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Calculate the limit of $lim_krightarrowinfty(k!)^frac1k$ [duplicate]
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$limlimits_n to+inftysqrt[n]n!$ is infinite
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$$lim_krightarrowinfty(k!)^frac1k$$
I want to determine the limit as formal as possible, that is, to prove, for example
$$lim_krightarrowinftyfracln(k!)k$$
I had the intention of using the L'Hopital rule and I do not know that it does exist, despite only representing a number. But it is not very justified.
Can you use sanduche theorem? Can I limit?
calculus real-analysis complex-analysis
edited Aug 5 at 23:38
Asaf Karagila♦
292k31403733
asked Aug 5 at 18:19
Santiago Seeker
597
marked as duplicate by Hans Lundmark, Claude Leibovici calculus
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Aug 6 at 4:49
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 |Â
up vote
1
down vote
favorite
This question already has an answer here:
$limlimits_n to+inftysqrt[n]n!$ is infinite
12 answers
$$lim_krightarrowinfty(k!)^frac1k$$
I want to determine the limit as formal as possible, that is, to prove, for example
$$lim_krightarrowinftyfracln(k!)k$$
I had the intention of using the L'Hopital rule and I do not know that it does exist, despite only representing a number. But it is not very justified.
Can you use sanduche theorem? Can I limit?
calculus real-analysis complex-analysis
edited Aug 5 at 23:38
Asaf Karagila♦
292k31403733
asked Aug 5 at 18:19
Santiago Seeker
597
marked as duplicate by Hans Lundmark, Claude Leibovici calculus
Users with the  calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Aug 6 at 4:49
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 |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
$limlimits_n to+inftysqrt[n]n!$ is infinite
12 answers
$$lim_krightarrowinfty(k!)^frac1k$$
I want to determine the limit as formal as possible, that is, to prove, for example
$$lim_krightarrowinftyfracln(k!)k$$
I had the intention of using the L'Hopital rule and I do not know that it does exist, despite only representing a number. But it is not very justified.
Can you use sanduche theorem? Can I limit?
calculus real-analysis complex-analysis
edited Aug 5 at 23:38
Asaf Karagila♦
292k31403733
asked Aug 5 at 18:19
Santiago Seeker
597
This question already has an answer here:
$limlimits_n to+inftysqrt[n]n!$ is infinite
12 answers
$$lim_krightarrowinfty(k!)^frac1k$$
I want to determine the limit as formal as possible, that is, to prove, for example
$$lim_krightarrowinftyfracln(k!)k$$
I had the intention of using the L'Hopital rule and I do not know that it does exist, despite only representing a number. But it is not very justified.
Can you use sanduche theorem? Can I limit?
This question already has an answer here:
$limlimits_n to+inftysqrt[n]n!$ is infinite
12 answers
calculus real-analysis complex-analysis
edited Aug 5 at 23:38
Asaf Karagila♦
292k31403733
asked Aug 5 at 18:19
Santiago Seeker
597
edited Aug 5 at 23:38
Asaf Karagila♦
292k31403733
edited Aug 5 at 23:38
Asaf Karagila♦
292k31403733
edited Aug 5 at 23:38
Asaf Karagila♦
292k31403733
292k31403733
asked Aug 5 at 18:19
Santiago Seeker
597
asked Aug 5 at 18:19
Santiago Seeker
597
asked Aug 5 at 18:19
Santiago Seeker
597
597
marked as duplicate by Hans Lundmark, Claude Leibovici calculus
Users with the  calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Aug 6 at 4:49
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 |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
As an alternative by Stolz-Cesaro
$$lim_krightarrowinftyfracln(k!)k=lim_krightarrowinftyfracln((k+1)!)-ln(k!)k+1-k=lim_krightarrowinfty log (k+1) =infty$$
answered Aug 5 at 19:09
gimusi
65.5k73684
add a comment |Â
up vote
2
down vote
Hint: for a sequence of positive numbers $a_n$, if $lim_nrightarrowinfty(fraca_n+1a_n)$ exists and equals $L$ ($L$ can be a real number of infinity, it doesn't matter) then $lim_nrightarrowinfty(sqrt[n]a_n)=L$.
answered Aug 5 at 18:27
Mark
93619
Thank you very much, I had not considered it.
