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











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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?










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  • 3




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






  • 1




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










  • 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?
    – JacksonFitzsimmons
    Nov 15 at 7:06















up vote
2
down vote

favorite
2












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?










share|cite|improve this question




















  • 3




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






  • 1




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










  • 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?
    – JacksonFitzsimmons
    Nov 15 at 7:06













up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





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?










share|cite|improve this question















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






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edited Nov 15 at 12:11









user21820

38k441148




38k441148










asked Nov 15 at 6:51









JacksonFitzsimmons

551212




551212








  • 3




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






  • 1




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










  • 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?
    – JacksonFitzsimmons
    Nov 15 at 7:06














  • 3




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






  • 1




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










  • 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?
    – JacksonFitzsimmons
    Nov 15 at 7:06








3




3




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




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




1




1




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




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












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?
– JacksonFitzsimmons
Nov 15 at 7:06




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?
– JacksonFitzsimmons
Nov 15 at 7:06










3 Answers
3






active

oldest

votes

















up vote
5
down vote



accepted










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





















  • Thanks, I never remember the squeeze theorem
    – JacksonFitzsimmons
    Nov 15 at 7:08


















up vote
4
down vote













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






























    up vote
    2
    down vote













    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








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    maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      5
      down vote



      accepted










      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





















      • Thanks, I never remember the squeeze theorem
        – JacksonFitzsimmons
        Nov 15 at 7:08















      up vote
      5
      down vote



      accepted










      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





















      • Thanks, I never remember the squeeze theorem
        – JacksonFitzsimmons
        Nov 15 at 7:08













      up vote
      5
      down vote



      accepted







      up vote
      5
      down vote



      accepted






      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












      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 at 7:05









      Tianlalu

      2,451632




      2,451632












      • Thanks, I never remember the squeeze theorem
        – JacksonFitzsimmons
        Nov 15 at 7:08


















      • Thanks, I never remember the squeeze theorem
        – JacksonFitzsimmons
        Nov 15 at 7:08
















      Thanks, I never remember the squeeze theorem
      – JacksonFitzsimmons
      Nov 15 at 7:08




      Thanks, I never remember the squeeze theorem
      – JacksonFitzsimmons
      Nov 15 at 7:08










      up vote
      4
      down vote













      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



























        up vote
        4
        down vote













        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

























          up vote
          4
          down vote










          up vote
          4
          down vote









          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














          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 at 7:31









          Tianlalu

          2,451632




          2,451632










          answered Nov 15 at 7:28









          J.G.

          18k11830




          18k11830






















              up vote
              2
              down vote













              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








              New contributor




              maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






















                up vote
                2
                down vote













                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








                New contributor




                maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.




















                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  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








                  New contributor




                  maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  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








                  New contributor




                  maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




                  maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered Nov 15 at 7:13









                  maxmilgram

                  3587




                  3587




                  New contributor




                  maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  New contributor





                  maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  maxmilgram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






























                       

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