Evaluate the limit: $limlimits_ntoinftyfraclog_a n!n^b, ninBbb N$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How can I prove that $log^k(n) = O(n^epsilon)$?Show that $p_n^1-epsilonle n$ using PNTLimit. $lim_n to inftyfrac1^p+2^p+ldots+n^pn^p+1$.Lim of $(n-1)!^1/n$ as $nrightarrow infty$Evaluate $limlimits_n toinfty left(fracn-1 2n+2right)^n$Limit $mathop lim limits_n to infty fracnleft( a_1…a_n right)^frac1na_1 + … + a_n$Evaluate limit $lim_n to infty 1 over n^k + 1left( k! + (k + 1)! over 1! + cdots + (k + n)! over n! right),k in mathbbN$Evaluate a limit involving powers of $2$Find $limlimits_n to infty fracx_nn$ when $limlimits_n to infty x_n+k-x_n$ existsCan't find a seemingly simple limit $lim_ntoinftyfrac(n+k)!n^n$Evaluate the limit $lim_ntoinftylog_aleft(frac4^nn!n^nright)$Evaluate the limit: $limlimits_ntoinfty frac4^nn!(3n)^n$
Keep at all times, the minus sign above aligned with minus sign below
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Evaluate the limit: $limlimits_ntoinftyfraclog_a n!n^b, ninBbb N$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How can I prove that $log^k(n) = O(n^epsilon)$?Show that $p_n^1-epsilonle n$ using PNTLimit. $lim_n to inftyfrac1^p+2^p+ldots+n^pn^p+1$.Lim of $(n-1)!^1/n$ as $nrightarrow infty$Evaluate $limlimits_n toinfty left(fracn-1 2n+2right)^n$Limit $mathop lim limits_n to infty fracnleft( a_1…a_n right)^frac1na_1 + … + a_n$Evaluate limit $lim_n to infty 1 over n^k + 1left( k! + (k + 1)! over 1! + cdots + (k + n)! over n! right),k in mathbbN$Evaluate a limit involving powers of $2$Find $limlimits_n to infty fracx_nn$ when $limlimits_n to infty x_n+k-x_n$ existsCan't find a seemingly simple limit $lim_ntoinftyfrac(n+k)!n^n$Evaluate the limit $lim_ntoinftylog_aleft(frac4^nn!n^nright)$Evaluate the limit: $limlimits_ntoinfty frac4^nn!(3n)^n$
$begingroup$
Evaluate the limit:
$$
lim_ntoinftyfraclog_a n!n^b, ninBbb N
$$
I've tried to consider two cases: $b < 0$, $b ge 0$. First $b < 0$. This case is simple since the limit becomes:
$$
lim_ntoinftyfraclog_a n!n^b = lim_ntoinftyn^blog_a n! = +infty
$$
Now consider the case when $b ge 0$, then we may apply Cesaro-Stolz theorem, then the limit is equal to the following limit:
$$
beginalign
lim_ntoinfty fraclog_a n!n^b &= lim_ntoinfty fraclog_a (n+1)! - log_an!(n+1)^b - n^b \
&= lim_ntoinfty fraclog_a(n+1)(n+1)^b - n^b
endalign
$$
I've shown earlier that:
$$
lim_ntoinftyleft((n+1)^b - n^bright) = 0, textif bin(0, 1)\
lim_ntoinftyleft((n+1)^b - n^bright) = +infty, textif b > 1\
$$
For $b = 1$ the limit becomes:
$$
lim_ntoinftylog_an!over n = lim_ntoinftylog_asqrt[n]n! = +infty
$$
So it looks like:
$$
b le 1 implies lim_ntoinftyfraclog_a n!n^b = +infty
$$
Here I'm not sure how to handle the case for $b > 1$. What are the steps to handle the case for $b > 1$? I know the limit is $0$, but want to justify that.
