Limit of $frac{2^n}{n!}$ [duplicate]
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
edited Dec 10 '18 at 21:17
Zach Langley
9691019
asked Dec 10 '18 at 20:46
Romeissa
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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.
add a comment |
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
edited Dec 10 '18 at 21:17
Zach Langley
9691019
asked Dec 10 '18 at 20:46
Romeissa
123
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
What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
– Arthur
Dec 10 '18 at 20:53
We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
– MathematicsStudent1122
Dec 10 '18 at 20:54
add a comment |
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
edited Dec 10 '18 at 21:17
Zach Langley
9691019
asked Dec 10 '18 at 20:46
Romeissa
123
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
edited Dec 10 '18 at 21:17
Zach Langley
9691019
asked Dec 10 '18 at 20:46
Romeissa
123
edited Dec 10 '18 at 21:17
Zach Langley
9691019
asked Dec 10 '18 at 20:46
Romeissa
123
edited Dec 10 '18 at 21:17
Zach Langley
9691019
edited Dec 10 '18 at 21:17
Zach Langley
9691019
edited Dec 10 '18 at 21:17
Zach Langley
9691019
9691019
asked Dec 10 '18 at 20:46
Romeissa
123
asked Dec 10 '18 at 20:46
Romeissa
123
asked Dec 10 '18 at 20:46
Romeissa
123
123
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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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
What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
– Arthur
Dec 10 '18 at 20:53
We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
– MathematicsStudent1122
Dec 10 '18 at 20:54
add a comment |
1
What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
– Arthur
Dec 10 '18 at 20:53
We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
– MathematicsStudent1122
Dec 10 '18 at 20:54
1
1
What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
– Arthur
Dec 10 '18 at 20:53
What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
– Arthur
Dec 10 '18 at 20:53
We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
– MathematicsStudent1122
Dec 10 '18 at 20:54
We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
– MathematicsStudent1122
Dec 10 '18 at 20:54
add a comment |
4 Answers
4
active
oldest
votes
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 $ntoinfty$.
answered Dec 10 '18 at 20:55
José Alejandro Aburto Araneda
756110
That's also very good.
– gimusi
Dec 10 '18 at 21:05
add a comment |
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 $0$.
answered Dec 10 '18 at 21:00
gimusi
1
@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
– gimusi
Dec 11 '18 at 7:20
add a comment |
Hint:
Let $u_n=dfrac{2^n}{n!}$.
Calculate $;dfrac{u_{n+1}}{u_n}$. What is its limit? What can you deduce for $u_n$?
answered Dec 10 '18 at 21:00
Bernard
118k639112
Same idea than mine! Of course the straightforward way,
– gimusi
Dec 10 '18 at 21:05
@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
– Bernard
Dec 10 '18 at 21:12
That's absolutely fine! 2 ratio tests never killed anyone :)
– gimusi
Dec 10 '18 at 21:13
add a comment |
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}
answered Dec 10 '18 at 21:07
GEdgar
61.9k267168
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $ntoinfty$.
answered Dec 10 '18 at 20:55
José Alejandro Aburto Araneda
756110
That's also very good.
– gimusi
Dec 10 '18 at 21:05
add a comment |
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 $ntoinfty$.
answered Dec 10 '18 at 20:55
José Alejandro Aburto Araneda
756110
That's also very good.
– gimusi
Dec 10 '18 at 21:05
add a comment |
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 $ntoinfty$.
answered Dec 10 '18 at 20:55
José Alejandro Aburto Araneda
756110
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 $ntoinfty$.
answered Dec 10 '18 at 20:55
José Alejandro Aburto Araneda
756110
answered Dec 10 '18 at 20:55
José Alejandro Aburto Araneda
756110
answered Dec 10 '18 at 20:55
José Alejandro Aburto Araneda
756110
answered Dec 10 '18 at 20:55
José Alejandro Aburto Araneda
756110
756110
That's also very good.
– gimusi
Dec 10 '18 at 21:05
add a comment |
That's also very good.
– gimusi
Dec 10 '18 at 21:05
That's also very good.
– gimusi
Dec 10 '18 at 21:05
That's also very good.
