Show that $lim_{nrightarrowinfty}e^{-n}=0$












2












$begingroup$


I wrote out the term from above and get



$text{Show that: }lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0$



I can use Leibniz' Criterion to find out that it converges but I don't know nothing about the value. Is there a $0$ convergent upper bound for this expression. Can somebody give me a hint how to find it please?



Edit: I have changed the fraction in the expression






sequences-and-series limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
    $endgroup$
    – Yanko
    Jan 23 at 14:27








  • 4




    $begingroup$
    This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
    $endgroup$
    – Lee Mosher
    Jan 23 at 14:39






  • 1




    $begingroup$
    Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
    $endgroup$
    – GEdgar
    Jan 23 at 15:04
















2












$begingroup$


I wrote out the term from above and get



$text{Show that: }lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0$



I can use Leibniz' Criterion to find out that it converges but I don't know nothing about the value. Is there a $0$ convergent upper bound for this expression. Can somebody give me a hint how to find it please?



Edit: I have changed the fraction in the expression






sequences-and-series limits







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
    $endgroup$
    – Yanko
    Jan 23 at 14:27








  • 4




    $begingroup$
    This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
    $endgroup$
    – Lee Mosher
    Jan 23 at 14:39






  • 1




    $begingroup$
    Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
    $endgroup$
    – GEdgar
    Jan 23 at 15:04














2












2








2





$begingroup$


I wrote out the term from above and get



$text{Show that: }lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0$



I can use Leibniz' Criterion to find out that it converges but I don't know nothing about the value. Is there a $0$ convergent upper bound for this expression. Can somebody give me a hint how to find it please?



Edit: I have changed the fraction in the expression






sequences-and-series limits







share|cite|improve this question











$endgroup$




I wrote out the term from above and get



$text{Show that: }lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0$



I can use Leibniz' Criterion to find out that it converges but I don't know nothing about the value. Is there a $0$ convergent upper bound for this expression. Can somebody give me a hint how to find it please?



Edit: I have changed the fraction in the expression






sequences-and-series limits




sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 14:42









Bernard

122k741116




122k741116










asked Jan 23 at 14:25









RM777RM777

38312




38312








  • 1




    $begingroup$
    $e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
    $endgroup$
    – Yanko
    Jan 23 at 14:27








  • 4




    $begingroup$
    This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
    $endgroup$
    – Lee Mosher
    Jan 23 at 14:39






  • 1




    $begingroup$
    Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
    $endgroup$
    – GEdgar
    Jan 23 at 15:04














  • 1




    $begingroup$
    $e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
    $endgroup$
    – Yanko
    Jan 23 at 14:27








  • 4




    $begingroup$
    This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
    $endgroup$
    – Lee Mosher
    Jan 23 at 14:39






  • 1




    $begingroup$
    Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
    $endgroup$
    – GEdgar
    Jan 23 at 15:04








1




1




$begingroup$
$e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
$endgroup$
– Yanko
Jan 23 at 14:27






$begingroup$
$e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
$endgroup$
– Yanko
Jan 23 at 14:27






4




4




$begingroup$
This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
$endgroup$
– Lee Mosher
Jan 23 at 14:39




$begingroup$
This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
$endgroup$
– Lee Mosher
Jan 23 at 14:39




1




1




$begingroup$
Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
$endgroup$
– GEdgar
Jan 23 at 15:04




$begingroup$
Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
$endgroup$
– GEdgar
Jan 23 at 15:04










3 Answers
3






active

oldest

votes


















2












$begingroup$

Here is the standard elementary proof that
$a^n to 0$
if $0 < a < 1$.



$a =1/(1+b)$ where $b = 1/a -1 > 0$.



By Bernoulli's inequality,
$(1+b)^n ge 1+bn > bn$
so
$a^n =1/(1+b)^n
< 1/(bn)
to 0$
as $n to infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
    $endgroup$
    – RM777
    Jan 27 at 13:51





















0












$begingroup$

For any $a$ such that $0 < a < 1$, you can find natural $n$ to make $a^n$ smaller than any positive number $x$:



$a^n < x iff n , log(a) < log(x) iff n > log(x)/log(a)$



Also, $a^n > 0$ for any natural $n$.



These 2 facts imply $lim_{n -> infty} , a^n = 0$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Do not expand, just use



    $$e^{-n}=(e^{-1})^n$$ which is quickly decreasing.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Here is the standard elementary proof that
      $a^n to 0$
      if $0 < a < 1$.



