Evaluate $lim_{ntoinfty}(1+frac{1}{a_1})cdots(1+frac{1}{a_n})$, where $a_1=1$, $a_n=n(1+a_{n-1})$












1












$begingroup$



Evaluate $lim_{ntoinfty}(1+frac{1}{a_1})(1+frac{1}{a_2})cdots(1+frac{1}{a_n})$,
where $a_1 = 1$, $a_n = n(1+a_{n-1})$




begin{align*}
&lim_{ntoinfty} left(1+frac{1}{a_1}right)left(1+frac{1}{a_2}right)cdotsleft(1+frac{1}{a_n}right) \
&= lim_{ntoinfty} frac{1}{a_1} cdot frac{1}{2} cdot frac{1}{3} cdots frac{1}{n} cdot (a_n + 1) \
&= lim_{ntoinfty} frac{a_n + 1}{n!}.
end{align*}



Then I'm stuck. How to proceed?






sequences-and-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    This is OEIS sequence A007526. The OEIS entry has much information including the limit you want to find.
    $endgroup$
    – Somos
    Jan 1 at 15:58
















1












$begingroup$



Evaluate $lim_{ntoinfty}(1+frac{1}{a_1})(1+frac{1}{a_2})cdots(1+frac{1}{a_n})$,
where $a_1 = 1$, $a_n = n(1+a_{n-1})$




begin{align*}
&lim_{ntoinfty} left(1+frac{1}{a_1}right)left(1+frac{1}{a_2}right)cdotsleft(1+frac{1}{a_n}right) \
&= lim_{ntoinfty} frac{1}{a_1} cdot frac{1}{2} cdot frac{1}{3} cdots frac{1}{n} cdot (a_n + 1) \
&= lim_{ntoinfty} frac{a_n + 1}{n!}.
end{align*}



Then I'm stuck. How to proceed?






sequences-and-series







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    This is OEIS sequence A007526. The OEIS entry has much information including the limit you want to find.
    $endgroup$
    – Somos
    Jan 1 at 15:58














1












1








1





$begingroup$



Evaluate $lim_{ntoinfty}(1+frac{1}{a_1})(1+frac{1}{a_2})cdots(1+frac{1}{a_n})$,
where $a_1 = 1$, $a_n = n(1+a_{n-1})$




begin{align*}
&lim_{ntoinfty} left(1+frac{1}{a_1}right)left(1+frac{1}{a_2}right)cdotsleft(1+frac{1}{a_n}right) \
&= lim_{ntoinfty} frac{1}{a_1} cdot frac{1}{2} cdot frac{1}{3} cdots frac{1}{n} cdot (a_n + 1) \
&= lim_{ntoinfty} frac{a_n + 1}{n!}.
end{align*}



Then I'm stuck. How to proceed?






sequences-and-series







share|cite|improve this question











$endgroup$





Evaluate $lim_{ntoinfty}(1+frac{1}{a_1})(1+frac{1}{a_2})cdots(1+frac{1}{a_n})$,
where $a_1 = 1$, $a_n = n(1+a_{n-1})$




begin{align*}
&lim_{ntoinfty} left(1+frac{1}{a_1}right)left(1+frac{1}{a_2}right)cdotsleft(1+frac{1}{a_n}right) \
&= lim_{ntoinfty} frac{1}{a_1} cdot frac{1}{2} cdot frac{1}{3} cdots frac{1}{n} cdot (a_n + 1) \
&= lim_{ntoinfty} frac{a_n + 1}{n!}.
end{align*}



Then I'm stuck. How to proceed?






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 1 at 15:43









Sangchul Lee

96.8k12173283




96.8k12173283










asked Jan 1 at 15:11









MathsaddictMathsaddict

3669




3669








  • 2




    $begingroup$
    This is OEIS sequence A007526. The OEIS entry has much information including the limit you want to find.
    $endgroup$
    – Somos
    Jan 1 at 15:58














  • 2




    $begingroup$
    This is OEIS sequence A007526. The OEIS entry has much information including the limit you want to find.
    $endgroup$
    – Somos
    Jan 1 at 15:58








2




2




$begingroup$
This is OEIS sequence A007526. The OEIS entry has much information including the limit you want to find.
$endgroup$
– Somos
Jan 1 at 15:58




$begingroup$
This is OEIS sequence A007526. The OEIS entry has much information including the limit you want to find.
$endgroup$
– Somos
Jan 1 at 15:58










3 Answers
3






active

oldest

votes


















5












$begingroup$

Observe that
$$a_n=n+na_{n-1}$$
$$a_n=n+n(n-1)+n(n-1)a_{n-2}$$
and so on, giving us eventually
$$a_n=sum_{k=1}^nfrac{n!}{(k-1)!}$$
Now we have
$$lim_{ntoinfty}frac{1}{n!}(a_n+1)=lim_{ntoinfty}frac{a_n}{n!}$$
$$=lim_{ntoinfty}sum_{k=1}^nfrac{1}{(k-1)!}$$
$$=sum_{k=0}^inftyfrac{1}{k!}=e$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Not $e+1$. The sum for $e$ includes the term with $0!$ as denominator.
    $endgroup$
    – Oscar Lanzi
    Jan 1 at 15:29






  • 1




    $begingroup$
    @OscarLanzi oops! Fixed it.
    $endgroup$
    – Ben W
    Jan 1 at 15:32



















3












$begingroup$

Show by induction that



$a_1=1!/0!$



$a_2=2!/0!+2!/1!$



$a_3=3!/0!+3!/1!+3!/2!$



...



