$lim_{n to infty}(1+frac{1}{n^2})(1+frac{2}{n^2})…(1+frac{n}{n^2})=e^{frac{1}{2}}$.












1












$begingroup$


Here is the beginning of a proof:
Suppose $0<k leq n$,
$1+frac{1}{n}<(1+frac{k}{n^2})(1+frac{n+1-k}{n^2})=1+frac{n+1}{n^2}+frac{k(n+1-k)}{n^4}leq 1+frac{1}{n}+frac{1}{n^2}+frac{(n+1)^2}{4n^4}$.
I'm confused by the second inequality above.






real-analysis sequences-and-series inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you interested in why $1+frac {n+1}{n^2}+frac {k(n+1-k)}{n^4} leq 1+frac 1n+frac 1{n^2}+frac {(n+1)^2}{4n^2}?$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 17:46












  • $begingroup$
    yes, that is what I'm confused about
    $endgroup$
    – nafhgood
    Dec 14 '18 at 17:47






  • 2




    $begingroup$
    Well, $1+frac{n+1}{n^2}=1+frac 1n+frac 1{n^2}$ so we are interested in proving that $frac {k(n+1-k)}{n^4} leq frac {(n+1)^2}{4n^2}.$ Now we get $4k(n+1-k)leq n^2(n+1)^2$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 17:50












  • $begingroup$
    The second inequality is just GM < AM with $k, n+1-k$
    $endgroup$
    – Naweed G. Seldon
    Dec 14 '18 at 17:50








  • 2




    $begingroup$
    Yes, that is what I meant. Sorry for the typo. Well, you can rearrange it to $k(n+1-k) leq frac {(n+1)^2}4$ and then we have $sqrt {k(n+1-k)} leq frac{n+1}2$. We can rewrite this as $sqrt {k(n+1-k)} leq frac{k+(n+1-k)}2$ and this can be proved by the AM-GM inequality by letting $k=a$ and $n+1-k=b$. According to AM-GM inequality, $frac {a+b}2 geq sqrt {ab}$. I basically paraphrased @NaweedG.Seldon .
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 18:10
















1












$begingroup$


Here is the beginning of a proof:
Suppose $0<k leq n$,
$1+frac{1}{n}<(1+frac{k}{n^2})(1+frac{n+1-k}{n^2})=1+frac{n+1}{n^2}+frac{k(n+1-k)}{n^4}leq 1+frac{1}{n}+frac{1}{n^2}+frac{(n+1)^2}{4n^4}$.
I'm confused by the second inequality above.






real-analysis sequences-and-series inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you interested in why $1+frac {n+1}{n^2}+frac {k(n+1-k)}{n^4} leq 1+frac 1n+frac 1{n^2}+frac {(n+1)^2}{4n^2}?$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 17:46












  • $begingroup$
    yes, that is what I'm confused about
    $endgroup$
    – nafhgood
    Dec 14 '18 at 17:47






  • 2




    $begingroup$
    Well, $1+frac{n+1}{n^2}=1+frac 1n+frac 1{n^2}$ so we are interested in proving that $frac {k(n+1-k)}{n^4} leq frac {(n+1)^2}{4n^2}.$ Now we get $4k(n+1-k)leq n^2(n+1)^2$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 17:50












  • $begingroup$
    The second inequality is just GM < AM with $k, n+1-k$
    $endgroup$
    – Naweed G. Seldon
    Dec 14 '18 at 17:50








  • 2




    $begingroup$
    Yes, that is what I meant. Sorry for the typo. Well, you can rearrange it to $k(n+1-k) leq frac {(n+1)^2}4$ and then we have $sqrt {k(n+1-k)} leq frac{n+1}2$. We can rewrite this as $sqrt {k(n+1-k)} leq frac{k+(n+1-k)}2$ and this can be proved by the AM-GM inequality by letting $k=a$ and $n+1-k=b$. According to AM-GM inequality, $frac {a+b}2 geq sqrt {ab}$. I basically paraphrased @NaweedG.Seldon .
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 18:10














1












1








1


1



$begingroup$


Here is the beginning of a proof:
Suppose $0<k leq n$,
$1+frac{1}{n}<(1+frac{k}{n^2})(1+frac{n+1-k}{n^2})=1+frac{n+1}{n^2}+frac{k(n+1-k)}{n^4}leq 1+frac{1}{n}+frac{1}{n^2}+frac{(n+1)^2}{4n^4}$.
I'm confused by the second inequality above.






real-analysis sequences-and-series inequality







share|cite|improve this question









$endgroup$




Here is the beginning of a proof:
Suppose $0<k leq n$,
$1+frac{1}{n}<(1+frac{k}{n^2})(1+frac{n+1-k}{n^2})=1+frac{n+1}{n^2}+frac{k(n+1-k)}{n^4}leq 1+frac{1}{n}+frac{1}{n^2}+frac{(n+1)^2}{4n^4}$.
I'm confused by the second inequality above.






