limit of $left(1+frac{1}{n!}right)^n$












4












$begingroup$


I'm having trouble resolving the following limit: $$ lim_{n to infty} left(1+frac{1}{n!}right)^n $$ Intuituvely the limit is equal to 1, but the exercises requires me to resolve via calculation and I have no idea how I can accomplish this.

Can someone please explain it to me?






limits factorial







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    I'm having trouble resolving the following limit: $$ lim_{n to infty} left(1+frac{1}{n!}right)^n $$ Intuituvely the limit is equal to 1, but the exercises requires me to resolve via calculation and I have no idea how I can accomplish this.

    Can someone please explain it to me?






    limits factorial







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      I'm having trouble resolving the following limit: $$ lim_{n to infty} left(1+frac{1}{n!}right)^n $$ Intuituvely the limit is equal to 1, but the exercises requires me to resolve via calculation and I have no idea how I can accomplish this.

      Can someone please explain it to me?






      limits factorial







      share|cite|improve this question











      $endgroup$




      I'm having trouble resolving the following limit: $$ lim_{n to infty} left(1+frac{1}{n!}right)^n $$ Intuituvely the limit is equal to 1, but the exercises requires me to resolve via calculation and I have no idea how I can accomplish this.

      Can someone please explain it to me?






      limits factorial




      limits factorial






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 29 '18 at 16:14







      Sceptwo

















      asked Nov 29 '18 at 15:51









      SceptwoSceptwo

      236




      236






















          5 Answers
          5






          active

          oldest

          votes


















          6












          $begingroup$

          HINT



          We have



          $$lim_{n to infty} left(1+frac{1}{n!}right)^n=lim_{n to infty} left[left(1+frac{1}{n!}right)^{n!}right]^{frac1{(n-1)!}}$$



          then refer to standard limit for $e$. How can we conclude form here?






          share|cite|improve this answer











          $endgroup$





















            9












            $begingroup$


            I thought it might be instructive to present a solution that relies on elementary inequalities and the squeeze theorem only. To that end, we now proceed.





            Using the inequality $1+xle frac1{1-x}$, for $x<1$, along with Bernoulli's Inequality, we have for $nge 2$



            $$1le left(1+frac1{n!}right)^nle frac1{left(1-frac1{n!}right)^n}le frac1{1-frac1{(n-1)!}}$$



            whereby application of the squeeze theorem yields the coveted limit.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
              $endgroup$
              – gimusi
              Nov 29 '18 at 17:00






            • 1




              $begingroup$
              Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
              $endgroup$
              – Mark Viola
              Nov 29 '18 at 17:04












            • $begingroup$
              Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
              $endgroup$
              – gimusi
              Nov 29 '18 at 17:07






            • 1




              $begingroup$
              Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
              $endgroup$
              – Mark Viola
              Nov 29 '18 at 17:09





















            5












            $begingroup$

            It suffices to compute the limit of the logarithm and show that it equals zero. Namely note that
            $$
            nlogleft(1+frac{1}{n!}right)=ntimes frac{1}{n!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}
            $$
            Now let $nto infty$. Using the fact that
            $$
            lim_{uto 0}frac{log(1+u)-log1}{u-0}=1
            $$
            by definition of the derivative we get that
            $$
            frac{1}{(n-1)!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}to0times 1=0
            $$
            as $nto infty$.






            share|cite|improve this answer









            $endgroup$





















              3












              $begingroup$

              Note that
              $$
              frac{1}{n!}<frac{1}{n(n-1)}<frac{1}{(n-1)^2}
              $$
              Now prove that
              $$
              lim_{tto0}frac{log(1+t^2)}{t}=0
              $$
              and observe that
              begin{align}
              left(1+frac{1}{(n-1)^2}right)^n&=
              left(1+frac{1}{(n-1)^2}right)left(1+frac{1}{(n-1)^2}right)^{n-1}
              \[6px]
              &=left(1+frac{1}{(n-1)^2}right)expleft(fleft(frac{1}{n-1}right)right)
              end{align}
              Finally note that
              $$
              1leleft(1+frac{1}{n!}right)^n
              $$






              share|cite|improve this answer











              $endgroup$





















                3












                $begingroup$

                Since Mark Viola has already used the elementary approach of Bernoulli inequality here is one more approach via binomial theorem.



                We use the following lemma of Thomas Andrews (proved using binomial theorem):




                Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$.