– Santiago Seeker
Aug 5 at 18:56
add a comment |Â
up vote
2
down vote
HINT:
Note that
$$beginalign
fraclog(k!)k&=frac1ksum_n=1^k log(n)\\
&=log(k)+frac1ksum_n=1^k log(n/k)tag1
endalign$$
The sum on the right-hand side of $(1)$ is the Riemann sum for the improper integral $lim_varepsilonto0int_varepsilon^1 log(x),dx$.
Can you finish this?
answered Aug 5 at 18:43
Mark Viola
126k1172167
Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
– Santiago Seeker
Aug 5 at 18:57
You're welcome. My pleasure.
– Mark Viola
Aug 5 at 19:11
Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
– zhw.
Aug 5 at 22:29
No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
– zhw.
Aug 5 at 22:40
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
Aug 5 at 22:54
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
As an alternative by Stolz-Cesaro
$$lim_krightarrowinftyfracln(k!)k=lim_krightarrowinftyfracln((k+1)!)-ln(k!)k+1-k=lim_krightarrowinfty log (k+1) =infty$$
answered Aug 5 at 19:09
gimusi
65.5k73684
add a comment |Â
up vote
2
down vote
accepted
As an alternative by Stolz-Cesaro
$$lim_krightarrowinftyfracln(k!)k=lim_krightarrowinftyfracln((k+1)!)-ln(k!)k+1-k=lim_krightarrowinfty log (k+1) =infty$$
answered Aug 5 at 19:09
gimusi
65.5k73684
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
As an alternative by Stolz-Cesaro
$$lim_krightarrowinftyfracln(k!)k=lim_krightarrowinftyfracln((k+1)!)-ln(k!)k+1-k=lim_krightarrowinfty log (k+1) =infty$$
answered Aug 5 at 19:09
gimusi
65.5k73684
As an alternative by Stolz-Cesaro
$$lim_krightarrowinftyfracln(k!)k=lim_krightarrowinftyfracln((k+1)!)-ln(k!)k+1-k=lim_krightarrowinfty log (k+1) =infty$$
answered Aug 5 at 19:09
gimusi
65.5k73684
answered Aug 5 at 19:09
gimusi
65.5k73684
answered Aug 5 at 19:09
gimusi
65.5k73684
answered Aug 5 at 19:09
gimusi
65.5k73684
65.5k73684
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint: for a sequence of positive numbers $a_n$, if $lim_nrightarrowinfty(fraca_n+1a_n)$ exists and equals $L$ ($L$ can be a real number of infinity, it doesn't matter) then $lim_nrightarrowinfty(sqrt[n]a_n)=L$.
answered Aug 5 at 18:27
Mark
93619
Thank you very much, I had not considered it.
– Santiago Seeker
Aug 5 at 18:56
add a comment |Â
up vote
2
down vote
Hint: for a sequence of positive numbers $a_n$, if $lim_nrightarrowinfty(fraca_n+1a_n)$ exists and equals $L$ ($L$ can be a real number of infinity, it doesn't matter) then $lim_nrightarrowinfty(sqrt[n]a_n)=L$.
answered Aug 5 at 18:27
Mark
93619
Thank you very much, I had not considered it.
– Santiago Seeker
Aug 5 at 18:56
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: for a sequence of positive numbers $a_n$, if $lim_nrightarrowinfty(fraca_n+1a_n)$ exists and equals $L$ ($L$ can be a real number of infinity, it doesn't matter) then $lim_nrightarrowinfty(sqrt[n]a_n)=L$.
answered Aug 5 at 18:27
Mark
93619
Hint: for a sequence of positive numbers $a_n$, if $lim_nrightarrowinfty(fraca_n+1a_n)$ exists and equals $L$ ($L$ can be a real number of infinity, it doesn't matter) then $lim_nrightarrowinfty(sqrt[n]a_n)=L$.
answered Aug 5 at 18:27
Mark
93619
answered Aug 5 at 18:27
Mark
93619
answered Aug 5 at 18:27
Mark
93619
answered Aug 5 at 18:27
Mark
93619
93619
Thank you very much, I had not considered it.
– Santiago Seeker
Aug 5 at 18:56
add a comment |Â
Thank you very much, I had not considered it.
– Santiago Seeker
Aug 5 at 18:56
Thank you very much, I had not considered it.
– Santiago Seeker
Aug 5 at 18:56
Thank you very much, I had not considered it.
