Euler's Number Identity












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In pre-calc, we learned that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(1+frac{1}{n}right)^{n}$$
I expanded the right hand side using Binomial Theorem. Here are my steps:
$$(x+y)^{n} = sum_{k=0}^{n} {n choose k} cdot x^{n-k}cdot y^{k}$$
$$x = 1, spacespacespacespacespacespacespace y = frac{1}{n}$$
Plugging that in gives:
$$lim_{ntoinfty}left(1+frac{1}{n}right)^{n} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot frac{1}{n^k}right)$$
That means that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot y^{k}right)$$
Could anyone help me finish this? Ideally, an explanation accessible to a middle school pre-calc student. Thanks.






algebra-precalculus







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    $begingroup$
    I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
    $endgroup$
    – Doug M
    Dec 12 '18 at 18:59
















2












$begingroup$


In pre-calc, we learned that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(1+frac{1}{n}right)^{n}$$
I expanded the right hand side using Binomial Theorem. Here are my steps:
$$(x+y)^{n} = sum_{k=0}^{n} {n choose k} cdot x^{n-k}cdot y^{k}$$
$$x = 1, spacespacespacespacespacespacespace y = frac{1}{n}$$
Plugging that in gives:
$$lim_{ntoinfty}left(1+frac{1}{n}right)^{n} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot frac{1}{n^k}right)$$
That means that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot y^{k}right)$$
Could anyone help me finish this? Ideally, an explanation accessible to a middle school pre-calc student. Thanks.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
    $endgroup$
    – Doug M
    Dec 12 '18 at 18:59














2












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$begingroup$


In pre-calc, we learned that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(1+frac{1}{n}right)^{n}$$
I expanded the right hand side using Binomial Theorem. Here are my steps:
$$(x+y)^{n} = sum_{k=0}^{n} {n choose k} cdot x^{n-k}cdot y^{k}$$
$$x = 1, spacespacespacespacespacespacespace y = frac{1}{n}$$
Plugging that in gives:
$$lim_{ntoinfty}left(1+frac{1}{n}right)^{n} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot frac{1}{n^k}right)$$
That means that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot y^{k}right)$$
Could anyone help me finish this? Ideally, an explanation accessible to a middle school pre-calc student. Thanks.






algebra-precalculus







share|cite|improve this question











$endgroup$




In pre-calc, we learned that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(1+frac{1}{n}right)^{n}$$
I expanded the right hand side using Binomial Theorem. Here are my steps:
$$(x+y)^{n} = sum_{k=0}^{n} {n choose k} cdot x^{n-k}cdot y^{k}$$
$$x = 1, spacespacespacespacespacespacespace y = frac{1}{n}$$
Plugging that in gives:
$$lim_{ntoinfty}left(1+frac{1}{n}right)^{n} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot frac{1}{n^k}right)$$
That means that
$$sum_{k=0}^{infty} frac{1}{k!} = lim_{ntoinfty}left(sum_{k=0}^{n} {n choose k} cdot y^{k}right)$$
Could anyone help me finish this? Ideally, an explanation accessible to a middle school pre-calc student. Thanks.






algebra-precalculus




algebra-precalculus






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edited Dec 12 '18 at 19:29









saulspatz

15.9k31331




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asked Dec 12 '18 at 18:45









IlikemathIlikemath

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  • 3




    $begingroup$
    I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
    $endgroup$
    – Doug M
    Dec 12 '18 at 18:59














  • 3




    $begingroup$
    I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
    $endgroup$
    – Doug M
    Dec 12 '18 at 18:59








3




3




$begingroup$
I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
$endgroup$
– Doug M
Dec 12 '18 at 18:59




$begingroup$
I am going to point you to "baby Rudin" theorem 3.31, page 64 notendur.hi.is/vae11/%C3%9Eekking/…
$endgroup$
– Doug M
Dec 12 '18 at 18:59










4 Answers
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$$begin{pmatrix}n\kend{pmatrix}.frac{1}{n^k}=frac{n!}{n^kk!(n-k)!}$$
but I believe that you have got confused here. The easiest way to do it is:
$$left(1+frac1nright)^n=1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...$$
and so we get that:
$$lim_{ntoinfty}left(1+frac1nright)^n=lim_{ntoinfty}left(1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...right)=1+1+frac{1}{2!}+frac{1}{3!}...$$
$$=sum_{n=0}^inftyfrac{1}{n!}$$






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    0












    $begingroup$

    Hint: You can easily find the value of the limit using $log$ technique. Also, you can find a solution using Taylor's series of $e^x$! To know more see here.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
      $endgroup$
      – Ilikemath
      Dec 12 '18 at 18:52



















