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











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












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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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








    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.


















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














       

      draft saved


      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999327%2fhow-to-compute-1n1n1-e-sum-k-0n-frac-1kk%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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.






























                       

                      draft saved


                      draft discarded



















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999327%2fhow-to-compute-1n1n1-e-sum-k-0n-frac-1kk%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      AnyDesk - Fatal Program Failure

                      How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

                      QoS: MAC-Priority for clients behind a repeater