– Santiago Seeker
Aug 5 at 18:56
add a comment |Â
up vote
2
down vote
HINT:
Note that
$$beginalign
fraclog(k!)k&=frac1ksum_n=1^k log(n)\\
&=log(k)+frac1ksum_n=1^k log(n/k)tag1
endalign$$
The sum on the right-hand side of $(1)$ is the Riemann sum for the improper integral $lim_varepsilonto0int_varepsilon^1 log(x),dx$.
Can you finish this?
answered Aug 5 at 18:43
Mark Viola
126k1172167
Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
– Santiago Seeker
Aug 5 at 18:57
You're welcome. My pleasure.
– Mark Viola
Aug 5 at 19:11
Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
– zhw.
Aug 5 at 22:29
No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
– zhw.
Aug 5 at 22:40
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
Aug 5 at 22:54
add a comment |Â
up vote
2
down vote
HINT:
Note that
$$beginalign
fraclog(k!)k&=frac1ksum_n=1^k log(n)\\
&=log(k)+frac1ksum_n=1^k log(n/k)tag1
endalign$$
The sum on the right-hand side of $(1)$ is the Riemann sum for the improper integral $lim_varepsilonto0int_varepsilon^1 log(x),dx$.
Can you finish this?
answered Aug 5 at 18:43
Mark Viola
126k1172167
Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
– Santiago Seeker
Aug 5 at 18:57
You're welcome. My pleasure.
– Mark Viola
Aug 5 at 19:11
Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
– zhw.
Aug 5 at 22:29
No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
– zhw.
Aug 5 at 22:40
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
Aug 5 at 22:54
add a comment |Â
up vote
2
down vote
up vote
2
down vote
HINT:
Note that
$$beginalign
fraclog(k!)k&=frac1ksum_n=1^k log(n)\\
&=log(k)+frac1ksum_n=1^k log(n/k)tag1
endalign$$
The sum on the right-hand side of $(1)$ is the Riemann sum for the improper integral $lim_varepsilonto0int_varepsilon^1 log(x),dx$.
Can you finish this?
answered Aug 5 at 18:43
Mark Viola
126k1172167
HINT:
Note that
$$beginalign
fraclog(k!)k&=frac1ksum_n=1^k log(n)\\
&=log(k)+frac1ksum_n=1^k log(n/k)tag1
endalign$$
The sum on the right-hand side of $(1)$ is the Riemann sum for the improper integral $lim_varepsilonto0int_varepsilon^1 log(x),dx$.
Can you finish this?
answered Aug 5 at 18:43
Mark Viola
126k1172167
answered Aug 5 at 18:43
Mark Viola
126k1172167
answered Aug 5 at 18:43
Mark Viola
126k1172167
answered Aug 5 at 18:43
Mark Viola
126k1172167
126k1172167
Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
– Santiago Seeker
Aug 5 at 18:57
You're welcome. My pleasure.
– Mark Viola
Aug 5 at 19:11
Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
– zhw.
Aug 5 at 22:29
No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
– zhw.
Aug 5 at 22:40
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
Aug 5 at 22:54
add a comment |Â
Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
– Santiago Seeker
Aug 5 at 18:57
You're welcome. My pleasure.
– Mark Viola
Aug 5 at 19:11
Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
– zhw.
Aug 5 at 22:29
No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
– zhw.
Aug 5 at 22:40
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
Aug 5 at 22:54
Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
– Santiago Seeker
Aug 5 at 18:57
Oh, wow! that great ... is the criterion of the integral and with that I show that it is divergent :) thank you very much !!
– Santiago Seeker
Aug 5 at 18:57
You're welcome. My pleasure.
– Mark Viola
Aug 5 at 19:11
You're welcome. My pleasure.
– Mark Viola
Aug 5 at 19:11
Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
– zhw.
Aug 5 at 22:29
Avoiding the problem of why those Riemann sums converge to the improper integral, we could estimate $$sum_j=1^kln j ge (k/2)ln (k/2),$$ from which the result follows.
– zhw.
Aug 5 at 22:29
No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
– zhw.
Aug 5 at 22:40
No concavity needed, just need $ln j$ increasing. And I'm doing fine thank you, and I hope the same is true your way.
– zhw.
Aug 5 at 22:40
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
Aug 5 at 22:54
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
Aug 5 at 22:54
add a comment |Â