≥2$



I couldn't find any pattern of terms to find the limit.
I applied $log$ both sides to get -
$log x_{n} = lim_{n to infty} 2sum_{k=1}^{n} log (1-(2/(k(k+1))$



Then I am stuck.



I also tried this -



$x_{n} = [(2/3)(5/6)(9/10) dots (frac{n(n+1)-2}{n(n+1)})]^{2}$
But, I am finding nothing.



How to solve this problem?






sequences-and-series







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Note that $$n(n+1)-2=(n-1)(n+2)$$ hence $$left(1-frac2{n(n+1)}right)^2=frac{(n-1)^2(n+2)^2}{n^2(n+1)^2}$$ In view of the above, I am surprised that you "couldn't find any pattern" in the partial products... Maybe a new look at these could help?
    $endgroup$
    – Did
    Jan 2 at 12:56










  • $begingroup$
    In an old notebook i have found this here $$frac{1}{9}left(frac{n+2}{n}right)^2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 13:03












  • $begingroup$
    we have proved it via induction.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 13:04






  • 1




    $begingroup$
    Who needs "old notebooks" here when the cancellations are so visible?
    $endgroup$
    – Did
    Jan 2 at 13:10










  • $begingroup$
    @Did I got the pattern $x_{n} = (frac{1(4)}{2(3)} frac{2(5)}{3(4)} frac{3(6)}{4(5)} dots )^{2}$ by applying limit $n to infty$, I got $1/9$
    $endgroup$
    – Mathsaddict
    Jan 2 at 13:21
















0












0








0





$begingroup$


Evaluate $lim_{n to infty} x_{n}$, where $x_{n} = (1-(1/3))^{2}(1-(1/6))^{2}(1-(1/10))^{2} dots (1-(2/(n(n+1)))^{2}, n≥2$



I couldn't find any pattern of terms to find the limit.
I applied $log$ both sides to get -
$log x_{n} = lim_{n to infty} 2sum_{k=1}^{n} log (1-(2/(k(k+1))$



Then I am stuck.



I also tried this -



$x_{n} = [(2/3)(5/6)(9/10) dots (frac{n(n+1)-2}{n(n+1)})]^{2}$
But, I am finding nothing.



How to solve this problem?






sequences-and-series







share|cite|improve this question











$endgroup$




Evaluate $lim_{n to infty} x_{n}$, where $x_{n} = (1-(1/3))^{2}(1-(1/6))^{2}(1-(1/10))^{2} dots (1-(2/(n(n+1)))^{2}, n≥2$



I couldn't find any pattern of terms to find the limit.
I applied $log$ both sides to get -
$log x_{n} = lim_{n to infty} 2sum_{k=1}^{n} log (1-(2/(k(k+1))$



Then I am stuck.



I also tried this -



$x_{n} = [(2/3)(5/6)(9/10) dots (frac{n(n+1)-2}{n(n+1)})]^{2}$
But, I am finding nothing.



How to solve this problem?






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 13:14









Henning Makholm

239k17303540




239k17303540










asked Jan 2 at 12:50









MathsaddictMathsaddict

3008




3008








  • 4




    $begingroup$
    Note that $$n(n+1)-2=(n-1)(n+2)$$ hence $$left(1-frac2{n(n+1)}right)^2=frac{(n-1)^2(n+2)^2}{n^2(n+1)^2}$$ In view of the above, I am surprised that you "couldn't find any pattern" in the partial products... Maybe a new look at these could help?
    $endgroup$
    – Did
    Jan 2 at 12:56










  • $begingroup$
    In an old notebook i have found this here $$frac{1}{9}left(frac{n+2}{n}right)^2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 13:03












  • $begingroup$
    we have proved it via induction.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 13:04






  • 1




    $begingroup$
    Who needs "old notebooks" here when the cancellations are so visible?
    $endgroup$
    – Did
    Jan 2 at 13:10










  • $begingroup$
    @Did I got the pattern $x_{n} = (frac{1(4)}{2(3)} frac{2(5)}{3(4)} frac{3(6)}{4(5)} dots )^{2}$ by applying limit $n to infty$, I got $1/9$
    $endgroup$
    – Mathsaddict
    Jan 2 at 13:21
















  • 4




    $begingroup$
    Note that $$n(n+1)-2=(n-1)(n+2)$$ hence $$left(1-frac2{n(n+1)}right)^2=frac{(n-1)^2(n+2)^2}{n^2(n+1)^2}$$ In view of the above, I am surprised that you "couldn't find any pattern" in the partial products... Maybe a new look at these could help?
    $endgroup$
    – Did
    Jan 2 at 12:56










  • $begingroup$
    In an old notebook i have found this here $$frac{1}{9}left(frac{n+2}{n}right)^2$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 13:03












  • $begingroup$
    we have proved it via induction.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 13:04






  • 1




    $begingroup$
    Who needs "old notebooks" here when the cancellations are so visible?
    $endgroup$
    – Did
    Jan 2 at 13:10










  • $begingroup$
    @Did I got the pattern $x_{n} = (frac{1(4)}{2(3)} frac{2(5)}{3(4)} frac{3(6)}{4(5)} dots )^{2}$ by applying limit $n to infty$, I got $1/9$
    $endgroup$
    – Mathsaddict
    Jan 2 at 13:21










4




4




$begingroup$
Note that $$n(n+1)-2=(n-1)(n+2)$$ hence $$left(1-frac2{n(n+1)}right)^2=frac{(n-1)^2(n+2)^2}{n^2(n+1)^2}$$ In view of the above, I am surprised that you "couldn't find any pattern" in the partial products... Maybe a new look at these could help?
$endgroup$
– Did
Jan 2 at 12:56




