Determine $lim_{nto infty}frac{1}{ (n!)!^frac{1}{n!} } $
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I am able to establish that $lim_{nto infty}frac{1}{ n! ^frac{1}{n} } $ converges to 0.
I suspect that $frac{1}{ (n!)!^frac{1}{n!} } $ is a subsequence of $ frac{1}{ n! ^frac{1}{n} } $ which would make the problem much easier to deal with. Is this a valid approach to the question?
real-analysis
asked Dec 4 '18 at 13:39
JamesJames
154
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add a comment |
$begingroup$
I am able to establish that $lim_{nto infty}frac{1}{ n! ^frac{1}{n} } $ converges to 0.
I suspect that $frac{1}{ (n!)!^frac{1}{n!} } $ is a subsequence of $ frac{1}{ n! ^frac{1}{n} } $ which would make the problem much easier to deal with. Is this a valid approach to the question?
real-analysis
asked Dec 4 '18 at 13:39
JamesJames
154
$endgroup$
add a comment |
$begingroup$
I am able to establish that $lim_{nto infty}frac{1}{ n! ^frac{1}{n} } $ converges to 0.
I suspect that $frac{1}{ (n!)!^frac{1}{n!} } $ is a subsequence of $ frac{1}{ n! ^frac{1}{n} } $ which would make the problem much easier to deal with. Is this a valid approach to the question?
real-analysis
asked Dec 4 '18 at 13:39
JamesJames
154
$endgroup$
I am able to establish that $lim_{nto infty}frac{1}{ n! ^frac{1}{n} } $ converges to 0.
I suspect that $frac{1}{ (n!)!^frac{1}{n!} } $ is a subsequence of $ frac{1}{ n! ^frac{1}{n} } $ which would make the problem much easier to deal with. Is this a valid approach to the question?
real-analysis
real-analysis
asked Dec 4 '18 at 13:39
JamesJames
154
asked Dec 4 '18 at 13:39
JamesJames
154
edited Dec 4 '18 at 13:43
James
asked Dec 4 '18 at 13:39
JamesJames
154
asked Dec 4 '18 at 13:39
JamesJames
154
asked Dec 4 '18 at 13:39
JamesJames
154
154
add a comment |
add a comment |
3 Answers
3
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$begingroup$
Hint: write down$$lim_{nto infty}frac{1}{ (n!)!^frac{1}{n!} }{=lim_{mto infty}frac{1}{ (m)!^frac{1}{m} } } $$and apply Stirling's approx. for $m!$.
answered Dec 4 '18 at 13:53
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
Think I got it. Thank you.
$endgroup$
– James
Dec 4 '18 at 13:57
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You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 13:57
1
$begingroup$
This doesn't actually answer the question...
$endgroup$
– TonyK
Dec 4 '18 at 14:10
add a comment |
$begingroup$
We have that
$$frac{1}{ (n!)!^frac{1}{n!} }=e^{frac{log (n!)!}{n!}}$$
and then by $m=n!$ and by $m!ge frac{m^m}{e^m}$
$$frac{log (n!)!}{n!}=frac{log m!}{m}ge log frac{m}{e}$$
answered Dec 4 '18 at 13:44
gimusigimusi
92.9k84494
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add a comment |
$begingroup$
Yes, this is a valid approach to the question, because $(n!)_{ninBbb N}$ is a subsequence of $(n)_{ninBbb N}$. And if a sequence converges, then all of its subsequences converge too.
answered Dec 4 '18 at 14:09
TonyKTonyK
42.8k355134
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint: write down$$lim_{nto infty}frac{1}{ (n!)!^frac{1}{n!} }{=lim_{mto infty}frac{1}{ (m)!^frac{1}{m} } } $$and apply Stirling's approx. for $m!$.
answered Dec 4 '18 at 13:53
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
Think I got it. Thank you.
$endgroup$
– James
Dec 4 '18 at 13:57
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 13:57
1
$begingroup$
This doesn't actually answer the question...
$endgroup$
– TonyK
Dec 4 '18 at 14:10
add a comment |
$begingroup$
Hint: write down$$lim_{nto infty}frac{1}{ (n!)!^frac{1}{n!} }{=lim_{mto infty}frac{1}{ (m)!^frac{1}{m} } } $$and apply Stirling's approx. for $m!$.
answered Dec 4 '18 at 13:53
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
Think I got it. Thank you.
$endgroup$
– James
Dec 4 '18 at 13:57
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 13:57
1
$begingroup$
This doesn't actually answer the question...
$endgroup$
– TonyK
Dec 4 '18 at 14:10
add a comment |
$begingroup$
Hint: write down$$lim_{nto infty}frac{1}{ (n!)!^frac{1}{n!} }{=lim_{mto infty}frac{1}{ (m)!^frac{1}{m} } } $$and apply Stirling's approx. for $m!$.
answered Dec 4 '18 at 13:53
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Hint: write down$$lim_{nto infty}frac{1}{ (n!)!^frac{1}{n!} }{=lim_{mto infty}frac{1}{ (m)!^frac{1}{m} } } $$and apply Stirling's approx. for $m!$.
answered Dec 4 '18 at 13:53
Mostafa AyazMostafa Ayaz
15.6k3939
answered Dec 4 '18 at 13:53
Mostafa AyazMostafa Ayaz
15.6k3939
answered Dec 4 '18 at 13:53
Mostafa AyazMostafa Ayaz
15.6k3939
answered Dec 4 '18 at 13:53
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
$begingroup$
Think I got it. Thank you.