real-analysis sequences-and-series inequality




real-analysis sequences-and-series inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 17:44









nafhgoodnafhgood

1,805422




1,805422












  • $begingroup$
    Are you interested in why $1+frac {n+1}{n^2}+frac {k(n+1-k)}{n^4} leq 1+frac 1n+frac 1{n^2}+frac {(n+1)^2}{4n^2}?$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 17:46












  • $begingroup$
    yes, that is what I'm confused about
    $endgroup$
    – nafhgood
    Dec 14 '18 at 17:47






  • 2




    $begingroup$
    Well, $1+frac{n+1}{n^2}=1+frac 1n+frac 1{n^2}$ so we are interested in proving that $frac {k(n+1-k)}{n^4} leq frac {(n+1)^2}{4n^2}.$ Now we get $4k(n+1-k)leq n^2(n+1)^2$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 17:50












  • $begingroup$
    The second inequality is just GM < AM with $k, n+1-k$
    $endgroup$
    – Naweed G. Seldon
    Dec 14 '18 at 17:50








  • 2




    $begingroup$
    Yes, that is what I meant. Sorry for the typo. Well, you can rearrange it to $k(n+1-k) leq frac {(n+1)^2}4$ and then we have $sqrt {k(n+1-k)} leq frac{n+1}2$. We can rewrite this as $sqrt {k(n+1-k)} leq frac{k+(n+1-k)}2$ and this can be proved by the AM-GM inequality by letting $k=a$ and $n+1-k=b$. According to AM-GM inequality, $frac {a+b}2 geq sqrt {ab}$. I basically paraphrased @NaweedG.Seldon .
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 18:10


















  • $begingroup$
    Are you interested in why $1+frac {n+1}{n^2}+frac {k(n+1-k)}{n^4} leq 1+frac 1n+frac 1{n^2}+frac {(n+1)^2}{4n^2}?$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 17:46












  • $begingroup$
    yes, that is what I'm confused about
    $endgroup$
    – nafhgood
    Dec 14 '18 at 17:47






  • 2




    $begingroup$
    Well, $1+frac{n+1}{n^2}=1+frac 1n+frac 1{n^2}$ so we are interested in proving that $frac {k(n+1-k)}{n^4} leq frac {(n+1)^2}{4n^2}.$ Now we get $4k(n+1-k)leq n^2(n+1)^2$
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 17:50












  • $begingroup$
    The second inequality is just GM < AM with $k, n+1-k$
    $endgroup$
    – Naweed G. Seldon
    Dec 14 '18 at 17:50








  • 2




    $begingroup$
    Yes, that is what I meant. Sorry for the typo. Well, you can rearrange it to $k(n+1-k) leq frac {(n+1)^2}4$ and then we have $sqrt {k(n+1-k)} leq frac{n+1}2$. We can rewrite this as $sqrt {k(n+1-k)} leq frac{k+(n+1-k)}2$ and this can be proved by the AM-GM inequality by letting $k=a$ and $n+1-k=b$. According to AM-GM inequality, $frac {a+b}2 geq sqrt {ab}$. I basically paraphrased @NaweedG.Seldon .
    $endgroup$
    – Mohammad Zuhair Khan
    Dec 14 '18 at 18:10
















$begingroup$
Are you interested in why $1+frac {n+1}{n^2}+frac {k(n+1-k)}{n^4} leq 1+frac 1n+frac 1{n^2}+frac {(n+1)^2}{4n^2}?$
$endgroup$
– Mohammad Zuhair Khan
Dec 14 '18 at 17:46






$begingroup$
Are you interested in why $1+frac {n+1}{n^2}+frac {k(n+1-k)}{n^4} leq 1+frac 1n+frac 1{n^2}+frac {(n+1)^2}{4n^2}?$
$endgroup$
– Mohammad Zuhair Khan
Dec 14 '18 at 17:46














$begingroup$
yes, that is what I'm confused about
$endgroup$
– nafhgood
Dec 14 '18 at 17:47




$begingroup$
yes, that is what I'm confused about
$endgroup$
– nafhgood
Dec 14 '18 at 17:47




2




2




$begingroup$
Well, $1+frac{n+1}{n^2}=1+frac 1n+frac 1{n^2}$ so we are interested in proving that $frac {k(n+1-k)}{n^4} leq frac {(n+1)^2}{4n^2}.$ Now we get $4k(n+1-k)leq n^2(n+1)^2$
$endgroup$
– Mohammad Zuhair Khan
Dec 14 '18 at 17:50






$begingroup$
Well, $1+frac{n+1}{n^2}=1+frac 1n+frac 1{n^2}$ so we are interested in proving that $frac {k(n+1-k)}{n^4} leq frac {(n+1)^2}{4n^2}.$ Now we get $4k(n+1-k)leq n^2(n+1)^2$
$endgroup$
– Mohammad Zuhair Khan
Dec 14 '18 at 17:50














