Limit of $frac{2^n}{n!}$ [duplicate]












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  • Alternative way to prove $lim_{ntoinfty}frac{2^n}{n!}=0$? [duplicate]

    7 answers




How can I prove that $lim_{ntoinfty} frac{2^n}{n!}=0$?






real-analysis real-numbers







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marked as duplicate by rtybase, T. Bongers, Shailesh, RRL real-analysis
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 Dec 10 '18 at 23:11


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
    $endgroup$
    – Arthur
    Dec 10 '18 at 20:53










  • $begingroup$
    We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
    $endgroup$
    – MathematicsStudent1122
    Dec 10 '18 at 20:54


















0












$begingroup$



This question already has an answer here:




  • Alternative way to prove $lim_{ntoinfty}frac{2^n}{n!}=0$? [duplicate]

    7 answers




How can I prove that $lim_{ntoinfty} frac{2^n}{n!}=0$?






real-analysis real-numbers







share|cite|improve this question











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 Dec 10 '18 at 23:11


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1




    $begingroup$
    What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
    $endgroup$
    – Arthur
    Dec 10 '18 at 20:53










  • $begingroup$
    We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
    $endgroup$
    – MathematicsStudent1122
    Dec 10 '18 at 20:54
















0












0








0





$begingroup$



This question already has an answer here:




  • Alternative way to prove $lim_{ntoinfty}frac{2^n}{n!}=0$? [duplicate]

    7 answers




How can I prove that $lim_{ntoinfty} frac{2^n}{n!}=0$?






real-analysis real-numbers







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Alternative way to prove $lim_{ntoinfty}frac{2^n}{n!}=0$? [duplicate]

    7 answers




How can I prove that $lim_{ntoinfty} frac{2^n}{n!}=0$?





This question already has an answer here:




  • Alternative way to prove $lim_{ntoinfty}frac{2^n}{n!}=0$? [duplicate]

    7 answers







real-analysis real-numbers




real-analysis real-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 21:17









Zach Langley

9731019




9731019










asked Dec 10 '18 at 20:46









RomeissaRomeissa

43




43




marked as duplicate by rtybase, T. Bongers, Shailesh, RRL real-analysis
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 Dec 10 '18 at 23:11


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marked as duplicate by rtybase, T. Bongers, Shailesh, RRL real-analysis
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 Dec 10 '18 at 23:11


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
    $endgroup$
    – Arthur
    Dec 10 '18 at 20:53










  • $begingroup$
    We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
    $endgroup$
    – MathematicsStudent1122
    Dec 10 '18 at 20:54
















  • 1




    $begingroup$
    What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
    $endgroup$
    – Arthur
    Dec 10 '18 at 20:53










  • $begingroup$
    We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
    $endgroup$
    – MathematicsStudent1122
    Dec 10 '18 at 20:54










1




1




$begingroup$
What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
$endgroup$
– Arthur
Dec 10 '18 at 20:53




$begingroup$
What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
$endgroup$
– Arthur
Dec 10 '18 at 20:53












$begingroup$
We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
$endgroup$
– MathematicsStudent1122
Dec 10 '18 at 20:54






$begingroup$
We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
$endgroup$
– MathematicsStudent1122
Dec 10 '18 at 20:54












4 Answers
4






active

oldest

votes


















1












$begingroup$

Try with this inequality:



$$frac{2^n}{n!}leq frac{2^n}{1cdot 2cdot 3cdot 4^{n-3}}=frac{4^3}{6} 4^{-n/2}.$$



Since $4^{-n/2}to 0$ as $ntoinfty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's also very good.
    $endgroup$
    – gimusi
    Dec 10 '18 at 21:05



















1












$begingroup$

Simplest way by ratio test



$$frac{2^{n+1}}{(n+1)!}frac{n!}{2^n}=frac2{n+1}to 0 <1$$



then the sequence converges to $0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @T.Bongers Sorry but I'm going to not reply anymore to such kind of unfair and OT comments. If you have something against an answer, please use the proper ways. I'm at disposal for discussions in chat. Thanks
    $endgroup$
    – gimusi
    Dec 11 '18 at 7:20



















1












$begingroup$

Hint:
Let $u_n=dfrac{2^n}{n!}$.



