How to compute $(-1)^{n+1}n!(1-esum_{k=0}^nfrac{(-1)^k}{k!})$?












2












$begingroup$


I was doing some work on the integral $int_0^1 x^ne^x dx$ and I eventually came to this expression in terms of n $$int_0^1 x^ne^x dx=(-1)^{n+1}n!biggl(1-esum_{k=0}^nfrac{(-1)^k}{k!}biggr)$$ Now my question is this, how do I compute the limit of this expression as $n$ goes to $infty$? I recognize that $$lim_{ntoinfty}esum_{k=0}^{n}frac{(-1)^k}{k!}=1$$ per the series expression of $e^x$ at $-1$. This gives me an indeterminate form, but I can't apply L'Hopital's rule because I can't differentiate $n!$ (or $1/n!$ as the case would be after re-arranging), and I also can't think of a variable substitution that would help me here. Common sense and wolfram definitely imply that the limit is in fact $0$, but how can I show this?






calculus integration limits definite-integrals exponential-function







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $n!=Gamma(n+1)$ is differentiable.
    $endgroup$
    – J.G.
    Nov 15 '18 at 7:01






  • 1




    $begingroup$
    perhaps some properties of the gamma function/complex analysis would be useful
    $endgroup$
    – rubikscube09
    Nov 15 '18 at 7:04










  • $begingroup$
    I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
    $endgroup$
    – JacksonFitzsimmons
    Nov 15 '18 at 7:06
















2












$begingroup$


I was doing some work on the integral $int_0^1 x^ne^x dx$ and I eventually came to this expression in terms of n $$int_0^1 x^ne^x dx=(-1)^{n+1}n!biggl(1-esum_{k=0}^nfrac{(-1)^k}{k!}biggr)$$ Now my question is this, how do I compute the limit of this expression as $n$ goes to $infty$? I recognize that $$lim_{ntoinfty}esum_{k=0}^{n}frac{(-1)^k}{k!}=1$$ per the series expression of $e^x$ at $-1$. This gives me an indeterminate form, but I can't apply L'Hopital's rule because I can't differentiate $n!$ (or $1/n!$ as the case would be after re-arranging), and I also can't think of a variable substitution that would help me here. Common sense and wolfram definitely imply that the limit is in fact $0$, but how can I show this?






calculus integration limits definite-integrals exponential-function







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $n!=Gamma(n+1)$ is differentiable.
    $endgroup$
    – J.G.
    Nov 15 '18 at 7:01






  • 1




    $begingroup$
    perhaps some properties of the gamma function/complex analysis would be useful
    $endgroup$
    – rubikscube09
    Nov 15 '18 at 7:04










  • $begingroup$
    I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
    $endgroup$
    – JacksonFitzsimmons
    Nov 15 '18 at 7:06














2












2








2


2



$begingroup$


I was doing some work on the integral $int_0^1 x^ne^x dx$ and I eventually came to this expression in terms of n $$int_0^1 x^ne^x dx=(-1)^{n+1}n!biggl(1-esum_{k=0}^nfrac{(-1)^k}{k!}biggr)$$ Now my question is this, how do I compute the limit of this expression as $n$ goes to $infty$? I recognize that $$lim_{ntoinfty}esum_{k=0}^{n}frac{(-1)^k}{k!}=1$$ per the series expression of $e^x$ at $-1$. This gives me an indeterminate form, but I can't apply L'Hopital's rule because I can't differentiate $n!$ (or $1/n!$ as the case would be after re-arranging), and I also can't think of a variable substitution that would help me here. Common sense and wolfram definitely imply that the limit is in fact $0$, but how can I show this?






calculus integration limits definite-integrals exponential-function







share|cite|improve this question











$endgroup$




I was doing some work on the integral $int_0^1 x^ne^x dx$ and I eventually came to this expression in terms of n $$int_0^1 x^ne^x dx=(-1)^{n+1}n!biggl(1-esum_{k=0}^nfrac{(-1)^k}{k!}biggr)$$ Now my question is this, how do I compute the limit of this expression as $n$ goes to $infty$? I recognize that $$lim_{ntoinfty}esum_{k=0}^{n}frac{(-1)^k}{k!}=1$$ per the series expression of $e^x$ at $-1$. This gives me an indeterminate form, but I can't apply L'Hopital's rule because I can't differentiate $n!$ (or $1/n!$ as the case would be after re-arranging), and I also can't think of a variable substitution that would help me here. Common sense and wolfram definitely imply that the limit is in fact $0$, but how can I show this?






