Mixing
Calculate the limit of $lim_krightarrowinfty(k!)^frac1k$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
$$lim_krightarrowinfty(k!)^frac1k$$
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 4 hours ago
Asaf Karagila
291k31400731
asked 9 hours ago
Santiago Seeker
577
add a comment |Â
up vote
1
down vote
favorite
$$lim_krightarrowinfty(k!)^frac1k$$
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 4 hours ago
Asaf Karagila
291k31400731
asked 9 hours ago
Santiago Seeker
577
Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
– Hans Lundmark
8 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$lim_krightarrowinfty(k!)^frac1k$$
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 4 hours ago
Asaf Karagila
291k31400731
asked 9 hours ago
Santiago Seeker
577
$$lim_krightarrowinfty(k!)^frac1k$$
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 4 hours ago
Asaf Karagila
291k31400731
asked 9 hours ago
Santiago Seeker
577
edited 4 hours ago
Asaf Karagila
291k31400731
edited 4 hours ago
Asaf Karagila
291k31400731
edited 4 hours ago
Asaf Karagila
291k31400731
291k31400731
asked 9 hours ago
Santiago Seeker
577
asked 9 hours ago
Santiago Seeker
577
asked 9 hours ago
Santiago Seeker
577
577
Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
– Hans Lundmark
8 hours ago
add a comment |Â
Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
– Hans Lundmark
8 hours ago
Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
– Hans Lundmark
8 hours ago
Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
– Hans Lundmark
8 hours ago
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 8 hours ago
gimusi
63.4k73379
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 9 hours ago
Mark
5849
Thank you very much, I had not considered it.
– Santiago Seeker
8 hours ago
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 9 hours ago
Mark Viola
126k1170165
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
8 hours ago
You're welcome. My pleasure.
– Mark Viola
8 hours ago
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.
5 hours ago
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.
5 hours ago
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
4 hours ago
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 8 hours ago
gimusi
63.4k73379
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 8 hours ago
gimusi
63.4k73379
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 8 hours ago
gimusi
63.4k73379
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 8 hours ago
gimusi
63.4k73379
answered 8 hours ago
gimusi
63.4k73379
answered 8 hours ago
gimusi
63.4k73379
answered 8 hours ago
gimusi
63.4k73379
63.4k73379
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 9 hours ago
Mark
5849
Thank you very much, I had not considered it.
– Santiago Seeker
8 hours ago
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 9 hours ago
Mark
5849
Thank you very much, I had not considered it.
– Santiago Seeker
8 hours ago
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 9 hours ago
Mark
5849
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 9 hours ago
Mark
5849
answered 9 hours ago
Mark
5849
answered 9 hours ago
Mark
5849
answered 9 hours ago
Mark
5849
5849
Thank you very much, I had not considered it.
– Santiago Seeker
8 hours ago
add a comment |Â
Thank you very much, I had not considered it.
– Santiago Seeker
8 hours ago
Thank you very much, I had not considered it.
– Santiago Seeker
8 hours ago
Thank you very much, I had not considered it.
– Santiago Seeker
8 hours ago
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 9 hours ago
Mark Viola
126k1170165
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
8 hours ago
You're welcome. My pleasure.
– Mark Viola
8 hours ago
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.
5 hours ago
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.
5 hours ago
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
4 hours ago
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 9 hours ago
Mark Viola
126k1170165
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
8 hours ago
You're welcome. My pleasure.
– Mark Viola
8 hours ago
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.
5 hours ago
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.
5 hours ago
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
4 hours ago
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 9 hours ago
Mark Viola
126k1170165
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 9 hours ago
Mark Viola
126k1170165
answered 9 hours ago
Mark Viola
126k1170165
answered 9 hours ago
Mark Viola
126k1170165
answered 9 hours ago
Mark Viola
126k1170165
126k1170165
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
8 hours ago
You're welcome. My pleasure.
– Mark Viola
8 hours ago
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.
5 hours ago
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.
5 hours ago
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
4 hours ago
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
8 hours ago
You're welcome. My pleasure.
– Mark Viola
8 hours ago
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.
5 hours ago
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.
5 hours ago
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
4 hours ago
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
8 hours ago
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
8 hours ago
You're welcome. My pleasure.
– Mark Viola
8 hours ago
You're welcome. My pleasure.
– Mark Viola
8 hours ago
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.
5 hours ago
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.
5 hours ago
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.
5 hours ago
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.
5 hours ago
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
4 hours ago
I see. You're using $n!ge (n/2)^n/2$ and $log(x)$ is increasing. I missed that upon first glance.
– Mark Viola
4 hours ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873237%2fcalculate-the-limit-of-lim-k-rightarrow-inftyk-frac1k%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Possible duplicate of $limlimits_n to+inftysqrt[n]n!$ is infinite
– Hans Lundmark
8 hours ago