$endgroup$
– James
Dec 4 '18 at 13:57
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 13:57
1
$begingroup$
This doesn't actually answer the question...
$endgroup$
– TonyK
Dec 4 '18 at 14:10
add a comment |
$begingroup$
Think I got it. Thank you.
$endgroup$
– James
Dec 4 '18 at 13:57
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 13:57
1
$begingroup$
This doesn't actually answer the question...
$endgroup$
– TonyK
Dec 4 '18 at 14:10
$begingroup$
Think I got it. Thank you.
$endgroup$
– James
Dec 4 '18 at 13:57
$begingroup$
Think I got it. Thank you.
$endgroup$
– James
Dec 4 '18 at 13:57
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 13:57
$begingroup$
You're welcome. Good luck!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 13:57
1
1
$begingroup$
This doesn't actually answer the question...
$endgroup$
– TonyK
Dec 4 '18 at 14:10
$begingroup$
This doesn't actually answer the question...
$endgroup$
– TonyK
Dec 4 '18 at 14:10
add a comment |
$begingroup$
We have that
$$frac{1}{ (n!)!^frac{1}{n!} }=e^{frac{log (n!)!}{n!}}$$
and then by $m=n!$ and by $m!ge frac{m^m}{e^m}$
$$frac{log (n!)!}{n!}=frac{log m!}{m}ge log frac{m}{e}$$
answered Dec 4 '18 at 13:44
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
We have that
$$frac{1}{ (n!)!^frac{1}{n!} }=e^{frac{log (n!)!}{n!}}$$
and then by $m=n!$ and by $m!ge frac{m^m}{e^m}$
$$frac{log (n!)!}{n!}=frac{log m!}{m}ge log frac{m}{e}$$
answered Dec 4 '18 at 13:44
gimusigimusi
92.9k84494
$endgroup$
add a comment |
$begingroup$
We have that
$$frac{1}{ (n!)!^frac{1}{n!} }=e^{frac{log (n!)!}{n!}}$$
and then by $m=n!$ and by $m!ge frac{m^m}{e^m}$
$$frac{log (n!)!}{n!}=frac{log m!}{m}ge log frac{m}{e}$$
answered Dec 4 '18 at 13:44
gimusigimusi
92.9k84494
$endgroup$
We have that
$$frac{1}{ (n!)!^frac{1}{n!} }=e^{frac{log (n!)!}{n!}}$$
and then by $m=n!$ and by $m!ge frac{m^m}{e^m}$
$$frac{log (n!)!}{n!}=frac{log m!}{m}ge log frac{m}{e}$$
answered Dec 4 '18 at 13:44
gimusigimusi
92.9k84494
edited Dec 4 '18 at 14:02
answered Dec 4 '18 at 13:44
gimusigimusi
92.9k84494
answered Dec 4 '18 at 13:44
gimusigimusi
92.9k84494
answered Dec 4 '18 at 13:44
gimusigimusi
92.9k84494
92.9k84494
add a comment |
add a comment |
$begingroup$
Yes, this is a valid approach to the question, because $(n!)_{ninBbb N}$ is a subsequence of $(n)_{ninBbb N}$. And if a sequence converges, then all of its subsequences converge too.
answered Dec 4 '18 at 14:09
TonyKTonyK
42.8k355134
$endgroup$
add a comment |
$begingroup$
Yes, this is a valid approach to the question, because $(n!)_{ninBbb N}$ is a subsequence of $(n)_{ninBbb N}$. And if a sequence converges, then all of its subsequences converge too.
answered Dec 4 '18 at 14:09
TonyKTonyK
42.8k355134
$endgroup$
add a comment |
$begingroup$
Yes, this is a valid approach to the question, because $(n!)_{ninBbb N}$ is a subsequence of $(n)_{ninBbb N}$. And if a sequence converges, then all of its subsequences converge too.
answered Dec 4 '18 at 14:09
TonyKTonyK
42.8k355134
$endgroup$
Yes, this is a valid approach to the question, because $(n!)_{ninBbb N}$ is a subsequence of $(n)_{ninBbb N}$. And if a sequence converges, then all of its subsequences converge too.
answered Dec 4 '18 at 14:09
TonyKTonyK
42.8k355134
answered Dec 4 '18 at 14:09
TonyKTonyK
42.8k355134
answered Dec 4 '18 at 14:09
TonyKTonyK
42.8k355134
answered Dec 4 '18 at 14:09
TonyKTonyK
42.8k355134
42.8k355134
add a comment |
add a comment |
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