calculus integration limits definite-integrals exponential-function




calculus integration limits definite-integrals exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 15 '18 at 12:11









user21820

39.3k543155




39.3k543155










asked Nov 15 '18 at 6:51









JacksonFitzsimmonsJacksonFitzsimmons

551212




551212








  • 3




    $begingroup$
    $n!=Gamma(n+1)$ is differentiable.
    $endgroup$
    – J.G.
    Nov 15 '18 at 7:01






  • 1




    $begingroup$
    perhaps some properties of the gamma function/complex analysis would be useful
    $endgroup$
    – rubikscube09
    Nov 15 '18 at 7:04










  • $begingroup$
    I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
    $endgroup$
    – JacksonFitzsimmons
    Nov 15 '18 at 7:06














  • 3




    $begingroup$
    $n!=Gamma(n+1)$ is differentiable.
    $endgroup$
    – J.G.
    Nov 15 '18 at 7:01






  • 1




    $begingroup$
    perhaps some properties of the gamma function/complex analysis would be useful
    $endgroup$
    – rubikscube09
    Nov 15 '18 at 7:04










  • $begingroup$
    I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
    $endgroup$
    – JacksonFitzsimmons
    Nov 15 '18 at 7:06








3




3




$begingroup$
$n!=Gamma(n+1)$ is differentiable.
$endgroup$
– J.G.
Nov 15 '18 at 7:01




$begingroup$
$n!=Gamma(n+1)$ is differentiable.
$endgroup$
– J.G.
Nov 15 '18 at 7:01




1




1




$begingroup$
perhaps some properties of the gamma function/complex analysis would be useful
$endgroup$
– rubikscube09
Nov 15 '18 at 7:04




$begingroup$
perhaps some properties of the gamma function/complex analysis would be useful
$endgroup$
– rubikscube09
Nov 15 '18 at 7:04












$begingroup$
I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
$endgroup$
– JacksonFitzsimmons
Nov 15 '18 at 7:06




$begingroup$
I'm not very familiar with the gamma function, is there any other way? And if not, how could I use the gamma function to do this? Also, would it be sufficient to show that the integrand approaches $0$ on the interval?
$endgroup$
– JacksonFitzsimmons
Nov 15 '18 at 7:06










3 Answers
3






active

oldest

votes


















5












$begingroup$

When $xin [0,1]$, $e^xle e$, by first mean value theorem for definite integrals,
begin{align*}
I_n&lemax_{xin[0,1]}{e^x}int_0^1x^{n}~mathrm dx\
&=frac{e}{n+1}to0.
end{align*}
Since $I_n>0$, the limit is $0$ by squeeze theorem.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, I never remember the squeeze theorem
    $endgroup$
    – JacksonFitzsimmons
    Nov 15 '18 at 7:08



















4












$begingroup$

Not as slick as Tianlalu's method, but for $yin (0,,1)$ we have
$$
lim_{ntoinfty}int_0^y x^n e^x dx le
lim_{ntoinfty}y^nint_0^y e^x dx=0.
$$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:



    $$exp(-1)=sum_{k=0}^nfrac{(-1)^k}{k!}+frac{e^xi}{(n+1)!}(-1)^{n+1}$$



    for some $-1<xi<0$. If you estimate $e^xi $ and replace the sum you get convergence to $0$.






    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












      $begingroup$

      When $xin [0,1]$, $e^xle e$, by first mean value theorem for definite integrals,
      begin{align*}
      I_n&lemax_{xin[0,1]}{e^x}int_0^1x^{n}~mathrm dx\
      &=frac{e}{n+1}to0.
      end{align*}
      Since $I_n>0$, the limit is $0$ by squeeze theorem.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks, I never remember the squeeze theorem
        $endgroup$
        – JacksonFitzsimmons
        Nov 15 '18 at 7:08
















      5












      $begingroup$

      When $xin [0,1]$, $e^xle e$, by first mean value theorem for definite integrals,
      begin{align*}
      I_n&lemax_{xin[0,1]}{e^x}int_0^1x^{n}~mathrm dx\
      &=frac{e}{n+1}to0.
      end{align*}
      Since $I_n>0$, the limit is $0$ by squeeze theorem.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks, I never remember the squeeze theorem
        $endgroup$
        – JacksonFitzsimmons
        Nov 15 '18 at 7:08