real-analysis sequences-and-series limits
edited Apr 8 at 18:36
rtybase
11.7k31534
asked Apr 2 at 16:24
romanroman
2,54721226
$endgroup$
add a comment |
$begingroup$
Evaluate the limit:
$$
lim_ntoinftyfraclog_a n!n^b, ninBbb N
$$
I've tried to consider two cases: $b < 0$, $b ge 0$. First $b < 0$. This case is simple since the limit becomes:
$$
lim_ntoinftyfraclog_a n!n^b = lim_ntoinftyn^blog_a n! = +infty
$$
Now consider the case when $b ge 0$, then we may apply Cesaro-Stolz theorem, then the limit is equal to the following limit:
$$
beginalign
lim_ntoinfty fraclog_a n!n^b &= lim_ntoinfty fraclog_a (n+1)! - log_an!(n+1)^b - n^b \
&= lim_ntoinfty fraclog_a(n+1)(n+1)^b - n^b
endalign
$$
I've shown earlier that:
$$
lim_ntoinftyleft((n+1)^b - n^bright) = 0, textif bin(0, 1)\
lim_ntoinftyleft((n+1)^b - n^bright) = +infty, textif b > 1\
$$
For $b = 1$ the limit becomes:
$$
lim_ntoinftylog_an!over n = lim_ntoinftylog_asqrt[n]n! = +infty
$$
So it looks like:
$$
b le 1 implies lim_ntoinftyfraclog_a n!n^b = +infty
$$
Here I'm not sure how to handle the case for $b > 1$. What are the steps to handle the case for $b > 1$? I know the limit is $0$, but want to justify that.
real-analysis sequences-and-series limits
edited Apr 8 at 18:36
rtybase
11.7k31534
asked Apr 2 at 16:24
romanroman
2,54721226
$endgroup$
1
$begingroup$
$log n! sim n log n$, so you get limit $0$ when $b>1$. So far, you have not completed the case $b=1$.
$endgroup$
– GEdgar
Apr 2 at 16:30
$begingroup$
If you vote down, please provide a reason for that, otherwise, it's not clear what's wrong with the question
$endgroup$
– roman
Apr 8 at 17:46
add a comment |
$begingroup$
Evaluate the limit:
$$
lim_ntoinftyfraclog_a n!n^b, ninBbb N
$$
I've tried to consider two cases: $b < 0$, $b ge 0$. First $b < 0$. This case is simple since the limit becomes:
$$
lim_ntoinftyfraclog_a n!n^b = lim_ntoinftyn^blog_a n! = +infty
$$
Now consider the case when $b ge 0$, then we may apply Cesaro-Stolz theorem, then the limit is equal to the following limit:
$$
beginalign
lim_ntoinfty fraclog_a n!n^b &= lim_ntoinfty fraclog_a (n+1)! - log_an!(n+1)^b - n^b \
&= lim_ntoinfty fraclog_a(n+1)(n+1)^b - n^b
endalign
$$
I've shown earlier that:
$$
lim_ntoinftyleft((n+1)^b - n^bright) = 0, textif bin(0, 1)\
lim_ntoinftyleft((n+1)^b - n^bright) = +infty, textif b > 1\
$$
For $b = 1$ the limit becomes:
$$
lim_ntoinftylog_an!over n = lim_ntoinftylog_asqrt[n]n! = +infty
$$
So it looks like:
$$
b le 1 implies lim_ntoinftyfraclog_a n!n^b = +infty
$$
Here I'm not sure how to handle the case for $b > 1$. What are the steps to handle the case for $b > 1$? I know the limit is $0$, but want to justify that.
real-analysis sequences-and-series limits
edited Apr 8 at 18:36
rtybase
11.7k31534
asked Apr 2 at 16:24
romanroman
2,54721226
$endgroup$
Evaluate the limit:
$$
lim_ntoinftyfraclog_a n!n^b, ninBbb N
$$
I've tried to consider two cases: $b < 0$, $b ge 0$. First $b < 0$. This case is simple since the limit becomes:
$$
lim_ntoinftyfraclog_a n!n^b = lim_ntoinftyn^blog_a n! = +infty
$$
Now consider the case when $b ge 0$, then we may apply Cesaro-Stolz theorem, then the limit is equal to the following limit:
$$
beginalign
lim_ntoinfty fraclog_a n!n^b &= lim_ntoinfty fraclog_a (n+1)! - log_an!(n+1)^b - n^b \
&= lim_ntoinfty fraclog_a(n+1)(n+1)^b - n^b
endalign
$$
I've shown earlier that:
$$
lim_ntoinftyleft((n+1)^b - n^bright) = 0, textif bin(0, 1)\
lim_ntoinftyleft((n+1)^b - n^bright) = +infty, textif b > 1\
$$
For $b = 1$ the limit becomes:
$$
lim_ntoinftylog_an!over n = lim_ntoinftylog_asqrt[n]n! = +infty
$$
So it looks like:
$$
b le 1 implies lim_ntoinftyfraclog_a n!n^b = +infty
$$
Here I'm not sure how to handle the case for $b > 1$. What are the steps to handle the case for $b > 1$? I know the limit is $0$, but want to justify that.