$color{blue}{a_n=n!/0!+n!/1!+n!/2!+...+n!/(n-1)!}$



Thereby the limit is $1/0!+1/1!+2/2!+...1/n!+...=e$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    From where you stuck you can proceed as
    begin{align}
    frac{a_n+1}{n!}&=frac{a_n}{n!}+frac{1}{n!}=frac{1}{(n-1)!}frac{a_n}{n}+frac{1}{n!}=frac{1}{(n-1)!}(a_{n-1}+1)+frac{1}{n!}=\
    &=frac{a_{n-1}}{(n-1)!}+frac{1}{(n-1)!}+frac{1}{n!}=ldots=
    1+frac{1}{1!}+ldots+frac{1}{n!}.
    end{align}






    share|cite|improve this answer









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Observe that
      $$a_n=n+na_{n-1}$$
      $$a_n=n+n(n-1)+n(n-1)a_{n-2}$$
      and so on, giving us eventually
      $$a_n=sum_{k=1}^nfrac{n!}{(k-1)!}$$
      Now we have
      $$lim_{ntoinfty}frac{1}{n!}(a_n+1)=lim_{ntoinfty}frac{a_n}{n!}$$
      $$=lim_{ntoinfty}sum_{k=1}^nfrac{1}{(k-1)!}$$
      $$=sum_{k=0}^inftyfrac{1}{k!}=e$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Not $e+1$. The sum for $e$ includes the term with $0!$ as denominator.
        $endgroup$
        – Oscar Lanzi
        Jan 1 at 15:29






      • 1




        $begingroup$
        @OscarLanzi oops! Fixed it.
        $endgroup$
        – Ben W
        Jan 1 at 15:32
















      5












      $begingroup$

      Observe that
      $$a_n=n+na_{n-1}$$
      $$a_n=n+n(n-1)+n(n-1)a_{n-2}$$
      and so on, giving us eventually
      $$a_n=sum_{k=1}^nfrac{n!}{(k-1)!}$$
      Now we have
      $$lim_{ntoinfty}frac{1}{n!}(a_n+1)=lim_{ntoinfty}frac{a_n}{n!}$$
      $$=lim_{ntoinfty}sum_{k=1}^nfrac{1}{(k-1)!}$$
      $$=sum_{k=0}^inftyfrac{1}{k!}=e$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Not $e+1$. The sum for $e$ includes the term with $0!$ as denominator.
        $endgroup$
        – Oscar Lanzi
        Jan 1 at 15:29






      • 1




        $begingroup$
        @OscarLanzi oops! Fixed it.
        $endgroup$
        – Ben W
        Jan 1 at 15:32














      5












      5








      5





      $begingroup$

      Observe that
      $$a_n=n+na_{n-1}$$
      $$a_n=n+n(n-1)+n(n-1)a_{n-2}$$
      and so on, giving us eventually
      $$a_n=sum_{k=1}^nfrac{n!}{(k-1)!}$$
      Now we have
      $$lim_{ntoinfty}frac{1}{n!}(a_n+1)=lim_{ntoinfty}frac{a_n}{n!}$$
      $$=lim_{ntoinfty}sum_{k=1}^nfrac{1}{(k-1)!}$$
      $$=sum_{k=0}^inftyfrac{1}{k!}=e$$






      share|cite|improve this answer











      $endgroup$



      Observe that
      $$a_n=n+na_{n-1}$$
      $$a_n=n+n(n-1)+n(n-1)a_{n-2}$$
      and so on, giving us eventually
      $$a_n=sum_{k=1}^nfrac{n!}{(k-1)!}$$
      Now we have
      $$lim_{ntoinfty}frac{1}{n!}(a_n+1)=lim_{ntoinfty}frac{a_n}{n!}$$
      $$=lim_{ntoinfty}sum_{k=1}^nfrac{1}{(k-1)!}$$
      $$=sum_{k=0}^inftyfrac{1}{k!}=e$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 1 at 15:32

























      answered Jan 1 at 15:27









      Ben WBen W

      2,734918




      2,734918












      • $begingroup$
        Not $e+1$. The sum for $e$ includes the term with $0!$ as denominator.
        $endgroup$
        – Oscar Lanzi
        Jan 1 at 15:29






      • 1




        $begingroup$
        @OscarLanzi oops! Fixed it.
        $endgroup$
        – Ben W
        Jan 1 at 15:32


















      • $begingroup$
        Not $e+1$. The sum for $e$ includes the term with $0!$ as denominator.
        $endgroup$
        – Oscar Lanzi
        Jan 1 at 15:29