$begingroup$
Note that $$n(n+1)-2=(n-1)(n+2)$$ hence $$left(1-frac2{n(n+1)}right)^2=frac{(n-1)^2(n+2)^2}{n^2(n+1)^2}$$ In view of the above, I am surprised that you "couldn't find any pattern" in the partial products... Maybe a new look at these could help?
$endgroup$
– Did
Jan 2 at 12:56












$begingroup$
In an old notebook i have found this here $$frac{1}{9}left(frac{n+2}{n}right)^2$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 13:03






$begingroup$
In an old notebook i have found this here $$frac{1}{9}left(frac{n+2}{n}right)^2$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 13:03














$begingroup$
we have proved it via induction.
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 13:04




$begingroup$
we have proved it via induction.
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 13:04




1




1




$begingroup$
Who needs "old notebooks" here when the cancellations are so visible?
$endgroup$
– Did
Jan 2 at 13:10




$begingroup$
Who needs "old notebooks" here when the cancellations are so visible?
$endgroup$
– Did
Jan 2 at 13:10












$begingroup$
@Did I got the pattern $x_{n} = (frac{1(4)}{2(3)} frac{2(5)}{3(4)} frac{3(6)}{4(5)} dots )^{2}$ by applying limit $n to infty$, I got $1/9$
$endgroup$
– Mathsaddict
Jan 2 at 13:21






$begingroup$
@Did I got the pattern $x_{n} = (frac{1(4)}{2(3)} frac{2(5)}{3(4)} frac{3(6)}{4(5)} dots )^{2}$ by applying limit $n to infty$, I got $1/9$
$endgroup$
– Mathsaddict
Jan 2 at 13:21












1 Answer
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$$prod_{k=2}^nleft(1-frac2{k(k+1)}right)^2{=prod_{k=2}^nleft({k(k+1)-2over k(k+1)}right)^2\=left(prod_{k=2}^n{k(k+1)-2over k(k+1)}right)^2\=left(prod_{k=2}^n{(k-1)(k+2)over k(k+1)}right)^2\=left(prod_{k=2}^n{k-1over k}right)^2left(prod_{k=2}^n{k+2over k+1}right)^2\=left({1over n}right)^2left({n+2over 3}right)^2\={n^2+4n+4over 9n^2}}$$therefore $$lim_{nto infty}prod_{k=2}^nleft(1-frac2{k(k+1)}right)^2={1over 9}$$






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    $begingroup$

    $$prod_{k=2}^nleft(1-frac2{k(k+1)}right)^2{=prod_{k=2}^nleft({k(k+1)-2over k(k+1)}right)^2\=left(prod_{k=2}^n{k(k+1)-2over k(k+1)}right)^2\=left(prod_{k=2}^n{(k-1)(k+2)over k(k+1)}right)^2\=left(prod_{k=2}^n{k-1over k}right)^2left(prod_{k=2}^n{k+2over k+1}right)^2\=left({1over n}right)^2left({n+2over 3}right)^2\={n^2+4n+4over 9n^2}}$$therefore $$lim_{nto infty}prod_{k=2}^nleft(1-frac2{k(k+1)}right)^2={1over 9}$$






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      $begingroup$

      $$prod_{k=2}^nleft(1-frac2{k(k+1)}right)^2{=prod_{k=2}^nleft({k(k+1)-2over k(k+1)}right)^2\=left(prod_{k=2}^n{k(k+1)-2over k(k+1)}right)^2\=left(prod_{k=2}^n{(k-1)(k+2)over k(k+1)}right)^2\=left(prod_{k=2}^n{k-1over k}right)^2left(prod_{k=2}^n{k+2over k+1}right)^2\=left({1over n}right)^2left({n+2over 3}right)^2\={n^2+4n+4over 9n^2}}$$therefore $$lim_{nto infty}prod_{k=2}^nleft(1-frac2{k(k+1)}right)^2={1over 9}$$






      share|cite|improve this answer









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        $begingroup$

        $$prod_{k=2}^nleft(1-frac2{k(k+1)}right)^2{=prod_{k=2}^nleft({k(k+1)-2over k(k+1)}right)^2\=left(prod_{k=2}^n{k(k+1)-2over k(k+1)}right)^2\=left(prod_{k=2}^n{(k-1)(k+2)over k(k+1)}right)^2\=left(prod_{k=2}^n{k-1over k}right)^2left(prod_{k=2}^n{k+2over k+1}right)^2\=left({1over n}right)^2left({n+2over 3}right)^2\={n^2+4n+4over 9n^2}}$$therefore $$lim_{nto infty}prod_{k=2}^nleft(1-frac2{k(k+1)}right)^2={1over 9}$$






        share|cite|improve this answer









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        $$prod_{k=2}^nleft(1-frac2{k(k+1)}right)^2{=prod_{k=2}^nleft({k(k+1)-2over k(k+1)}right)^2\=left(prod_{k=2}^n{k(k+1)-2over k(k+1)}right)^2\=left(prod_{k=2}^n{(k-1)(k+2)over k(k+1)}right)^2\=left(prod_{k=2}^n{k-1over k}right)^2left(prod_{k=2}^n{k+2over k+1}right)^2\=left({1over n}right)^2left({n+2over 3}right)^2\={n^2+4n+4over 9n^2}}$$therefore $$lim_{nto infty}prod_{k=2}^nleft(1-frac2{k(k+1)}right)^2={1over 9}$$







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        share|cite|improve this answer



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        answered Jan 5 at 7:10









        Mostafa AyazMostafa Ayaz

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