                Now use $a_n=1+(1/n!)$ and check that $n(a_n-1)=1/(n-1)!to 0$. The desired limit is $1$ by the above lemma.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  (+1) Nice use of that "lemma."
                  $endgroup$
                  – Mark Viola
                  Nov 30 '18 at 15:23











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                5 Answers
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                active

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                5 Answers
                5






                active

                oldest

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                active

                oldest

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                active

                oldest

                votes









                6












                $begingroup$

                HINT



                We have



                $$lim_{n to infty} left(1+frac{1}{n!}right)^n=lim_{n to infty} left[left(1+frac{1}{n!}right)^{n!}right]^{frac1{(n-1)!}}$$



                then refer to standard limit for $e$. How can we conclude form here?






                share|cite|improve this answer











                $endgroup$


















                  6












                  $begingroup$

                  HINT



                  We have



                  $$lim_{n to infty} left(1+frac{1}{n!}right)^n=lim_{n to infty} left[left(1+frac{1}{n!}right)^{n!}right]^{frac1{(n-1)!}}$$



                  then refer to standard limit for $e$. How can we conclude form here?






                  share|cite|improve this answer











                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$

                    HINT



                    We have



                    $$lim_{n to infty} left(1+frac{1}{n!}right)^n=lim_{n to infty} left[left(1+frac{1}{n!}right)^{n!}right]^{frac1{(n-1)!}}$$



                    then refer to standard limit for $e$. How can we conclude form here?






                    share|cite|improve this answer











                    $endgroup$



                    HINT



                    We have



                    $$lim_{n to infty} left(1+frac{1}{n!}right)^n=lim_{n to infty} left[left(1+frac{1}{n!}right)^{n!}right]^{frac1{(n-1)!}}$$



                    then refer to standard limit for $e$. How can we conclude form here?







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 29 '18 at 15:54

























                    answered Nov 29 '18 at 15:53









                    gimusigimusi

                    1




                    1























                        9












                        $begingroup$


                        I thought it might be instructive to present a solution that relies on elementary inequalities and the squeeze theorem only. To that end, we now proceed.





                        Using the inequality $1+xle frac1{1-x}$, for $x<1$, along with Bernoulli's Inequality, we have for $nge 2$



                        $$1le left(1+frac1{n!}right)^nle frac1{left(1-frac1{n!}right)^n}le frac1{1-frac1{(n-1)!}}$$



                        whereby application of the squeeze theorem yields the coveted limit.






                        share|cite|improve this answer











                        $endgroup$









                        • 1




                          $begingroup$
                          It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
                          $endgroup$
                          – gimusi
                          Nov 29 '18 at 17:00






                        • 1




                          $begingroup$
                          Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
                          $endgroup$
                          – Mark Viola
                          Nov 29 '18 at 17:04












                        • $begingroup$
                          Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
                          $endgroup$
                          – gimusi
                          Nov 29 '18 at 17:07






                        • 1




                          $begingroup$
                          Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
                          $endgroup$
                          – Mark Viola
                          Nov 29 '18 at 17:09


















                        9












                        $begingroup$


                        I thought it might be instructive to present a solution that relies on elementary inequalities and the squeeze theorem only. To that end, we now proceed.





                        Using the inequality $1+xle frac1{1-x}$, for $x<1$, along with Bernoulli's Inequality, we have for $nge 2$



                        $$1le left(1+frac1{n!}right)^nle frac1{left(1-frac1{n!}right)^n}le frac1{1-frac1{(n-1)!}}$$



                        whereby application of the squeeze theorem yields the coveted limit.






                        share|cite|improve this answer











                        $endgroup$









                        • 1




                          $begingroup$
                          It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
                          $endgroup$
                          – gimusi
                          Nov 29 '18 at 17:00






                        • 1




                          $begingroup$
                          Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
                          $endgroup$
                          – Mark Viola
                          Nov 29 '18 at 17:04












                        • $begingroup$
                          Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
                          $endgroup$
                          – gimusi
                          Nov 29 '18 at 17:07






                        • 1




                          $begingroup$
                          Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
                          $endgroup$
                          – Mark Viola
                          Nov 29 '18 at 17:09
















                        9












                        9








                        9





                        $begingroup$


                        I thought it might be instructive to present a solution that relies on elementary inequalities and the squeeze theorem only. To that end, we now proceed.