Calculate $;dfrac{u_{n+1}}{u_n}$. What is its limit? What can you deduce for $u_n$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Same idea than mine! Of course the straightforward way,
    $endgroup$
    – gimusi
    Dec 10 '18 at 21:05










  • $begingroup$
    @gimusi: Great minds think together… :o) You seem to have posted while I was typing, and I didn't think of checking whether another answer had been posted in the meantime
    $endgroup$
    – Bernard
    Dec 10 '18 at 21:12










  • $begingroup$
    That's absolutely fine! 2 ratio tests never killed anyone :)
    $endgroup$
    – gimusi
    Dec 10 '18 at 21:13



















0












$begingroup$

begin{align}
frac{2^n}{n!} &= frac{2}{1} frac{2}{2} frac{2}{3} cdots frac{2}{n-1} frac{2}{n}
\ &< 2cdot 1cdot 1 cdot 1dots 1cdot frac{2}{n}
\ &= frac{4}{n} to 0
end{align}






share|cite|improve this answer









$endgroup$




















    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Try with this inequality:



    $$frac{2^n}{n!}leq frac{2^n}{1cdot 2cdot 3cdot 4^{n-3}}=frac{4^3}{6} 4^{-n/2}.$$



    Since $4^{-n/2}to 0$ as $ntoinfty$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      That's also very good.
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:05
















    1












    $begingroup$

    Try with this inequality:



    $$frac{2^n}{n!}leq frac{2^n}{1cdot 2cdot 3cdot 4^{n-3}}=frac{4^3}{6} 4^{-n/2}.$$



    Since $4^{-n/2}to 0$ as $ntoinfty$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      That's also very good.
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:05














    1












    1








    1





    $begingroup$

    Try with this inequality:



    $$frac{2^n}{n!}leq frac{2^n}{1cdot 2cdot 3cdot 4^{n-3}}=frac{4^3}{6} 4^{-n/2}.$$



    Since $4^{-n/2}to 0$ as $ntoinfty$.






    share|cite|improve this answer









    $endgroup$



    Try with this inequality:



    $$frac{2^n}{n!}leq frac{2^n}{1cdot 2cdot 3cdot 4^{n-3}}=frac{4^3}{6} 4^{-n/2}.$$



    Since $4^{-n/2}to 0$ as $ntoinfty$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 10 '18 at 20:55









    José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

    802110




    802110












    • $begingroup$
      That's also very good.
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:05


















    • $begingroup$
      That's also very good.
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:05
















    $begingroup$
    That's also very good.
    $endgroup$
    – gimusi
    Dec 10 '18 at 21:05




    $begingroup$
    That's also very good.
    $endgroup$
    – gimusi
    Dec 10 '18 at 21:05











    1












    $begingroup$

    Simplest way by ratio test



    $$frac{2^{n+1}}{(n+1)!}frac{n!}{2^n}=frac2{n+1}to 0 <1$$



    then the sequence converges to $0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @T.Bongers Sorry but I'm going to not reply anymore to such kind of unfair and OT comments. If you have something against an answer, please use the proper ways. I'm at disposal for discussions in chat. Thanks
      $endgroup$
      – gimusi
      Dec 11 '18 at 7:20
















    1












    $begingroup$

    Simplest way by ratio test



    $$frac{2^{n+1}}{(n+1)!}frac{n!}{2^n}=frac2{n+1}to 0 <1$$



    then the sequence converges to $0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @T.Bongers Sorry but I'm going to not reply anymore to such kind of unfair and OT comments. If you have something against an answer, please use the proper ways. I'm at disposal for discussions in chat. Thanks
      $endgroup$
      – gimusi
      Dec 11 '18 at 7:20














    1












    1








    1





    $begingroup$

    Simplest way by ratio test



    $$frac{2^{n+1}}{(n+1)!}frac{n!}{2^n}=frac2{n+1}to 0 <1$$



    then the sequence converges to $0$.






    share|cite|improve this answer









    $endgroup$



    Simplest way by ratio test



    $$frac{2^{n+1}}{(n+1)!}frac{n!}{2^n}=frac2{n+1}to 0 <1$$



    then the sequence converges to $0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 10 '18 at 21:00









    gimusigimusi

    92.8k84494




    92.8k84494












    • $begingroup$
      @T.Bongers Sorry but I'm going to not reply anymore to such kind of unfair and OT comments. If you have something against an answer, please use the proper ways. I'm at disposal for discussions in chat. Thanks
      $endgroup$
      – gimusi
      Dec 11 '18 at 7:20


















    • $begingroup$
      @T.Bongers Sorry but I'm going to not reply anymore to such kind of unfair and OT comments. If you have something against an answer, please use the proper ways. I'm at disposal for discussions in chat. Thanks
      $endgroup$
      – gimusi
      Dec 11 '18 at 7:20
















    $begingroup$
    @T.Bongers Sorry but I'm going to not reply anymore to such kind of unfair and OT comments. If you have something against an answer, please use the proper ways. I'm at disposal for discussions in chat. Thanks
    $endgroup$
    – gimusi
    Dec 11 '18 at 7:20




    $begingroup$
    @T.Bongers Sorry but I'm going to not reply anymore to such kind of unfair and OT comments. If you have something against an answer, please use the proper ways. I'm at disposal for discussions in chat. Thanks
    $endgroup$
    – gimusi
    Dec 11 '18 at 7:20











    1












    $begingroup$

    Hint:
    Let $u_n=dfrac{2^n}{n!}$.