      $a =1/(1+b)$ where $b = 1/a -1 > 0$.



      By Bernoulli's inequality,
      $(1+b)^n ge 1+bn > bn$
      so
      $a^n =1/(1+b)^n
      < 1/(bn)
      to 0$
      as $n to infty$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
        $endgroup$
        – RM777
        Jan 27 at 13:51


















      2












      $begingroup$

      Here is the standard elementary proof that
      $a^n to 0$
      if $0 < a < 1$.



      $a =1/(1+b)$ where $b = 1/a -1 > 0$.



      By Bernoulli's inequality,
      $(1+b)^n ge 1+bn > bn$
      so
      $a^n =1/(1+b)^n
      < 1/(bn)
      to 0$
      as $n to infty$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
        $endgroup$
        – RM777
        Jan 27 at 13:51
















      2












      2








      2





      $begingroup$

      Here is the standard elementary proof that
      $a^n to 0$
      if $0 < a < 1$.



      $a =1/(1+b)$ where $b = 1/a -1 > 0$.



      By Bernoulli's inequality,
      $(1+b)^n ge 1+bn > bn$
      so
      $a^n =1/(1+b)^n
      < 1/(bn)
      to 0$
      as $n to infty$.






      share|cite|improve this answer









      $endgroup$



      Here is the standard elementary proof that
      $a^n to 0$
      if $0 < a < 1$.



      $a =1/(1+b)$ where $b = 1/a -1 > 0$.



      By Bernoulli's inequality,
      $(1+b)^n ge 1+bn > bn$
      so
      $a^n =1/(1+b)^n
      < 1/(bn)
      to 0$
      as $n to infty$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 23 at 14:53









      marty cohenmarty cohen

      74.3k549128




      74.3k549128












      • $begingroup$
        Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
        $endgroup$
        – RM777
        Jan 27 at 13:51




















      • $begingroup$
        Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
        $endgroup$
        – RM777
        Jan 27 at 13:51


















      $begingroup$
      Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
      $endgroup$
      – RM777
      Jan 27 at 13:51






      $begingroup$
      Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
      $endgroup$
      – RM777
      Jan 27 at 13:51













      0












      $begingroup$

      For any $a$ such that $0 < a < 1$, you can find natural $n$ to make $a^n$ smaller than any positive number $x$:



      $a^n < x iff n , log(a) < log(x) iff n > log(x)/log(a)$



      Also, $a^n > 0$ for any natural $n$.



      These 2 facts imply $lim_{n -> infty} , a^n = 0$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        For any $a$ such that $0 < a < 1$, you can find natural $n$ to make $a^n$ smaller than any positive number $x$:



        $a^n < x iff n , log(a) < log(x) iff n > log(x)/log(a)$



        Also, $a^n > 0$ for any natural $n$.



        These 2 facts imply $lim_{n -> infty} , a^n = 0$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          For any $a$ such that $0 < a < 1$, you can find natural $n$ to make $a^n$ smaller than any positive number $x$:



          $a^n < x iff n , log(a) < log(x) iff n > log(x)/log(a)$



          Also, $a^n > 0$ for any natural $n$.



          These 2 facts imply $lim_{n -> infty} , a^n = 0$






          share|cite|improve this answer









          $endgroup$



          For any $a$ such that $0 < a < 1$, you can find natural $n$ to make $a^n$ smaller than any positive number $x$:



          $a^n < x iff n , log(a) < log(x) iff n > log(x)/log(a)$



          Also, $a^n > 0$ for any natural $n$.



          These 2 facts imply $lim_{n -> infty} , a^n = 0$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 14:46









          JaviJavi

          3979




          3979























              0












              $begingroup$

              Do not expand, just use



              $$e^{-n}=(e^{-1})^n$$ which is quickly decreasing.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Do not expand, just use



                $$e^{-n}=(e^{-1})^n$$ which is quickly decreasing.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Do not expand, just use



                  $$e^{-n}=(e^{-1})^n$$ which is quickly decreasing.






                  share|cite|improve this answer









                  $endgroup$



                  Do not expand, just use



                  $$e^{-n}=(e^{-1})^n$$ which is quickly decreasing.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 23 at 14:57









                  Yves DaoustYves Daoust

                  130k676227




                  130k676227






























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