$begingroup$
The second inequality is just GM < AM with $k, n+1-k$
$endgroup$
– Naweed G. Seldon
Dec 14 '18 at 17:50






$begingroup$
The second inequality is just GM < AM with $k, n+1-k$
$endgroup$
– Naweed G. Seldon
Dec 14 '18 at 17:50






2




2




$begingroup$
Yes, that is what I meant. Sorry for the typo. Well, you can rearrange it to $k(n+1-k) leq frac {(n+1)^2}4$ and then we have $sqrt {k(n+1-k)} leq frac{n+1}2$. We can rewrite this as $sqrt {k(n+1-k)} leq frac{k+(n+1-k)}2$ and this can be proved by the AM-GM inequality by letting $k=a$ and $n+1-k=b$. According to AM-GM inequality, $frac {a+b}2 geq sqrt {ab}$. I basically paraphrased @NaweedG.Seldon .
$endgroup$
– Mohammad Zuhair Khan
Dec 14 '18 at 18:10




$begingroup$
Yes, that is what I meant. Sorry for the typo. Well, you can rearrange it to $k(n+1-k) leq frac {(n+1)^2}4$ and then we have $sqrt {k(n+1-k)} leq frac{n+1}2$. We can rewrite this as $sqrt {k(n+1-k)} leq frac{k+(n+1-k)}2$ and this can be proved by the AM-GM inequality by letting $k=a$ and $n+1-k=b$. According to AM-GM inequality, $frac {a+b}2 geq sqrt {ab}$. I basically paraphrased @NaweedG.Seldon .
$endgroup$
– Mohammad Zuhair Khan
Dec 14 '18 at 18:10










3 Answers
3






active

oldest

votes


















2












$begingroup$

Hint: By AM-GM Inequality
$$frac{k+ n+1-k}{2} geq sqrt{k(n+1-k)}$$






share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    As an alternative, we have that



    $$left(1+frac{1}{n^2}right)left(1+frac{2}{n^2}right)ldotsleft(1+frac{n}{n^2}right)=prod_{k=1}^{n}left(1+frac{k}{n^2}right)=e^{sum_{k=1}^{n} logleft(1+frac{k}{n^2}right) }$$



    and



    $$sum_{k=1}^{n} logleft(1+frac{k}{n^2}right)=sum_{k=1}^{n}left(frac{k}{n^2}+k^2Oleft(frac{1}{n^4}right)right)=frac1{n^2}sum_{k=1}^{n}k+Oleft(frac{1}{n^4}right)sum_{k=1}^{n}k^2to frac12$$



    indeed




    • $sum_{k=1}^{n}k=frac{n(n+1)}{2}implies frac1{n^2}sum_{k=1}^{n}k=frac{n(n+1)}{2n^2}to frac12$

    • $sum_{k=1}^{n}k^2=frac{n(n+1)(2n+1)}{6}implies Oleft(frac{1}{n^4}right)sum_{k=1}^{n}k^2=Oleft(frac{1}{n}right)to 0$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry, how do you get the first equality on the second line?
      $endgroup$
      – nafhgood
      Dec 14 '18 at 18:17










    • $begingroup$
      @mathnoob We have $A=e^{log A}$ and $log prod a_i = sum log a_i$.
      $endgroup$
      – gimusi
      Dec 14 '18 at 18:32












    • $begingroup$
      @mathnoob the series for logarithm: $log(1+x) = sum_{j=1}^infty frac{(-1)^{j+1}}{j} x^j$
      $endgroup$
      – ploosu2
      Dec 14 '18 at 19:41





















    3












    $begingroup$

    $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
    newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
    newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
    newcommand{dd}{mathrm{d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,mathrm{e}^{#1},}
    newcommand{ic}{mathrm{i}}
    newcommand{mc}[1]{mathcal{#1}}
    newcommand{mrm}[1]{mathrm{#1}}
    newcommand{pars}[1]{left(,{#1},right)}
    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{root}[2]{,sqrt[#1]{,{#2},},}
    newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
    newcommand{verts}[1]{leftvert,{#1},rightvert}$




    $ds{lim_{n to infty}pars{1 + {1 over n^{2}}}
    pars{1 + {2 over n^{2}}}cdotspars{1 + {n over n^{2}}} = expo{1/2}: {LARGE ?}}$    .