– gimusi
Dec 10 '18 at 21:05
add a comment |
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 $0$.
answered Dec 10 '18 at 21:00
gimusi
1
@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
– gimusi
Dec 11 '18 at 7:20
add a comment |
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 $0$.
answered Dec 10 '18 at 21:00
gimusi
1
@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
– gimusi
Dec 11 '18 at 7:20
add a comment |
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 $0$.
answered Dec 10 '18 at 21:00
gimusi
1
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 $0$.
answered Dec 10 '18 at 21:00
gimusi
1
answered Dec 10 '18 at 21:00
gimusi
1
answered Dec 10 '18 at 21:00
gimusi
1
answered Dec 10 '18 at 21:00
gimusi
1
1
@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
– gimusi
Dec 11 '18 at 7:20
add a comment |
@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
– gimusi
Dec 11 '18 at 7:20
@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
– gimusi
Dec 11 '18 at 7:20
@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
– gimusi
Dec 11 '18 at 7:20
add a comment |
Hint:
Let $u_n=dfrac{2^n}{n!}$.
Calculate $;dfrac{u_{n+1}}{u_n}$. What is its limit? What can you deduce for $u_n$?
answered Dec 10 '18 at 21:00
Bernard
118k639112
Same idea than mine! Of course the straightforward way,
– gimusi
Dec 10 '18 at 21:05
@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
– Bernard
Dec 10 '18 at 21:12
That's absolutely fine! 2 ratio tests never killed anyone :)
– gimusi
Dec 10 '18 at 21:13
add a comment |
Hint:
Let $u_n=dfrac{2^n}{n!}$.
Calculate $;dfrac{u_{n+1}}{u_n}$. What is its limit? What can you deduce for $u_n$?
answered Dec 10 '18 at 21:00
Bernard
118k639112
Same idea than mine! Of course the straightforward way,
– gimusi
Dec 10 '18 at 21:05
@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
– Bernard
Dec 10 '18 at 21:12
That's absolutely fine! 2 ratio tests never killed anyone :)
– gimusi
Dec 10 '18 at 21:13
add a comment |
Hint:
Let $u_n=dfrac{2^n}{n!}$.
Calculate $;dfrac{u_{n+1}}{u_n}$. What is its limit? What can you deduce for $u_n$?
answered Dec 10 '18 at 21:00
Bernard
118k639112
Hint:
Let $u_n=dfrac{2^n}{n!}$.
Calculate $;dfrac{u_{n+1}}{u_n}$. What is its limit? What can you deduce for $u_n$?
answered Dec 10 '18 at 21:00
Bernard
118k639112
answered Dec 10 '18 at 21:00
Bernard
118k639112
answered Dec 10 '18 at 21:00
Bernard
118k639112
answered Dec 10 '18 at 21:00
Bernard
118k639112
118k639112
Same idea than mine! Of course the straightforward way,
– gimusi
Dec 10 '18 at 21:05
@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
– Bernard
Dec 10 '18 at 21:12
That's absolutely fine! 2 ratio tests never killed anyone :)
– gimusi
Dec 10 '18 at 21:13
add a comment |
Same idea than mine! Of course the straightforward way,
– gimusi
Dec 10 '18 at 21:05
@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
– Bernard
Dec 10 '18 at 21:12
That's absolutely fine! 2 ratio tests never killed anyone :)
– gimusi
Dec 10 '18 at 21:13
Same idea than mine! Of course the straightforward way,
– gimusi
Dec 10 '18 at 21:05
Same idea than mine! Of course the straightforward way,
– gimusi
Dec 10 '18 at 21:05
@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
– Bernard
Dec 10 '18 at 21:12
@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
– Bernard
Dec 10 '18 at 21:12
That's absolutely fine! 2 ratio tests never killed anyone :)
– gimusi
Dec 10 '18 at 21:13
That's absolutely fine! 2 ratio tests never killed anyone :)
– gimusi
Dec 10 '18 at 21:13
add a comment |
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}
answered Dec 10 '18 at 21:07
GEdgar
61.9k267168
add a comment |
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}
answered Dec 10 '18 at 21:07
GEdgar
61.9k267168
add a comment |
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}
answered Dec 10 '18 at 21:07
GEdgar
61.9k267168
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}
answered Dec 10 '18 at 21:07
GEdgar
61.9k267168
answered Dec 10 '18 at 21:07
GEdgar
61.9k267168
answered Dec 10 '18 at 21:07
GEdgar
61.9k267168
answered Dec 10 '18 at 21:07
GEdgar
61.9k267168
61.9k267168
add a comment |
add a comment |
1
What do you know about limits? Which techniques do you know? What have you tried? Where are you stuck?
– Arthur
Dec 10 '18 at 20:53
We have that $$frac{2^n}{n!} = 2prod_{3 leq i leq n} frac{2}{i} leq 2left(frac{2}{3}right)^{n-2} xrightarrow[ntoinfty]{} 0$$
– MathematicsStudent1122
Dec 10 '18 at 20:54