                        Using the inequality $1+xle frac1{1-x}$, for $x<1$, along with Bernoulli's Inequality, we have for $nge 2$



                        $$1le left(1+frac1{n!}right)^nle frac1{left(1-frac1{n!}right)^n}le frac1{1-frac1{(n-1)!}}$$



                        whereby application of the squeeze theorem yields the coveted limit.






                        share|cite|improve this answer











                        $endgroup$




                        I thought it might be instructive to present a solution that relies on elementary inequalities and the squeeze theorem only. To that end, we now proceed.





                        Using the inequality $1+xle frac1{1-x}$, for $x<1$, along with Bernoulli's Inequality, we have for $nge 2$



                        $$1le left(1+frac1{n!}right)^nle frac1{left(1-frac1{n!}right)^n}le frac1{1-frac1{(n-1)!}}$$



                        whereby application of the squeeze theorem yields the coveted limit.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Nov 29 '18 at 17:07

























                        answered Nov 29 '18 at 16:00









                        Mark ViolaMark Viola

                        131k1275171




                        131k1275171








                        • 1




                          $begingroup$
                          It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
                          $endgroup$
                          – gimusi
                          Nov 29 '18 at 17:00






                        • 1




                          $begingroup$
                          Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
                          $endgroup$
                          – Mark Viola
                          Nov 29 '18 at 17:04












                        • $begingroup$
                          Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
                          $endgroup$
                          – gimusi
                          Nov 29 '18 at 17:07






                        • 1




                          $begingroup$
                          Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
                          $endgroup$
                          – Mark Viola
                          Nov 29 '18 at 17:09
















                        • 1




                          $begingroup$
                          It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
                          $endgroup$
                          – gimusi
                          Nov 29 '18 at 17:00






                        • 1




                          $begingroup$
                          Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
                          $endgroup$
                          – Mark Viola
                          Nov 29 '18 at 17:04












                        • $begingroup$
                          Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
                          $endgroup$
                          – gimusi
                          Nov 29 '18 at 17:07






                        • 1




                          $begingroup$
                          Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
                          $endgroup$
                          – Mark Viola
                          Nov 29 '18 at 17:09










                        1




                        1




                        $begingroup$
                        It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
                        $endgroup$
                        – gimusi
                        Nov 29 '18 at 17:00




                        $begingroup$
                        It is a very elegant way which only make use of simple results, the only issue is to have great the idea to start with the inequality $1+xle frac1{1-x}$ as $(x<1)$ and take is as a given.
                        $endgroup$
                        – gimusi
                        Nov 29 '18 at 17:00




                        1




                        1




                        $begingroup$
                        Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
                        $endgroup$
                        – Mark Viola
                        Nov 29 '18 at 17:04






                        $begingroup$
                        Well, for all $x$, $x^2ge 0$. Hence, $(1+x)(1-x)=1-x^2le 1$. For $x<1$, $1+xle frac1{1-x}$. So, that is a fairly standard "trick."
                        $endgroup$
                        – Mark Viola
                        Nov 29 '18 at 17:04














                        $begingroup$
                        Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
                        $endgroup$
                        – gimusi
                        Nov 29 '18 at 17:07




                        $begingroup$
                        Yes indeed the proof is simple maybe it is less obviuos have the idea to strat with it. Of course it is a good trick to keep in mind :)
                        $endgroup$
                        – gimusi
                        Nov 29 '18 at 17:07




                        1




                        1




                        $begingroup$
                        Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
                        $endgroup$
                        – Mark Viola
                        Nov 29 '18 at 17:09






                        $begingroup$
                        Yes, this "trick" can be used, for example, to show that $exp(x)exp(y)=exp(x+y)$ where we use the limit characterization $exp(x)=lim_{ntoinfty}left(1+frac xnright)^n$ of the exponential function.
                        $endgroup$
                        – Mark Viola
                        Nov 29 '18 at 17:09













                        5












                        $begingroup$

                        It suffices to compute the limit of the logarithm and show that it equals zero. Namely note that
                        $$
                        nlogleft(1+frac{1}{n!}right)=ntimes frac{1}{n!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}
                        $$
                        Now let $nto infty$. Using the fact that
                        $$
                        lim_{uto 0}frac{log(1+u)-log1}{u-0}=1
                        $$
                        by definition of the derivative we get that
                        $$
                        frac{1}{(n-1)!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}to0times 1=0
                        $$
                        as $nto infty$.