      5












      5








      5





      $begingroup$

      When $xin [0,1]$, $e^xle e$, by first mean value theorem for definite integrals,
      begin{align*}
      I_n&lemax_{xin[0,1]}{e^x}int_0^1x^{n}~mathrm dx\
      &=frac{e}{n+1}to0.
      end{align*}
      Since $I_n>0$, the limit is $0$ by squeeze theorem.






      share|cite|improve this answer









      $endgroup$



      When $xin [0,1]$, $e^xle e$, by first mean value theorem for definite integrals,
      begin{align*}
      I_n&lemax_{xin[0,1]}{e^x}int_0^1x^{n}~mathrm dx\
      &=frac{e}{n+1}to0.
      end{align*}
      Since $I_n>0$, the limit is $0$ by squeeze theorem.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 15 '18 at 7:05









      TianlaluTianlalu

      3,08421138




      3,08421138












      • $begingroup$
        Thanks, I never remember the squeeze theorem
        $endgroup$
        – JacksonFitzsimmons
        Nov 15 '18 at 7:08


















      • $begingroup$
        Thanks, I never remember the squeeze theorem
        $endgroup$
        – JacksonFitzsimmons
        Nov 15 '18 at 7:08
















      $begingroup$
      Thanks, I never remember the squeeze theorem
      $endgroup$
      – JacksonFitzsimmons
      Nov 15 '18 at 7:08




      $begingroup$
      Thanks, I never remember the squeeze theorem
      $endgroup$
      – JacksonFitzsimmons
      Nov 15 '18 at 7:08











      4












      $begingroup$

      Not as slick as Tianlalu's method, but for $yin (0,,1)$ we have
      $$
      lim_{ntoinfty}int_0^y x^n e^x dx le
      lim_{ntoinfty}y^nint_0^y e^x dx=0.
      $$






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        Not as slick as Tianlalu's method, but for $yin (0,,1)$ we have
        $$
        lim_{ntoinfty}int_0^y x^n e^x dx le
        lim_{ntoinfty}y^nint_0^y e^x dx=0.
        $$






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          Not as slick as Tianlalu's method, but for $yin (0,,1)$ we have
          $$
          lim_{ntoinfty}int_0^y x^n e^x dx le
          lim_{ntoinfty}y^nint_0^y e^x dx=0.
          $$






          share|cite|improve this answer











          $endgroup$



          Not as slick as Tianlalu's method, but for $yin (0,,1)$ we have
          $$
          lim_{ntoinfty}int_0^y x^n e^x dx le
          lim_{ntoinfty}y^nint_0^y e^x dx=0.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 15 '18 at 7:31









          Tianlalu

          3,08421138




          3,08421138










          answered Nov 15 '18 at 7:28









          J.G.J.G.

          28.8k22845




          28.8k22845























              2












              $begingroup$

              I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:



              $$exp(-1)=sum_{k=0}^nfrac{(-1)^k}{k!}+frac{e^xi}{(n+1)!}(-1)^{n+1}$$



              for some $-1<xi<0$. If you estimate $e^xi $ and replace the sum you get convergence to $0$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:



                $$exp(-1)=sum_{k=0}^nfrac{(-1)^k}{k!}+frac{e^xi}{(n+1)!}(-1)^{n+1}$$



                for some $-1<xi<0$. If you estimate $e^xi $ and replace the sum you get convergence to $0$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:



                  $$exp(-1)=sum_{k=0}^nfrac{(-1)^k}{k!}+frac{e^xi}{(n+1)!}(-1)^{n+1}$$



                  for some $-1<xi<0$. If you estimate $e^xi $ and replace the sum you get convergence to $0$.






                  share|cite|improve this answer









                  $endgroup$



                  I don't know what you are 'allowed' to use but you could use the lagrange remainer for the exponential function:



                  $$exp(-1)=sum_{k=0}^nfrac{(-1)^k}{k!}+frac{e^xi}{(n+1)!}(-1)^{n+1}$$



                  for some $-1<xi<0$. If you estimate $e^xi $ and replace the sum you get convergence to $0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 15 '18 at 7:13









                  maxmilgrammaxmilgram

                  7007




                  7007






























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