    begin{align}
    &bbox[#ffd,10px]{lim_{n to infty}pars{1 + {1 over n^{2}}}
    pars{1 + {2 over n^{2}}}cdotspars{1 + {n over n^{2}}}} =
    lim_{n to infty}prod_{k = 1}^{n}pars{1 + {k over n^{2}}}
    \[5mm] = &
    lim_{n to infty}{prod_{k = 1}^{n}pars{k + n^{2}} over
    prod_{k = 1}^{n}n^{2}} =
    lim_{n to infty}{pars{1 + n^{2}}^{largeoverline{n}} over pars{n^{2}}^{n}}
    \[5mm] = &
    lim_{n to infty}{Gammapars{1 + n^{2} + n}/Gammapars{1 + n^{2}} over n^{2n}} =
    lim_{n to infty}{pars{n^{2} + n}! over n^{2n}pars{n^{2}}!}
    \[5mm] = &
    lim_{n to infty}
    {root{2pi}pars{n^{2} + n}^{n^{2} + n + 1/2},
    expo{-pars{n^{2} + n}} over
    n^{2n}bracks{root{2pi}pars{n^{2}}^{n^{2} + 1/2}expo{-n^{2}}}} \[5mm] = &
    lim_{n to infty}
    {pars{n^{2}}^{n^{2} + n + 1/2}pars{1 + 1/n}^{n^{2} + n + 1/2},
    expo{-n} over
    n^{2n}pars{n^{2n^{2} + 1}}}
    \[5mm] = &
    lim_{n to infty}exppars{bracks{n^{2} + n + {1 over 2}}
    lnpars{1 + {1 over n}} - n}
    \[5mm] = &
    lim_{n to infty}exppars{bracks{n^{2} + n + {1 over 2}}
    bracks{{1 over n} - {1 over 2n^{2}}} - n} = bbx{expo{1/2}}
    approx 1.6487
    end{align}




    Note that


    $ds{pars{n^{2} + n + {1 over 2}}
    lnpars{1 + {1 over n}} - n
    ,,,stackrel{mrm{as} n to infty}{sim},,,
    {1 over 2} + {1 over 3n} - {1 over 6n^{2}}}$    .







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry, What does $a^{overline{n}}$ mean?
      $endgroup$
      – nafhgood
      Dec 14 '18 at 22:57






    • 1




      $begingroup$
      @mathnoob It's the Raising Factorial. A shortcut notation for $displaystyle x^{largeoverline{n}} equiv xleft(,{x},right)left(,{x + 1},right)left(,{x + n - 1},right)$.
      $endgroup$
      – Felix Marin
      Dec 15 '18 at 0:35













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Hint: By AM-GM Inequality
    $$frac{k+ n+1-k}{2} geq sqrt{k(n+1-k)}$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint: By AM-GM Inequality
      $$frac{k+ n+1-k}{2} geq sqrt{k(n+1-k)}$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint: By AM-GM Inequality
        $$frac{k+ n+1-k}{2} geq sqrt{k(n+1-k)}$$






        share|cite|improve this answer









        $endgroup$



        Hint: By AM-GM Inequality
        $$frac{k+ n+1-k}{2} geq sqrt{k(n+1-k)}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 17:52









        Naweed G. SeldonNaweed G. Seldon

        1,304419




        1,304419























            5












            $begingroup$

            As an alternative, we have that



            $$left(1+frac{1}{n^2}right)left(1+frac{2}{n^2}right)ldotsleft(1+frac{n}{n^2}right)=prod_{k=1}^{n}left(1+frac{k}{n^2}right)=e^{sum_{k=1}^{n} logleft(1+frac{k}{n^2}right) }$$



            and



            $$sum_{k=1}^{n} logleft(1+frac{k}{n^2}right)=sum_{k=1}^{n}left(frac{k}{n^2}+k^2Oleft(frac{1}{n^4}right)right)=frac1{n^2}sum_{k=1}^{n}k+Oleft(frac{1}{n^4}right)sum_{k=1}^{n}k^2to frac12$$



            indeed




            • $sum_{k=1}^{n}k=frac{n(n+1)}{2}implies frac1{n^2}sum_{k=1}^{n}k=frac{n(n+1)}{2n^2}to frac12$

            • $sum_{k=1}^{n}k^2=frac{n(n+1)(2n+1)}{6}implies Oleft(frac{1}{n^4}right)sum_{k=1}^{n}k^2=Oleft(frac{1}{n}right)to 0$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Sorry, how do you get the first equality on the second line?
              $endgroup$
              – nafhgood
              Dec 14 '18 at 18:17










            • $begingroup$
              @mathnoob We have $A=e^{log A}$ and $log prod a_i = sum log a_i$.
              $endgroup$
              – gimusi
              Dec 14 '18 at 18:32












            • $begingroup$
              @mathnoob the series for logarithm: $log(1+x) = sum_{j=1}^infty frac{(-1)^{j+1}}{j} x^j$
              $endgroup$
              – ploosu2
              Dec 14 '18 at 19:41


