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    First note
    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}lesum_{k=0}^nfrac{1}{k!}. $$
    For the fixed $m<n$,
    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}>sum_{k=0}^mfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}gesum_{k=0}^mleft(1-frac{1}{n}right)left(1-frac{2}{n}right)cdotsleft(1-frac{k-1}{n}right)frac{1}{k!}. $$
    Let $ntoinfty$, one has
    $$ lim_{ntoinfty}left(1+frac1nright)^ngesum_{k=0}^mfrac{1}{k!}. $$
    So for any $m$,
    $$ sum_{k=0}^mfrac{1}{k!}lelim_{ntoinfty}left(1+frac1nright)^nle e. $$
    Let $mtoinfty$, one has
    $$ sum_{k=0}^inftyfrac{1}{k!}=lim_{ntoinfty}left(1+frac1nright)^n= e. $$






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      Let’s extend this to a more general form$$e^x=limlimits_{ntoinfty}left(1+frac xnright)^n$$Expand the right - hand side with the binomial theorem to get that$$begin{align*}e^x & =limlimits_{ntoinfty}left[1+ncdotfrac {x}{n}+frac {n(n-1)}{2!}left(frac xnright)^2+frac {n(n-1)(n-2)}{3!}left(frac xnright)^3+cdotsright]\ & =limlimits_{ntoinfty}left[1+x+frac {(n^2-n)x^2}{2!n^2}+frac {(n^3-3n^2+2n)x^3}{3!n^3}+cdotsright]\ & =1+x+frac {x^2}{2!}+frac {x^3}{3!}+cdotsend{align*}$$Therefore$$e^xcolor{blue}{=sumlimits_{ngeq0}frac {x^n}{n!}}$$






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        4 Answers
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        4 Answers
        4






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        1












        $begingroup$

        $$begin{pmatrix}n\kend{pmatrix}.frac{1}{n^k}=frac{n!}{n^kk!(n-k)!}$$
        but I believe that you have got confused here. The easiest way to do it is:
        $$left(1+frac1nright)^n=1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...$$
        and so we get that:
        $$lim_{ntoinfty}left(1+frac1nright)^n=lim_{ntoinfty}left(1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...right)=1+1+frac{1}{2!}+frac{1}{3!}...$$
        $$=sum_{n=0}^inftyfrac{1}{n!}$$






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          1












          $begingroup$

          $$begin{pmatrix}n\kend{pmatrix}.frac{1}{n^k}=frac{n!}{n^kk!(n-k)!}$$
          but I believe that you have got confused here. The easiest way to do it is:
          $$left(1+frac1nright)^n=1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...$$
          and so we get that:
          $$lim_{ntoinfty}left(1+frac1nright)^n=lim_{ntoinfty}left(1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...right)=1+1+frac{1}{2!}+frac{1}{3!}...$$
          $$=sum_{n=0}^inftyfrac{1}{n!}$$






          share|cite|improve this answer









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            1












            1








            1





            $begingroup$

            $$begin{pmatrix}n\kend{pmatrix}.frac{1}{n^k}=frac{n!}{n^kk!(n-k)!}$$
            but I believe that you have got confused here. The easiest way to do it is:
            $$left(1+frac1nright)^n=1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...$$
            and so we get that:
            $$lim_{ntoinfty}left(1+frac1nright)^n=lim_{ntoinfty}left(1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...right)=1+1+frac{1}{2!}+frac{1}{3!}...$$
            $$=sum_{n=0}^inftyfrac{1}{n!}$$






            share|cite|improve this answer









            $endgroup$



            $$begin{pmatrix}n\kend{pmatrix}.frac{1}{n^k}=frac{n!}{n^kk!(n-k)!}$$
            but I believe that you have got confused here. The easiest way to do it is:
            $$left(1+frac1nright)^n=1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...$$
            and so we get that:
            $$lim_{ntoinfty}left(1+frac1nright)^n=lim_{ntoinfty}left(1+n.frac1n+frac{n(n-1)}{2!times n^2}+frac{n(n-1)(n-2)}{3!times n^3}...right)=1+1+frac{1}{2!}+frac{1}{3!}...$$
            $$=sum_{n=0}^inftyfrac{1}{n!}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 12 '18 at 19:12









            Henry LeeHenry Lee

            2,054219




            2,054219























                0












                $begingroup$

                Hint: You can easily find the value of the limit using $log$ technique. Also, you can find a solution using Taylor's series of $e^x$! To know more see here.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                  $endgroup$
                  – Ilikemath
                  Dec 12 '18 at 18:52
