      • 1




        $begingroup$
        @OscarLanzi oops! Fixed it.
        $endgroup$
        – Ben W
        Jan 1 at 15:32
















      $begingroup$
      Not $e+1$. The sum for $e$ includes the term with $0!$ as denominator.
      $endgroup$
      – Oscar Lanzi
      Jan 1 at 15:29




      $begingroup$
      Not $e+1$. The sum for $e$ includes the term with $0!$ as denominator.
      $endgroup$
      – Oscar Lanzi
      Jan 1 at 15:29




      1




      1




      $begingroup$
      @OscarLanzi oops! Fixed it.
      $endgroup$
      – Ben W
      Jan 1 at 15:32




      $begingroup$
      @OscarLanzi oops! Fixed it.
      $endgroup$
      – Ben W
      Jan 1 at 15:32











      3












      $begingroup$

      Show by induction that



      $a_1=1!/0!$



      $a_2=2!/0!+2!/1!$



      $a_3=3!/0!+3!/1!+3!/2!$



      ...



      $color{blue}{a_n=n!/0!+n!/1!+n!/2!+...+n!/(n-1)!}$



      Thereby the limit is $1/0!+1/1!+2/2!+...1/n!+...=e$.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Show by induction that



        $a_1=1!/0!$



        $a_2=2!/0!+2!/1!$



        $a_3=3!/0!+3!/1!+3!/2!$



        ...



        $color{blue}{a_n=n!/0!+n!/1!+n!/2!+...+n!/(n-1)!}$



        Thereby the limit is $1/0!+1/1!+2/2!+...1/n!+...=e$.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Show by induction that



          $a_1=1!/0!$



          $a_2=2!/0!+2!/1!$



          $a_3=3!/0!+3!/1!+3!/2!$



          ...



          $color{blue}{a_n=n!/0!+n!/1!+n!/2!+...+n!/(n-1)!}$



          Thereby the limit is $1/0!+1/1!+2/2!+...1/n!+...=e$.






          share|cite|improve this answer











          $endgroup$



          Show by induction that



          $a_1=1!/0!$



          $a_2=2!/0!+2!/1!$



          $a_3=3!/0!+3!/1!+3!/2!$



          ...



          $color{blue}{a_n=n!/0!+n!/1!+n!/2!+...+n!/(n-1)!}$



          Thereby the limit is $1/0!+1/1!+2/2!+...1/n!+...=e$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 1 at 22:53

























          answered Jan 1 at 15:28









          Oscar LanziOscar Lanzi

          13.7k12136




          13.7k12136























              1












              $begingroup$

              From where you stuck you can proceed as
              begin{align}
              frac{a_n+1}{n!}&=frac{a_n}{n!}+frac{1}{n!}=frac{1}{(n-1)!}frac{a_n}{n}+frac{1}{n!}=frac{1}{(n-1)!}(a_{n-1}+1)+frac{1}{n!}=\
              &=frac{a_{n-1}}{(n-1)!}+frac{1}{(n-1)!}+frac{1}{n!}=ldots=
              1+frac{1}{1!}+ldots+frac{1}{n!}.
              end{align}






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                From where you stuck you can proceed as
                begin{align}
                frac{a_n+1}{n!}&=frac{a_n}{n!}+frac{1}{n!}=frac{1}{(n-1)!}frac{a_n}{n}+frac{1}{n!}=frac{1}{(n-1)!}(a_{n-1}+1)+frac{1}{n!}=\
                &=frac{a_{n-1}}{(n-1)!}+frac{1}{(n-1)!}+frac{1}{n!}=ldots=
                1+frac{1}{1!}+ldots+frac{1}{n!}.
                end{align}






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  From where you stuck you can proceed as
                  begin{align}
                  frac{a_n+1}{n!}&=frac{a_n}{n!}+frac{1}{n!}=frac{1}{(n-1)!}frac{a_n}{n}+frac{1}{n!}=frac{1}{(n-1)!}(a_{n-1}+1)+frac{1}{n!}=\
                  &=frac{a_{n-1}}{(n-1)!}+frac{1}{(n-1)!}+frac{1}{n!}=ldots=
                  1+frac{1}{1!}+ldots+frac{1}{n!}.
                  end{align}






                  share|cite|improve this answer









                  $endgroup$



                  From where you stuck you can proceed as
                  begin{align}
                  frac{a_n+1}{n!}&=frac{a_n}{n!}+frac{1}{n!}=frac{1}{(n-1)!}frac{a_n}{n}+frac{1}{n!}=frac{1}{(n-1)!}(a_{n-1}+1)+frac{1}{n!}=\
                  &=frac{a_{n-1}}{(n-1)!}+frac{1}{(n-1)!}+frac{1}{n!}=ldots=
                  1+frac{1}{1!}+ldots+frac{1}{n!}.
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 1 at 15:49









                  A.Γ.A.Γ.

                  22.9k32656




                  22.9k32656






























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