            5












            $begingroup$

            As an alternative, we have that



            $$left(1+frac{1}{n^2}right)left(1+frac{2}{n^2}right)ldotsleft(1+frac{n}{n^2}right)=prod_{k=1}^{n}left(1+frac{k}{n^2}right)=e^{sum_{k=1}^{n} logleft(1+frac{k}{n^2}right) }$$



            and



            $$sum_{k=1}^{n} logleft(1+frac{k}{n^2}right)=sum_{k=1}^{n}left(frac{k}{n^2}+k^2Oleft(frac{1}{n^4}right)right)=frac1{n^2}sum_{k=1}^{n}k+Oleft(frac{1}{n^4}right)sum_{k=1}^{n}k^2to frac12$$



            indeed




            • $sum_{k=1}^{n}k=frac{n(n+1)}{2}implies frac1{n^2}sum_{k=1}^{n}k=frac{n(n+1)}{2n^2}to frac12$

            • $sum_{k=1}^{n}k^2=frac{n(n+1)(2n+1)}{6}implies Oleft(frac{1}{n^4}right)sum_{k=1}^{n}k^2=Oleft(frac{1}{n}right)to 0$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Sorry, how do you get the first equality on the second line?
              $endgroup$
              – nafhgood
              Dec 14 '18 at 18:17










            • $begingroup$
              @mathnoob We have $A=e^{log A}$ and $log prod a_i = sum log a_i$.
              $endgroup$
              – gimusi
              Dec 14 '18 at 18:32












            • $begingroup$
              @mathnoob the series for logarithm: $log(1+x) = sum_{j=1}^infty frac{(-1)^{j+1}}{j} x^j$
              $endgroup$
              – ploosu2
              Dec 14 '18 at 19:41
















            5












            5








            5





            $begingroup$

            As an alternative, we have that



            $$left(1+frac{1}{n^2}right)left(1+frac{2}{n^2}right)ldotsleft(1+frac{n}{n^2}right)=prod_{k=1}^{n}left(1+frac{k}{n^2}right)=e^{sum_{k=1}^{n} logleft(1+frac{k}{n^2}right) }$$



            and



            $$sum_{k=1}^{n} logleft(1+frac{k}{n^2}right)=sum_{k=1}^{n}left(frac{k}{n^2}+k^2Oleft(frac{1}{n^4}right)right)=frac1{n^2}sum_{k=1}^{n}k+Oleft(frac{1}{n^4}right)sum_{k=1}^{n}k^2to frac12$$



            indeed




            • $sum_{k=1}^{n}k=frac{n(n+1)}{2}implies frac1{n^2}sum_{k=1}^{n}k=frac{n(n+1)}{2n^2}to frac12$

            • $sum_{k=1}^{n}k^2=frac{n(n+1)(2n+1)}{6}implies Oleft(frac{1}{n^4}right)sum_{k=1}^{n}k^2=Oleft(frac{1}{n}right)to 0$






            share|cite|improve this answer











            $endgroup$



            As an alternative, we have that



            $$left(1+frac{1}{n^2}right)left(1+frac{2}{n^2}right)ldotsleft(1+frac{n}{n^2}right)=prod_{k=1}^{n}left(1+frac{k}{n^2}right)=e^{sum_{k=1}^{n} logleft(1+frac{k}{n^2}right) }$$



            and



            $$sum_{k=1}^{n} logleft(1+frac{k}{n^2}right)=sum_{k=1}^{n}left(frac{k}{n^2}+k^2Oleft(frac{1}{n^4}right)right)=frac1{n^2}sum_{k=1}^{n}k+Oleft(frac{1}{n^4}right)sum_{k=1}^{n}k^2to frac12$$



            indeed




            • $sum_{k=1}^{n}k=frac{n(n+1)}{2}implies frac1{n^2}sum_{k=1}^{n}k=frac{n(n+1)}{2n^2}to frac12$

            • $sum_{k=1}^{n}k^2=frac{n(n+1)(2n+1)}{6}implies Oleft(frac{1}{n^4}right)sum_{k=1}^{n}k^2=Oleft(frac{1}{n}right)to 0$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 15 '18 at 10:16

























            answered Dec 14 '18 at 17:56









            gimusigimusi

            92.8k84494




            92.8k84494












            • $begingroup$
              Sorry, how do you get the first equality on the second line?
              $endgroup$
              – nafhgood
              Dec 14 '18 at 18:17










            • $begingroup$
              @mathnoob We have $A=e^{log A}$ and $log prod a_i = sum log a_i$.
              $endgroup$
              – gimusi
              Dec 14 '18 at 18:32












            • $begingroup$
              @mathnoob the series for logarithm: $log(1+x) = sum_{j=1}^infty frac{(-1)^{j+1}}{j} x^j$
              $endgroup$
              – ploosu2
              Dec 14 '18 at 19:41




















            • $begingroup$
              Sorry, how do you get the first equality on the second line?
              $endgroup$
              – nafhgood
              Dec 14 '18 at 18:17










            • $begingroup$
              @mathnoob We have $A=e^{log A}$ and $log prod a_i = sum log a_i$.
              $endgroup$
              – gimusi
              Dec 14 '18 at 18:32












