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












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











$endgroup$



marked as duplicate by rtybase, T. Bongers, Shailesh, RRL real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 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











$endgroup$



marked as duplicate by rtybase, T. Bongers, Shailesh, RRL real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 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
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 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.









marked as duplicate by rtybase, T. Bongers, Shailesh, RRL real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
 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

    120k740115




    120k740115












    • $begingroup$
      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

        62.4k267170




        62.4k267170















            -

            Popular posts from this blog

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more