                        share|cite|improve this answer









                        $endgroup$


















                          5












                          $begingroup$

                          It suffices to compute the limit of the logarithm and show that it equals zero. Namely note that
                          $$
                          nlogleft(1+frac{1}{n!}right)=ntimes frac{1}{n!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}
                          $$
                          Now let $nto infty$. Using the fact that
                          $$
                          lim_{uto 0}frac{log(1+u)-log1}{u-0}=1
                          $$
                          by definition of the derivative we get that
                          $$
                          frac{1}{(n-1)!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}to0times 1=0
                          $$
                          as $nto infty$.






                          share|cite|improve this answer









                          $endgroup$
















                            5












                            5








                            5





                            $begingroup$

                            It suffices to compute the limit of the logarithm and show that it equals zero. Namely note that
                            $$
                            nlogleft(1+frac{1}{n!}right)=ntimes frac{1}{n!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}
                            $$
                            Now let $nto infty$. Using the fact that
                            $$
                            lim_{uto 0}frac{log(1+u)-log1}{u-0}=1
                            $$
                            by definition of the derivative we get that
                            $$
                            frac{1}{(n-1)!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}to0times 1=0
                            $$
                            as $nto infty$.






                            share|cite|improve this answer









                            $endgroup$



                            It suffices to compute the limit of the logarithm and show that it equals zero. Namely note that
                            $$
                            nlogleft(1+frac{1}{n!}right)=ntimes frac{1}{n!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}
                            $$
                            Now let $nto infty$. Using the fact that
                            $$
                            lim_{uto 0}frac{log(1+u)-log1}{u-0}=1
                            $$
                            by definition of the derivative we get that
                            $$
                            frac{1}{(n-1)!}times frac{logleft(1+frac{1}{n!}right)}{1/n!}to0times 1=0
                            $$
                            as $nto infty$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 29 '18 at 16:01









                            Foobaz JohnFoobaz John

                            21.6k41352




                            21.6k41352























                                3












                                $begingroup$

                                Note that
                                $$
                                frac{1}{n!}<frac{1}{n(n-1)}<frac{1}{(n-1)^2}
                                $$
                                Now prove that
                                $$
                                lim_{tto0}frac{log(1+t^2)}{t}=0
                                $$
                                and observe that
                                begin{align}
                                left(1+frac{1}{(n-1)^2}right)^n&=
                                left(1+frac{1}{(n-1)^2}right)left(1+frac{1}{(n-1)^2}right)^{n-1}
                                \[6px]
                                &=left(1+frac{1}{(n-1)^2}right)expleft(fleft(frac{1}{n-1}right)right)
                                end{align}
                                Finally note that
                                $$
                                1leleft(1+frac{1}{n!}right)^n
                                $$






                                share|cite|improve this answer











                                $endgroup$


















                                  3












                                  $begingroup$

                                  Note that
                                  $$
                                  frac{1}{n!}<frac{1}{n(n-1)}<frac{1}{(n-1)^2}
                                  $$
                                  Now prove that
                                  $$
                                  lim_{tto0}frac{log(1+t^2)}{t}=0
                                  $$
                                  and observe that
                                  begin{align}
                                  left(1+frac{1}{(n-1)^2}right)^n&=
                                  left(1+frac{1}{(n-1)^2}right)left(1+frac{1}{(n-1)^2}right)^{n-1}
                                  \[6px]
                                  &=left(1+frac{1}{(n-1)^2}right)expleft(fleft(frac{1}{n-1}right)right)
                                  end{align}
                                  Finally note that
                                  $$
                                  1leleft(1+frac{1}{n!}right)^n
                                  $$






                                  share|cite|improve this answer











                                  $endgroup$
















                                    3












                                    3








                                    3





                                    $begingroup$

                                    Note that
                                    $$
                                    frac{1}{n!}<frac{1}{n(n-1)}<frac{1}{(n-1)^2}
                                    $$
                                    Now prove that
                                    $$
                                    lim_{tto0}frac{log(1+t^2)}{t}=0
                                    $$
                                    and observe that
                                    begin{align}
                                    left(1+frac{1}{(n-1)^2}right)^n&=
                                    left(1+frac{1}{(n-1)^2}right)left(1+frac{1}{(n-1)^2}right)^{n-1}
                                    \[6px]
                                    &=left(1+frac{1}{(n-1)^2}right)expleft(fleft(frac{1}{n-1}right)right)
                                    end{align}
                                    Finally note that
                                    $$
                                    1leleft(1+frac{1}{n!}right)^n
                                    $$