            • $begingroup$
              @mathnoob the series for logarithm: $log(1+x) = sum_{j=1}^infty frac{(-1)^{j+1}}{j} x^j$
              $endgroup$
              – ploosu2
              Dec 14 '18 at 19:41


















            $begingroup$
            Sorry, how do you get the first equality on the second line?
            $endgroup$
            – nafhgood
            Dec 14 '18 at 18:17




            $begingroup$
            Sorry, how do you get the first equality on the second line?
            $endgroup$
            – nafhgood
            Dec 14 '18 at 18:17












            $begingroup$
            @mathnoob We have $A=e^{log A}$ and $log prod a_i = sum log a_i$.
            $endgroup$
            – gimusi
            Dec 14 '18 at 18:32






            $begingroup$
            @mathnoob We have $A=e^{log A}$ and $log prod a_i = sum log a_i$.
            $endgroup$
            – gimusi
            Dec 14 '18 at 18:32














            $begingroup$
            @mathnoob the series for logarithm: $log(1+x) = sum_{j=1}^infty frac{(-1)^{j+1}}{j} x^j$
            $endgroup$
            – ploosu2
            Dec 14 '18 at 19:41






            $begingroup$
            @mathnoob the series for logarithm: $log(1+x) = sum_{j=1}^infty frac{(-1)^{j+1}}{j} x^j$
            $endgroup$
            – ploosu2
            Dec 14 '18 at 19:41













            3












            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            $ds{lim_{n to infty}pars{1 + {1 over n^{2}}}
            pars{1 + {2 over n^{2}}}cdotspars{1 + {n over n^{2}}} = expo{1/2}: {LARGE ?}}$            .




            begin{align}
            &bbox[#ffd,10px]{lim_{n to infty}pars{1 + {1 over n^{2}}}
            pars{1 + {2 over n^{2}}}cdotspars{1 + {n over n^{2}}}} =
            lim_{n to infty}prod_{k = 1}^{n}pars{1 + {k over n^{2}}}
            \[5mm] = &
            lim_{n to infty}{prod_{k = 1}^{n}pars{k + n^{2}} over
            prod_{k = 1}^{n}n^{2}} =
            lim_{n to infty}{pars{1 + n^{2}}^{largeoverline{n}} over pars{n^{2}}^{n}}
            \[5mm] = &
            lim_{n to infty}{Gammapars{1 + n^{2} + n}/Gammapars{1 + n^{2}} over n^{2n}} =
            lim_{n to infty}{pars{n^{2} + n}! over n^{2n}pars{n^{2}}!}
            \[5mm] = &
            lim_{n to infty}
            {root{2pi}pars{n^{2} + n}^{n^{2} + n + 1/2},
            expo{-pars{n^{2} + n}} over
            n^{2n}bracks{root{2pi}pars{n^{2}}^{n^{2} + 1/2}expo{-n^{2}}}} \[5mm] = &
            lim_{n to infty}
            {pars{n^{2}}^{n^{2} + n + 1/2}pars{1 + 1/n}^{n^{2} + n + 1/2},
            expo{-n} over
            n^{2n}pars{n^{2n^{2} + 1}}}
            \[5mm] = &
            lim_{n to infty}exppars{bracks{n^{2} + n + {1 over 2}}
            lnpars{1 + {1 over n}} - n}
            \[5mm] = &
            lim_{n to infty}exppars{bracks{n^{2} + n + {1 over 2}}
            bracks{{1 over n} - {1 over 2n^{2}}} - n} = bbx{expo{1/2}}
            approx 1.6487
            end{align}




            Note that


            $ds{pars{n^{2} + n + {1 over 2}}
            lnpars{1 + {1 over n}} - n
            ,,,stackrel{mrm{as} n to infty}{sim},,,
            {1 over 2} + {1 over 3n} - {1 over 6n^{2}}}$            .







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Sorry, What does $a^{overline{n}}$ mean?
              $endgroup$
              – nafhgood
              Dec 14 '18 at 22:57






            • 1




              $begingroup$
              @mathnoob It's the Raising Factorial. A shortcut notation for $displaystyle x^{largeoverline{n}} equiv xleft(,{x},right)left(,{x + 1},right)left(,{x + n - 1},right)$.
              $endgroup$
              – Felix Marin
              Dec 15 '18 at 0:35


















            3












            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            $ds{lim_{n to infty}pars{1 + {1 over n^{2}}}
            pars{1 + {2 over n^{2}}}cdotspars{1 + {n over n^{2}}} = expo{1/2}: {LARGE ?}}$            .