    Calculate $;dfrac{u_{n+1}}{u_n}$. What is its limit? What can you deduce for $u_n$?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Same idea than mine! Of course the straightforward way,
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:05










    • $begingroup$
      @gimusi: Great minds think together… :o) You seem to have posted while I was typing, and I didn't think of checking whether another answer had been posted in the meantime
      $endgroup$
      – Bernard
      Dec 10 '18 at 21:12










    • $begingroup$
      That's absolutely fine! 2 ratio tests never killed anyone :)
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:13
















    1












    $begingroup$

    Hint:
    Let $u_n=dfrac{2^n}{n!}$.



    Calculate $;dfrac{u_{n+1}}{u_n}$. What is its limit? What can you deduce for $u_n$?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Same idea than mine! Of course the straightforward way,
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:05










    • $begingroup$
      @gimusi: Great minds think together… :o) You seem to have posted while I was typing, and I didn't think of checking whether another answer had been posted in the meantime
      $endgroup$
      – Bernard
      Dec 10 '18 at 21:12










    • $begingroup$
      That's absolutely fine! 2 ratio tests never killed anyone :)
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:13














    1












    1








    1





    $begingroup$

    Hint:
    Let $u_n=dfrac{2^n}{n!}$.



    Calculate $;dfrac{u_{n+1}}{u_n}$. What is its limit? What can you deduce for $u_n$?






    share|cite|improve this answer









    $endgroup$



    Hint:
    Let $u_n=dfrac{2^n}{n!}$.



    Calculate $;dfrac{u_{n+1}}{u_n}$. What is its limit? What can you deduce for $u_n$?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 10 '18 at 21:00









    BernardBernard

    120k740116




    120k740116












    • $begingroup$
      Same idea than mine! Of course the straightforward way,
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:05










    • $begingroup$
      @gimusi: Great minds think together… :o) You seem to have posted while I was typing, and I didn't think of checking whether another answer had been posted in the meantime
      $endgroup$
      – Bernard
      Dec 10 '18 at 21:12










    • $begingroup$
      That's absolutely fine! 2 ratio tests never killed anyone :)
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:13


















    • $begingroup$
      Same idea than mine! Of course the straightforward way,
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:05










    • $begingroup$
      @gimusi: Great minds think together… :o) You seem to have posted while I was typing, and I didn't think of checking whether another answer had been posted in the meantime
      $endgroup$
      – Bernard
      Dec 10 '18 at 21:12










    • $begingroup$
      That's absolutely fine! 2 ratio tests never killed anyone :)
      $endgroup$
      – gimusi
      Dec 10 '18 at 21:13
















    $begingroup$
    Same idea than mine! Of course the straightforward way,
    $endgroup$
    – gimusi
    Dec 10 '18 at 21:05




    $begingroup$
    Same idea than mine! Of course the straightforward way,
    $endgroup$
    – gimusi
    Dec 10 '18 at 21:05












    $begingroup$
    @gimusi: Great minds think together… :o) You seem to have posted while I was typing, and I didn't think of checking whether another answer had been posted in the meantime
    $endgroup$
    – Bernard
    Dec 10 '18 at 21:12




    $begingroup$
    @gimusi: Great minds think together… :o) You seem to have posted while I was typing, and I didn't think of checking whether another answer had been posted in the meantime
    $endgroup$
    – Bernard
    Dec 10 '18 at 21:12












    $begingroup$
    That's absolutely fine! 2 ratio tests never killed anyone :)
    $endgroup$
    – gimusi
    Dec 10 '18 at 21:13




    $begingroup$
    That's absolutely fine! 2 ratio tests never killed anyone :)
    $endgroup$
    – gimusi
    Dec 10 '18 at 21:13











    0












    $begingroup$

    begin{align}
    frac{2^n}{n!} &= frac{2}{1} frac{2}{2} frac{2}{3} cdots frac{2}{n-1} frac{2}{n}
    \ &< 2cdot 1cdot 1 cdot 1dots 1cdot frac{2}{n}
    \ &= frac{4}{n} to 0
    end{align}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      begin{align}
      frac{2^n}{n!} &= frac{2}{1} frac{2}{2} frac{2}{3} cdots frac{2}{n-1} frac{2}{n}
      \ &< 2cdot 1cdot 1 cdot 1dots 1cdot frac{2}{n}
      \ &= frac{4}{n} to 0
      end{align}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        begin{align}
        frac{2^n}{n!} &= frac{2}{1} frac{2}{2} frac{2}{3} cdots frac{2}{n-1} frac{2}{n}
        \ &< 2cdot 1cdot 1 cdot 1dots 1cdot frac{2}{n}
        \ &= frac{4}{n} to 0
        end{align}






        share|cite|improve this answer









        $endgroup$



        begin{align}
        frac{2^n}{n!} &= frac{2}{1} frac{2}{2} frac{2}{3} cdots frac{2}{n-1} frac{2}{n}
        \ &< 2cdot 1cdot 1 cdot 1dots 1cdot frac{2}{n}
        \ &= frac{4}{n} to 0
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 21:07









        GEdgarGEdgar

        62.4k267171




        62.4k267171















            -

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