show this inequality to find simple proof $prod_{i=1}^{n}(1+frac{1}{i^2})<4$












0












$begingroup$


show this
$$prod_{n=1}^{N}left(1+dfrac{1}{n^2}right)<4tag{1}$$



I have know this can find the value
$$prod_{n=1}^{+infty}left(1+dfrac{1}{n^2}right)=dfrac{sinh{pi}}{pi}=3.67<4$$Prove that $prod_{k=2}^{+infty} (1+1/k^2) = sinh(pi)/(2 pi)$.



But I think the problem $(1)$ have Simple proof?That is, we avoid calculating the result to give proof with inequality? Thanks






inequality







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  • $begingroup$
    @RossMillikan,the links is $prod_{k=2}^{+infty}$ and my problem is $prod_{k=1}^{+infty}$
    $endgroup$
    – function sug
    Dec 20 '18 at 6:10


















0












$begingroup$


show this
$$prod_{n=1}^{N}left(1+dfrac{1}{n^2}right)<4tag{1}$$



I have know this can find the value
$$prod_{n=1}^{+infty}left(1+dfrac{1}{n^2}right)=dfrac{sinh{pi}}{pi}=3.67<4$$Prove that $prod_{k=2}^{+infty} (1+1/k^2) = sinh(pi)/(2 pi)$.



But I think the problem $(1)$ have Simple proof?That is, we avoid calculating the result to give proof with inequality? Thanks






inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    @RossMillikan,the links is $prod_{k=2}^{+infty}$ and my problem is $prod_{k=1}^{+infty}$
    $endgroup$
    – function sug
    Dec 20 '18 at 6:10
















0












0








0


1



$begingroup$


show this
$$prod_{n=1}^{N}left(1+dfrac{1}{n^2}right)<4tag{1}$$



I have know this can find the value
$$prod_{n=1}^{+infty}left(1+dfrac{1}{n^2}right)=dfrac{sinh{pi}}{pi}=3.67<4$$Prove that $prod_{k=2}^{+infty} (1+1/k^2) = sinh(pi)/(2 pi)$.



But I think the problem $(1)$ have Simple proof?That is, we avoid calculating the result to give proof with inequality? Thanks






inequality







share|cite|improve this question











$endgroup$




show this
$$prod_{n=1}^{N}left(1+dfrac{1}{n^2}right)<4tag{1}$$



I have know this can find the value
$$prod_{n=1}^{+infty}left(1+dfrac{1}{n^2}right)=dfrac{sinh{pi}}{pi}=3.67<4$$Prove that $prod_{k=2}^{+infty} (1+1/k^2) = sinh(pi)/(2 pi)$.



But I think the problem $(1)$ have Simple proof?That is, we avoid calculating the result to give proof with inequality? Thanks






inequality




inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 6:09







function sug

















asked Dec 20 '18 at 6:05









function sugfunction sug

3061438




3061438












  • $begingroup$
    @RossMillikan,the links is $prod_{k=2}^{+infty}$ and my problem is $prod_{k=1}^{+infty}$
    $endgroup$
    – function sug
    Dec 20 '18 at 6:10




















  • $begingroup$
    @RossMillikan,the links is $prod_{k=2}^{+infty}$ and my problem is $prod_{k=1}^{+infty}$
    $endgroup$
    – function sug
    Dec 20 '18 at 6:10


















$begingroup$
@RossMillikan,the links is $prod_{k=2}^{+infty}$ and my problem is $prod_{k=1}^{+infty}$
$endgroup$
– function sug
Dec 20 '18 at 6:10






$begingroup$
@RossMillikan,the links is $prod_{k=2}^{+infty}$ and my problem is $prod_{k=1}^{+infty}$
$endgroup$
– function sug
Dec 20 '18 at 6:10












1 Answer
1






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$begingroup$

Note that
$$prod^N_{n=2} left(1-frac1{n^2}right) = prod^N_{n=2} frac{(n-1)(n+1)}{n^2} = frac12cdotfrac{N+1}{N} > frac12,$$
and that



$$prod^N_{n=2}left(1+frac1{n^2}right)prod^N_{n=2}left(1-frac1{n^2}right) =prod^N_{n=2}left(1-frac1{n^4}right)< 1.$$
So
$$prod^N_{n=2}left(1+frac1{n^2}right) < 2,$$
and we are done.






share|cite|improve this answer









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    $begingroup$

    Note that
    $$prod^N_{n=2} left(1-frac1{n^2}right) = prod^N_{n=2} frac{(n-1)(n+1)}{n^2} = frac12cdotfrac{N+1}{N} > frac12,$$
    and that



    $$prod^N_{n=2}left(1+frac1{n^2}right)prod^N_{n=2}left(1-frac1{n^2}right) =prod^N_{n=2}left(1-frac1{n^4}right)< 1.$$
    So
    $$prod^N_{n=2}left(1+frac1{n^2}right) < 2,$$
    and we are done.






    share|cite|improve this answer









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      5












      $begingroup$

      Note that
      $$prod^N_{n=2} left(1-frac1{n^2}right) = prod^N_{n=2} frac{(n-1)(n+1)}{n^2} = frac12cdotfrac{N+1}{N} > frac12,$$
      and that



      $$prod^N_{n=2}left(1+frac1{n^2}right)prod^N_{n=2}left(1-frac1{n^2}right) =prod^N_{n=2}left(1-frac1{n^4}right)< 1.$$
      So
      $$prod^N_{n=2}left(1+frac1{n^2}right) < 2,$$
      and we are done.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        Note that
        $$prod^N_{n=2} left(1-frac1{n^2}right) = prod^N_{n=2} frac{(n-1)(n+1)}{n^2} = frac12cdotfrac{N+1}{N} > frac12,$$
        and that



        $$prod^N_{n=2}left(1+frac1{n^2}right)prod^N_{n=2}left(1-frac1{n^2}right) =prod^N_{n=2}left(1-frac1{n^4}right)< 1.$$
        So
        $$prod^N_{n=2}left(1+frac1{n^2}right) < 2,$$
        and we are done.






        share|cite|improve this answer









        $endgroup$



        Note that
        $$prod^N_{n=2} left(1-frac1{n^2}right) = prod^N_{n=2} frac{(n-1)(n+1)}{n^2} = frac12cdotfrac{N+1}{N} > frac12,$$
        and that



        $$prod^N_{n=2}left(1+frac1{n^2}right)prod^N_{n=2}left(1-frac1{n^2}right) =prod^N_{n=2}left(1-frac1{n^4}right)< 1.$$
        So
        $$prod^N_{n=2}left(1+frac1{n^2}right) < 2,$$
        and we are done.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 6:15









        Quang HoangQuang Hoang

        13.2k1233




        13.2k1233






























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