            begin{align}
            &bbox[#ffd,10px]{lim_{n to infty}pars{1 + {1 over n^{2}}}
            pars{1 + {2 over n^{2}}}cdotspars{1 + {n over n^{2}}}} =
            lim_{n to infty}prod_{k = 1}^{n}pars{1 + {k over n^{2}}}
            \[5mm] = &
            lim_{n to infty}{prod_{k = 1}^{n}pars{k + n^{2}} over
            prod_{k = 1}^{n}n^{2}} =
            lim_{n to infty}{pars{1 + n^{2}}^{largeoverline{n}} over pars{n^{2}}^{n}}
            \[5mm] = &
            lim_{n to infty}{Gammapars{1 + n^{2} + n}/Gammapars{1 + n^{2}} over n^{2n}} =
            lim_{n to infty}{pars{n^{2} + n}! over n^{2n}pars{n^{2}}!}
            \[5mm] = &
            lim_{n to infty}
            {root{2pi}pars{n^{2} + n}^{n^{2} + n + 1/2},
            expo{-pars{n^{2} + n}} over
            n^{2n}bracks{root{2pi}pars{n^{2}}^{n^{2} + 1/2}expo{-n^{2}}}} \[5mm] = &
            lim_{n to infty}
            {pars{n^{2}}^{n^{2} + n + 1/2}pars{1 + 1/n}^{n^{2} + n + 1/2},
            expo{-n} over
            n^{2n}pars{n^{2n^{2} + 1}}}
            \[5mm] = &
            lim_{n to infty}exppars{bracks{n^{2} + n + {1 over 2}}
            lnpars{1 + {1 over n}} - n}
            \[5mm] = &
            lim_{n to infty}exppars{bracks{n^{2} + n + {1 over 2}}
            bracks{{1 over n} - {1 over 2n^{2}}} - n} = bbx{expo{1/2}}
            approx 1.6487
            end{align}




            Note that


            $ds{pars{n^{2} + n + {1 over 2}}
            lnpars{1 + {1 over n}} - n
            ,,,stackrel{mrm{as} n to infty}{sim},,,
            {1 over 2} + {1 over 3n} - {1 over 6n^{2}}}$            .







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Sorry, What does $a^{overline{n}}$ mean?
              $endgroup$
              – nafhgood
              Dec 14 '18 at 22:57






            • 1




              $begingroup$
              @mathnoob It's the Raising Factorial. A shortcut notation for $displaystyle x^{largeoverline{n}} equiv xleft(,{x},right)left(,{x + 1},right)left(,{x + n - 1},right)$.
              $endgroup$
              – Felix Marin
              Dec 15 '18 at 0:35
















            3












            3








            3





            $begingroup$

            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            $ds{lim_{n to infty}pars{1 + {1 over n^{2}}}
            pars{1 + {2 over n^{2}}}cdotspars{1 + {n over n^{2}}} = expo{1/2}: {LARGE ?}}$            .




            begin{align}
            &bbox[#ffd,10px]{lim_{n to infty}pars{1 + {1 over n^{2}}}
            pars{1 + {2 over n^{2}}}cdotspars{1 + {n over n^{2}}}} =
            lim_{n to infty}prod_{k = 1}^{n}pars{1 + {k over n^{2}}}
            \[5mm] = &
            lim_{n to infty}{prod_{k = 1}^{n}pars{k + n^{2}} over
            prod_{k = 1}^{n}n^{2}} =
            lim_{n to infty}{pars{1 + n^{2}}^{largeoverline{n}} over pars{n^{2}}^{n}}
            \[5mm] = &
            lim_{n to infty}{Gammapars{1 + n^{2} + n}/Gammapars{1 + n^{2}} over n^{2n}} =
            lim_{n to infty}{pars{n^{2} + n}! over n^{2n}pars{n^{2}}!}
            \[5mm] = &
            lim_{n to infty}
            {root{2pi}pars{n^{2} + n}^{n^{2} + n + 1/2},
            expo{-pars{n^{2} + n}} over
            n^{2n}bracks{root{2pi}pars{n^{2}}^{n^{2} + 1/2}expo{-n^{2}}}} \[5mm] = &
            lim_{n to infty}
            {pars{n^{2}}^{n^{2} + n + 1/2}pars{1 + 1/n}^{n^{2} + n + 1/2},
            expo{-n} over
            n^{2n}pars{n^{2n^{2} + 1}}}
            \[5mm] = &
            lim_{n to infty}exppars{bracks{n^{2} + n + {1 over 2}}
            lnpars{1 + {1 over n}} - n}
            \[5mm] = &
            lim_{n to infty}exppars{bracks{n^{2} + n + {1 over 2}}
            bracks{{1 over n} - {1 over 2n^{2}}} - n} = bbx{expo{1/2}}
            approx 1.6487
            end{align}




            Note that


            $ds{pars{n^{2} + n + {1 over 2}}
            lnpars{1 + {1 over n}} - n
            ,,,stackrel{mrm{as} n to infty}{sim},,,
            {1 over 2} + {1 over 3n} - {1 over 6n^{2}}}$            .