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
edited Apr 8 at 18:36
rtybase
11.7k31534
asked Apr 2 at 16:24
romanroman
2,54721226
edited Apr 8 at 18:36
rtybase
11.7k31534
asked Apr 2 at 16:24
romanroman
2,54721226
edited Apr 8 at 18:36
rtybase
11.7k31534
edited Apr 8 at 18:36
rtybase
11.7k31534
edited Apr 8 at 18:36
rtybase
11.7k31534
11.7k31534
asked Apr 2 at 16:24
romanroman
2,54721226
asked Apr 2 at 16:24
romanroman
2,54721226
asked Apr 2 at 16:24
romanroman
2,54721226
2,54721226
1
$begingroup$
$log n! sim n log n$, so you get limit $0$ when $b>1$. So far, you have not completed the case $b=1$.
$endgroup$
– GEdgar
Apr 2 at 16:30
$begingroup$
If you vote down, please provide a reason for that, otherwise, it's not clear what's wrong with the question
$endgroup$
– roman
Apr 8 at 17:46
add a comment |
1
$begingroup$
$log n! sim n log n$, so you get limit $0$ when $b>1$. So far, you have not completed the case $b=1$.
$endgroup$
– GEdgar
Apr 2 at 16:30
$begingroup$
If you vote down, please provide a reason for that, otherwise, it's not clear what's wrong with the question
$endgroup$
– roman
Apr 8 at 17:46
1
1
$begingroup$
$log n! sim n log n$, so you get limit $0$ when $b>1$. So far, you have not completed the case $b=1$.
$endgroup$
– GEdgar
Apr 2 at 16:30
$begingroup$
$log n! sim n log n$, so you get limit $0$ when $b>1$. So far, you have not completed the case $b=1$.
$endgroup$
– GEdgar
Apr 2 at 16:30
$begingroup$
If you vote down, please provide a reason for that, otherwise, it's not clear what's wrong with the question
$endgroup$
– roman
Apr 8 at 17:46
$begingroup$
If you vote down, please provide a reason for that, otherwise, it's not clear what's wrong with the question
$endgroup$
– roman
Apr 8 at 17:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For $b>1$, use MVT for $f(x)=x^b$, i.e. $exists cin(n,n+1)$ and $n>0$ s.t.
$$(n+1)^b - n^b=bc^b-1 Rightarrow \
b(n+1)^b-1>(n+1)^b - n^b > bn^b-1$$
and from some $n$ onwards, assuming $lna>0$:
$$0<fraclog_a(n+1)b(n+1)^b-1=
fracln(n+1)b(n+1)^b-1lna <
fraclog_a(n+1)(n+1)^b - n^b <
fraclog_a(n+1)bn^b-1=
fracln(n+1)bn^b-1lna$$
because $b-1>0$, RHS goes to $0$ and by squeezing, the limit is $0$. There are quite a few proofs for RHS going to $0$, for example here (proposition 2.2) and here (proposition 2).
For $lna<0$ we have
$$0>fracln(n+1)b(n+1)^b-1lna>
fraclog_a(n+1)(n+1)^b - n^b >
fracln(n+1)bn^b-1lna$$
we the same result.
answered Apr 2 at 20:03
rtybasertybase
11.7k31534
$endgroup$
add a comment |
$begingroup$
Hint:
Use Stirling approximation.
answered Apr 2 at 16:30
EurekaEureka
1,061115
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $b>1$, use MVT for $f(x)=x^b$, i.e. $exists cin(n,n+1)$ and $n>0$ s.t.
$$(n+1)^b - n^b=bc^b-1 Rightarrow \
b(n+1)^b-1>(n+1)^b - n^b > bn^b-1$$
and from some $n$ onwards, assuming $lna>0$:
$$0<fraclog_a(n+1)b(n+1)^b-1=
fracln(n+1)b(n+1)^b-1lna <
fraclog_a(n+1)(n+1)^b - n^b <
fraclog_a(n+1)bn^b-1=
fracln(n+1)bn^b-1lna$$
because $b-1>0$, RHS goes to $0$ and by squeezing, the limit is $0$. There are quite a few proofs for RHS going to $0$, for example here (proposition 2.2) and here (proposition 2).
For $lna<0$ we have
$$0>fracln(n+1)b(n+1)^b-1lna>
fraclog_a(n+1)(n+1)^b - n^b >
fracln(n+1)bn^b-1lna$$
we the same result.
answered Apr 2 at 20:03
rtybasertybase
11.7k31534
$endgroup$
add a comment |
$begingroup$
For $b>1$, use MVT for $f(x)=x^b$, i.e. $exists cin(n,n+1)$ and $n>0$ s.t.
$$(n+1)^b - n^b=bc^b-1 Rightarrow \
b(n+1)^b-1>(n+1)^b - n^b > bn^b-1$$
and from some $n$ onwards, assuming $lna>0$:
$$0<fraclog_a(n+1)b(n+1)^b-1=
fracln(n+1)b(n+1)^b-1lna <
fraclog_a(n+1)(n+1)^b - n^b <
fraclog_a(n+1)bn^b-1=
fracln(n+1)bn^b-1lna$$
because $b-1>0$, RHS goes to $0$ and by squeezing, the limit is $0$. There are quite a few proofs for RHS going to $0$, for example here (proposition 2.2) and here (proposition 2).