                0












                $begingroup$

                Hint: You can easily find the value of the limit using $log$ technique. Also, you can find a solution using Taylor's series of $e^x$! To know more see here.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                  $endgroup$
                  – Ilikemath
                  Dec 12 '18 at 18:52














                0












                0








                0





                $begingroup$

                Hint: You can easily find the value of the limit using $log$ technique. Also, you can find a solution using Taylor's series of $e^x$! To know more see here.






                share|cite|improve this answer











                $endgroup$



                Hint: You can easily find the value of the limit using $log$ technique. Also, you can find a solution using Taylor's series of $e^x$! To know more see here.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 12 '18 at 18:54

























                answered Dec 12 '18 at 18:47









                OmGOmG

                2,502822




                2,502822












                • $begingroup$
                  Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                  $endgroup$
                  – Ilikemath
                  Dec 12 '18 at 18:52


















                • $begingroup$
                  Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                  $endgroup$
                  – Ilikemath
                  Dec 12 '18 at 18:52
















                $begingroup$
                Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                $endgroup$
                – Ilikemath
                Dec 12 '18 at 18:52




                $begingroup$
                Thanks. I will look up taylor series. It looks like it has calculus involved. en.wikipedia.org/wiki/Taylor_series. Maybe i can just use logs?
                $endgroup$
                – Ilikemath
                Dec 12 '18 at 18:52











                0












                $begingroup$

                First note
                $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}lesum_{k=0}^nfrac{1}{k!}. $$
                For the fixed $m<n$,
                $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}>sum_{k=0}^mfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}gesum_{k=0}^mleft(1-frac{1}{n}right)left(1-frac{2}{n}right)cdotsleft(1-frac{k-1}{n}right)frac{1}{k!}. $$
                Let $ntoinfty$, one has
                $$ lim_{ntoinfty}left(1+frac1nright)^ngesum_{k=0}^mfrac{1}{k!}. $$
                So for any $m$,
                $$ sum_{k=0}^mfrac{1}{k!}lelim_{ntoinfty}left(1+frac1nright)^nle e. $$
                Let $mtoinfty$, one has
                $$ sum_{k=0}^inftyfrac{1}{k!}=lim_{ntoinfty}left(1+frac1nright)^n= e. $$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  First note
                  $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}lesum_{k=0}^nfrac{1}{k!}. $$
                  For the fixed $m<n$,
                  $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}>sum_{k=0}^mfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}gesum_{k=0}^mleft(1-frac{1}{n}right)left(1-frac{2}{n}right)cdotsleft(1-frac{k-1}{n}right)frac{1}{k!}. $$
                  Let $ntoinfty$, one has
                  $$ lim_{ntoinfty}left(1+frac1nright)^ngesum_{k=0}^mfrac{1}{k!}. $$
                  So for any $m$,
                  $$ sum_{k=0}^mfrac{1}{k!}lelim_{ntoinfty}left(1+frac1nright)^nle e. $$
                  Let $mtoinfty$, one has
                  $$ sum_{k=0}^inftyfrac{1}{k!}=lim_{ntoinfty}left(1+frac1nright)^n= e. $$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    First note
                    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}lesum_{k=0}^nfrac{1}{k!}. $$
                    For the fixed $m<n$,
                    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}>sum_{k=0}^mfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}gesum_{k=0}^mleft(1-frac{1}{n}right)left(1-frac{2}{n}right)cdotsleft(1-frac{k-1}{n}right)frac{1}{k!}. $$
                    Let $ntoinfty$, one has
                    $$ lim_{ntoinfty}left(1+frac1nright)^ngesum_{k=0}^mfrac{1}{k!}. $$
                    So for any $m$,
                    $$ sum_{k=0}^mfrac{1}{k!}lelim_{ntoinfty}left(1+frac1nright)^nle e. $$
                    Let $mtoinfty$, one has
                    $$ sum_{k=0}^inftyfrac{1}{k!}=lim_{ntoinfty}left(1+frac1nright)^n= e. $$