            share|cite|improve this answer











            $endgroup$



            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
            newcommand{dd}{mathrm{d}}
            newcommand{ds}[1]{displaystyle{#1}}
            newcommand{expo}[1]{,mathrm{e}^{#1},}
            newcommand{ic}{mathrm{i}}
            newcommand{mc}[1]{mathcal{#1}}
            newcommand{mrm}[1]{mathrm{#1}}
            newcommand{pars}[1]{left(,{#1},right)}
            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
            newcommand{verts}[1]{leftvert,{#1},rightvert}$




            $ds{lim_{n to infty}pars{1 + {1 over n^{2}}}
            pars{1 + {2 over n^{2}}}cdotspars{1 + {n over n^{2}}} = expo{1/2}: {LARGE ?}}$            .




            begin{align}
            &bbox[#ffd,10px]{lim_{n to infty}pars{1 + {1 over n^{2}}}
            pars{1 + {2 over n^{2}}}cdotspars{1 + {n over n^{2}}}} =
            lim_{n to infty}prod_{k = 1}^{n}pars{1 + {k over n^{2}}}
            \[5mm] = &
            lim_{n to infty}{prod_{k = 1}^{n}pars{k + n^{2}} over
            prod_{k = 1}^{n}n^{2}} =
            lim_{n to infty}{pars{1 + n^{2}}^{largeoverline{n}} over pars{n^{2}}^{n}}
            \[5mm] = &
            lim_{n to infty}{Gammapars{1 + n^{2} + n}/Gammapars{1 + n^{2}} over n^{2n}} =
            lim_{n to infty}{pars{n^{2} + n}! over n^{2n}pars{n^{2}}!}
            \[5mm] = &
            lim_{n to infty}
            {root{2pi}pars{n^{2} + n}^{n^{2} + n + 1/2},
            expo{-pars{n^{2} + n}} over
            n^{2n}bracks{root{2pi}pars{n^{2}}^{n^{2} + 1/2}expo{-n^{2}}}} \[5mm] = &
            lim_{n to infty}
            {pars{n^{2}}^{n^{2} + n + 1/2}pars{1 + 1/n}^{n^{2} + n + 1/2},
            expo{-n} over
            n^{2n}pars{n^{2n^{2} + 1}}}
            \[5mm] = &
            lim_{n to infty}exppars{bracks{n^{2} + n + {1 over 2}}
            lnpars{1 + {1 over n}} - n}
            \[5mm] = &
            lim_{n to infty}exppars{bracks{n^{2} + n + {1 over 2}}
            bracks{{1 over n} - {1 over 2n^{2}}} - n} = bbx{expo{1/2}}
            approx 1.6487
            end{align}




            Note that


            $ds{pars{n^{2} + n + {1 over 2}}
            lnpars{1 + {1 over n}} - n
            ,,,stackrel{mrm{as} n to infty}{sim},,,
            {1 over 2} + {1 over 3n} - {1 over 6n^{2}}}$            .








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 14 '18 at 22:09

























            answered Dec 14 '18 at 22:03









            Felix MarinFelix Marin

            68.1k7109142




            68.1k7109142












            • $begingroup$
              Sorry, What does $a^{overline{n}}$ mean?
              $endgroup$
              – nafhgood
              Dec 14 '18 at 22:57






            • 1




              $begingroup$
              @mathnoob It's the Raising Factorial. A shortcut notation for $displaystyle x^{largeoverline{n}} equiv xleft(,{x},right)left(,{x + 1},right)left(,{x + n - 1},right)$.
              $endgroup$
              – Felix Marin
              Dec 15 '18 at 0:35




















            • $begingroup$
              Sorry, What does $a^{overline{n}}$ mean?
              $endgroup$
              – nafhgood
              Dec 14 '18 at 22:57






            • 1




              $begingroup$
              @mathnoob It's the Raising Factorial. A shortcut notation for $displaystyle x^{largeoverline{n}} equiv xleft(,{x},right)left(,{x + 1},right)left(,{x + n - 1},right)$.
              $endgroup$
              – Felix Marin
              Dec 15 '18 at 0:35


















            $begingroup$
            Sorry, What does $a^{overline{n}}$ mean?
            $endgroup$
            – nafhgood
            Dec 14 '18 at 22:57




            $begingroup$
            Sorry, What does $a^{overline{n}}$ mean?
            $endgroup$
            – nafhgood
            Dec 14 '18 at 22:57




            1




            1




            $begingroup$
            @mathnoob It's the Raising Factorial. A shortcut notation for $displaystyle x^{largeoverline{n}} equiv xleft(,{x},right)left(,{x + 1},right)left(,{x + n - 1},right)$.
            $endgroup$
            – Felix Marin
            Dec 15 '18 at 0:35






            $begingroup$
            @mathnoob It's the Raising Factorial. A shortcut notation for $displaystyle x^{largeoverline{n}} equiv xleft(,{x},right)left(,{x + 1},right)left(,{x + n - 1},right)$.
            $endgroup$
            – Felix Marin
            Dec 15 '18 at 0:35




















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