For $lna<0$ we have
$$0>fracln(n+1)b(n+1)^b-1lna>
fraclog_a(n+1)(n+1)^b - n^b >
fracln(n+1)bn^b-1lna$$
we the same result.
answered Apr 2 at 20:03
rtybasertybase
11.7k31534
$endgroup$
add a comment |
$begingroup$
For $b>1$, use MVT for $f(x)=x^b$, i.e. $exists cin(n,n+1)$ and $n>0$ s.t.
$$(n+1)^b - n^b=bc^b-1 Rightarrow \
b(n+1)^b-1>(n+1)^b - n^b > bn^b-1$$
and from some $n$ onwards, assuming $lna>0$:
$$0<fraclog_a(n+1)b(n+1)^b-1=
fracln(n+1)b(n+1)^b-1lna <
fraclog_a(n+1)(n+1)^b - n^b <
fraclog_a(n+1)bn^b-1=
fracln(n+1)bn^b-1lna$$
because $b-1>0$, RHS goes to $0$ and by squeezing, the limit is $0$. There are quite a few proofs for RHS going to $0$, for example here (proposition 2.2) and here (proposition 2).
For $lna<0$ we have
$$0>fracln(n+1)b(n+1)^b-1lna>
fraclog_a(n+1)(n+1)^b - n^b >
fracln(n+1)bn^b-1lna$$
we the same result.
answered Apr 2 at 20:03
rtybasertybase
11.7k31534
$endgroup$
For $b>1$, use MVT for $f(x)=x^b$, i.e. $exists cin(n,n+1)$ and $n>0$ s.t.
$$(n+1)^b - n^b=bc^b-1 Rightarrow \
b(n+1)^b-1>(n+1)^b - n^b > bn^b-1$$
and from some $n$ onwards, assuming $lna>0$:
$$0<fraclog_a(n+1)b(n+1)^b-1=
fracln(n+1)b(n+1)^b-1lna <
fraclog_a(n+1)(n+1)^b - n^b <
fraclog_a(n+1)bn^b-1=
fracln(n+1)bn^b-1lna$$
because $b-1>0$, RHS goes to $0$ and by squeezing, the limit is $0$. There are quite a few proofs for RHS going to $0$, for example here (proposition 2.2) and here (proposition 2).
For $lna<0$ we have
$$0>fracln(n+1)b(n+1)^b-1lna>
fraclog_a(n+1)(n+1)^b - n^b >
fracln(n+1)bn^b-1lna$$
we the same result.
answered Apr 2 at 20:03
rtybasertybase
11.7k31534
edited Apr 2 at 20:21
answered Apr 2 at 20:03
rtybasertybase
11.7k31534
answered Apr 2 at 20:03
rtybasertybase
11.7k31534
answered Apr 2 at 20:03
rtybasertybase
11.7k31534
11.7k31534
add a comment |
add a comment |
$begingroup$
Hint:
Use Stirling approximation.
answered Apr 2 at 16:30
EurekaEureka
1,061115
$endgroup$
add a comment |
$begingroup$
Hint:
Use Stirling approximation.
answered Apr 2 at 16:30
EurekaEureka
1,061115
$endgroup$
add a comment |
$begingroup$
Hint:
Use Stirling approximation.
answered Apr 2 at 16:30
EurekaEureka
1,061115
$endgroup$
Hint:
Use Stirling approximation.
answered Apr 2 at 16:30
EurekaEureka
1,061115
answered Apr 2 at 16:30
EurekaEureka
1,061115
answered Apr 2 at 16:30
EurekaEureka
1,061115
answered Apr 2 at 16:30
EurekaEureka
1,061115
1,061115
add a comment |
add a comment |
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$begingroup$
$log n! sim n log n$, so you get limit $0$ when $b>1$. So far, you have not completed the case $b=1$.
$endgroup$
– GEdgar
Apr 2 at 16:30
$begingroup$
If you vote down, please provide a reason for that, otherwise, it's not clear what's wrong with the question
$endgroup$
– roman
Apr 8 at 17:46