                    share|cite|improve this answer









                    $endgroup$



                    First note
                    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}=sum_{k=0}^nfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}lesum_{k=0}^nfrac{1}{k!}. $$
                    For the fixed $m<n$,
                    $$left(1+frac1nright)^n=sum_{k=0}^nfrac{n!}{k!(n-k)!}frac{1}{n^{n-k}}>sum_{k=0}^mfrac{n(n-1)cdots(n-k+1)}{n^{n-k}}frac{1}{k!}gesum_{k=0}^mleft(1-frac{1}{n}right)left(1-frac{2}{n}right)cdotsleft(1-frac{k-1}{n}right)frac{1}{k!}. $$
                    Let $ntoinfty$, one has
                    $$ lim_{ntoinfty}left(1+frac1nright)^ngesum_{k=0}^mfrac{1}{k!}. $$
                    So for any $m$,
                    $$ sum_{k=0}^mfrac{1}{k!}lelim_{ntoinfty}left(1+frac1nright)^nle e. $$
                    Let $mtoinfty$, one has
                    $$ sum_{k=0}^inftyfrac{1}{k!}=lim_{ntoinfty}left(1+frac1nright)^n= e. $$







                    share|cite|improve this answer












                    share|cite|improve this answer



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                    answered Dec 12 '18 at 20:34









                    xpaulxpaul

                    23k24455




                    23k24455























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                        $begingroup$

                        Let’s extend this to a more general form$$e^x=limlimits_{ntoinfty}left(1+frac xnright)^n$$Expand the right - hand side with the binomial theorem to get that$$begin{align*}e^x & =limlimits_{ntoinfty}left[1+ncdotfrac {x}{n}+frac {n(n-1)}{2!}left(frac xnright)^2+frac {n(n-1)(n-2)}{3!}left(frac xnright)^3+cdotsright]\ & =limlimits_{ntoinfty}left[1+x+frac {(n^2-n)x^2}{2!n^2}+frac {(n^3-3n^2+2n)x^3}{3!n^3}+cdotsright]\ & =1+x+frac {x^2}{2!}+frac {x^3}{3!}+cdotsend{align*}$$Therefore$$e^xcolor{blue}{=sumlimits_{ngeq0}frac {x^n}{n!}}$$






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                          0












                          $begingroup$

                          Let’s extend this to a more general form$$e^x=limlimits_{ntoinfty}left(1+frac xnright)^n$$Expand the right - hand side with the binomial theorem to get that$$begin{align*}e^x & =limlimits_{ntoinfty}left[1+ncdotfrac {x}{n}+frac {n(n-1)}{2!}left(frac xnright)^2+frac {n(n-1)(n-2)}{3!}left(frac xnright)^3+cdotsright]\ & =limlimits_{ntoinfty}left[1+x+frac {(n^2-n)x^2}{2!n^2}+frac {(n^3-3n^2+2n)x^3}{3!n^3}+cdotsright]\ & =1+x+frac {x^2}{2!}+frac {x^3}{3!}+cdotsend{align*}$$Therefore$$e^xcolor{blue}{=sumlimits_{ngeq0}frac {x^n}{n!}}$$






                          share|cite|improve this answer









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                            0





                            $begingroup$

                            Let’s extend this to a more general form$$e^x=limlimits_{ntoinfty}left(1+frac xnright)^n$$Expand the right - hand side with the binomial theorem to get that$$begin{align*}e^x & =limlimits_{ntoinfty}left[1+ncdotfrac {x}{n}+frac {n(n-1)}{2!}left(frac xnright)^2+frac {n(n-1)(n-2)}{3!}left(frac xnright)^3+cdotsright]\ & =limlimits_{ntoinfty}left[1+x+frac {(n^2-n)x^2}{2!n^2}+frac {(n^3-3n^2+2n)x^3}{3!n^3}+cdotsright]\ & =1+x+frac {x^2}{2!}+frac {x^3}{3!}+cdotsend{align*}$$Therefore$$e^xcolor{blue}{=sumlimits_{ngeq0}frac {x^n}{n!}}$$






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                            $endgroup$



                            Let’s extend this to a more general form$$e^x=limlimits_{ntoinfty}left(1+frac xnright)^n$$Expand the right - hand side with the binomial theorem to get that$$begin{align*}e^x & =limlimits_{ntoinfty}left[1+ncdotfrac {x}{n}+frac {n(n-1)}{2!}left(frac xnright)^2+frac {n(n-1)(n-2)}{3!}left(frac xnright)^3+cdotsright]\ & =limlimits_{ntoinfty}left[1+x+frac {(n^2-n)x^2}{2!n^2}+frac {(n^3-3n^2+2n)x^3}{3!n^3}+cdotsright]\ & =1+x+frac {x^2}{2!}+frac {x^3}{3!}+cdotsend{align*}$$Therefore$$e^xcolor{blue}{=sumlimits_{ngeq0}frac {x^n}{n!}}$$







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                            answered Dec 12 '18 at 20:54









                            Frank W.Frank W.

                            3,7351321




                            3,7351321






























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