                                    share|cite|improve this answer











                                    $endgroup$



                                    Note that
                                    $$
                                    frac{1}{n!}<frac{1}{n(n-1)}<frac{1}{(n-1)^2}
                                    $$
                                    Now prove that
                                    $$
                                    lim_{tto0}frac{log(1+t^2)}{t}=0
                                    $$
                                    and observe that
                                    begin{align}
                                    left(1+frac{1}{(n-1)^2}right)^n&=
                                    left(1+frac{1}{(n-1)^2}right)left(1+frac{1}{(n-1)^2}right)^{n-1}
                                    \[6px]
                                    &=left(1+frac{1}{(n-1)^2}right)expleft(fleft(frac{1}{n-1}right)right)
                                    end{align}
                                    Finally note that
                                    $$
                                    1leleft(1+frac{1}{n!}right)^n
                                    $$







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Nov 29 '18 at 16:45

























                                    answered Nov 29 '18 at 16:05









                                    egregegreg

                                    180k1485202




                                    180k1485202























                                        3












                                        $begingroup$

                                        Since Mark Viola has already used the elementary approach of Bernoulli inequality here is one more approach via binomial theorem.



                                        We use the following lemma of Thomas Andrews (proved using binomial theorem):




                                        Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$.




                                        Now use $a_n=1+(1/n!)$ and check that $n(a_n-1)=1/(n-1)!to 0$. The desired limit is $1$ by the above lemma.






                                        share|cite|improve this answer









                                        $endgroup$













                                        • $begingroup$
                                          (+1) Nice use of that "lemma."
                                          $endgroup$
                                          – Mark Viola
                                          Nov 30 '18 at 15:23
















                                        3












                                        $begingroup$

                                        Since Mark Viola has already used the elementary approach of Bernoulli inequality here is one more approach via binomial theorem.



                                        We use the following lemma of Thomas Andrews (proved using binomial theorem):




                                        Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$.




                                        Now use $a_n=1+(1/n!)$ and check that $n(a_n-1)=1/(n-1)!to 0$. The desired limit is $1$ by the above lemma.






                                        share|cite|improve this answer









                                        $endgroup$













                                        • $begingroup$
                                          (+1) Nice use of that "lemma."
                                          $endgroup$
                                          – Mark Viola
                                          Nov 30 '18 at 15:23














                                        3












                                        3








                                        3





                                        $begingroup$

                                        Since Mark Viola has already used the elementary approach of Bernoulli inequality here is one more approach via binomial theorem.



                                        We use the following lemma of Thomas Andrews (proved using binomial theorem):




                                        Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$.




                                        Now use $a_n=1+(1/n!)$ and check that $n(a_n-1)=1/(n-1)!to 0$. The desired limit is $1$ by the above lemma.






                                        share|cite|improve this answer









                                        $endgroup$



                                        Since Mark Viola has already used the elementary approach of Bernoulli inequality here is one more approach via binomial theorem.



                                        We use the following lemma of Thomas Andrews (proved using binomial theorem):




                                        Lemma: If ${a_n} $ is a sequence of real or complex terms such that $n(a_n-1)to 0$ then $a_n^nto 1$.




                                        Now use $a_n=1+(1/n!)$ and check that $n(a_n-1)=1/(n-1)!to 0$. The desired limit is $1$ by the above lemma.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Nov 30 '18 at 2:52









                                        Paramanand SinghParamanand Singh

                                        49.4k555162




                                        49.4k555162












                                        • $begingroup$
                                          (+1) Nice use of that "lemma."
                                          $endgroup$
                                          – Mark Viola
                                          Nov 30 '18 at 15:23


















                                        • $begingroup$
                                          (+1) Nice use of that "lemma."
                                          $endgroup$
                                          – Mark Viola
                                          Nov 30 '18 at 15:23
















                                        $begingroup$
                                        (+1) Nice use of that "lemma."
                                        $endgroup$
                                        – Mark Viola
                                        Nov 30 '18 at 15:23




                                        $begingroup$
                                        (+1) Nice use of that "lemma."
                                        $endgroup$
                                        – Mark Viola
                                        Nov 30 '18 at 15:23


















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