Integral sequence
up vote
1
down vote
favorite
So we have this sequence, $$I_{n}=int_0^1ln(1+x^n):dx$$ and we have to show that this is monotone and bounded.
My solution is this: $$xin[0,1]implies 0leqln(1+x^n)leqln(2)$$ and if we integrate from $0$ to $1$ we get $$0leq I_{n}leqln(2)$$ which proves that $I_{n}$ is bounded. Is this correct?
To show that the sequence is monotone I was thinking about the difference $I_{n}-I_{n+1}$. Is this correct too?$$ $$
!!!EDIT!!! Now we have to show that $I_{n}leqfrac{1}{n+1},::forall:ninmathbb{N}$ and to find $limlimits_{xtoinfty}I_n$.$$$$So, I guess is obvious that $I_nleqint_{0}^{1}x^n:dx$, right???
integration
asked Nov 20 at 20:44
Numbers
1045
add a comment |
up vote
1
down vote
favorite
So we have this sequence, $$I_{n}=int_0^1ln(1+x^n):dx$$ and we have to show that this is monotone and bounded.
My solution is this: $$xin[0,1]implies 0leqln(1+x^n)leqln(2)$$ and if we integrate from $0$ to $1$ we get $$0leq I_{n}leqln(2)$$ which proves that $I_{n}$ is bounded. Is this correct?
To show that the sequence is monotone I was thinking about the difference $I_{n}-I_{n+1}$. Is this correct too?$$ $$
!!!EDIT!!! Now we have to show that $I_{n}leqfrac{1}{n+1},::forall:ninmathbb{N}$ and to find $limlimits_{xtoinfty}I_n$.$$$$So, I guess is obvious that $I_nleqint_{0}^{1}x^n:dx$, right???
integration
asked Nov 20 at 20:44
Numbers
1045
For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 at 20:48
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So we have this sequence, $$I_{n}=int_0^1ln(1+x^n):dx$$ and we have to show that this is monotone and bounded.
My solution is this: $$xin[0,1]implies 0leqln(1+x^n)leqln(2)$$ and if we integrate from $0$ to $1$ we get $$0leq I_{n}leqln(2)$$ which proves that $I_{n}$ is bounded. Is this correct?
To show that the sequence is monotone I was thinking about the difference $I_{n}-I_{n+1}$. Is this correct too?$$ $$
!!!EDIT!!! Now we have to show that $I_{n}leqfrac{1}{n+1},::forall:ninmathbb{N}$ and to find $limlimits_{xtoinfty}I_n$.$$$$So, I guess is obvious that $I_nleqint_{0}^{1}x^n:dx$, right???
integration
asked Nov 20 at 20:44
Numbers
1045
So we have this sequence, $$I_{n}=int_0^1ln(1+x^n):dx$$ and we have to show that this is monotone and bounded.
My solution is this: $$xin[0,1]implies 0leqln(1+x^n)leqln(2)$$ and if we integrate from $0$ to $1$ we get $$0leq I_{n}leqln(2)$$ which proves that $I_{n}$ is bounded. Is this correct?
To show that the sequence is monotone I was thinking about the difference $I_{n}-I_{n+1}$. Is this correct too?$$ $$
!!!EDIT!!! Now we have to show that $I_{n}leqfrac{1}{n+1},::forall:ninmathbb{N}$ and to find $limlimits_{xtoinfty}I_n$.$$$$So, I guess is obvious that $I_nleqint_{0}^{1}x^n:dx$, right???
integration
integration
asked Nov 20 at 20:44
Numbers
1045
asked Nov 20 at 20:44
Numbers
1045
edited Nov 20 at 21:35
asked Nov 20 at 20:44
Numbers
1045
asked Nov 20 at 20:44
Numbers
1045
asked Nov 20 at 20:44
Numbers
1045
1045
For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 at 20:48
add a comment |
For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 at 20:48
For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 at 20:48
For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 at 20:48
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
For the last question. you can derivate $ln(1+x^n)-x^n$ prove it's decreasing and then for x=0 it's equal 0 so in [0;1] the difference is negative.
then you need to integrate $x^n$ and for the limite use sandwich theorem( also called gent d'arme)
answered Nov 20 at 21:38
Amin Sassi
162
New contributor
Amin Sassi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
actually, this is what I did !! thx
– Numbers
Nov 20 at 22:13
add a comment |
up vote
1
down vote
What you did is correct. In order to show that $(I_n)_{ninmathbb N}$ is monotonic, you do$$bigl(forall xin[0,1]bigr):1+x^{n+1}leqslant1+x^nimpliesint_0^1ln(1+x^{n+1}),mathrm dxleqslantint_0^1ln(1+x^n),mathrm dx,$$since $ln$ is increasing.
answered Nov 20 at 20:49
José Carlos Santos
141k19111207
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 at 21:19
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
For the last question. you can derivate $ln(1+x^n)-x^n$ prove it's decreasing and then for x=0 it's equal 0 so in [0;1] the difference is negative.
then you need to integrate $x^n$ and for the limite use sandwich theorem( also called gent d'arme)
answered Nov 20 at 21:38
Amin Sassi
162
New contributor
Amin Sassi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
actually, this is what I did !! thx
– Numbers
Nov 20 at 22:13
add a comment |
up vote
0
down vote
accepted
For the last question. you can derivate $ln(1+x^n)-x^n$ prove it's decreasing and then for x=0 it's equal 0 so in [0;1] the difference is negative.
then you need to integrate $x^n$ and for the limite use sandwich theorem( also called gent d'arme)
answered Nov 20 at 21:38
Amin Sassi
162
New contributor
Amin Sassi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
actually, this is what I did !! thx
– Numbers
Nov 20 at 22:13
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
For the last question. you can derivate $ln(1+x^n)-x^n$ prove it's decreasing and then for x=0 it's equal 0 so in [0;1] the difference is negative.
then you need to integrate $x^n$ and for the limite use sandwich theorem( also called gent d'arme)
answered Nov 20 at 21:38
Amin Sassi
162
New contributor
Amin Sassi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For the last question. you can derivate $ln(1+x^n)-x^n$ prove it's decreasing and then for x=0 it's equal 0 so in [0;1] the difference is negative.
then you need to integrate $x^n$ and for the limite use sandwich theorem( also called gent d'arme)
answered Nov 20 at 21:38
Amin Sassi
162
New contributor
Amin Sassi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 20 at 21:55
answered Nov 20 at 21:38
Amin Sassi
162
New contributor
Amin Sassi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 20 at 21:38
Amin Sassi
162
answered Nov 20 at 21:38
Amin Sassi
162
162
New contributor
Amin Sassi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Amin Sassi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Amin Sassi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
actually, this is what I did !! thx
– Numbers
Nov 20 at 22:13
add a comment |
actually, this is what I did !! thx
– Numbers
Nov 20 at 22:13
actually, this is what I did !! thx
– Numbers
Nov 20 at 22:13
actually, this is what I did !! thx
– Numbers
Nov 20 at 22:13
add a comment |
up vote
1
down vote
What you did is correct. In order to show that $(I_n)_{ninmathbb N}$ is monotonic, you do$$bigl(forall xin[0,1]bigr):1+x^{n+1}leqslant1+x^nimpliesint_0^1ln(1+x^{n+1}),mathrm dxleqslantint_0^1ln(1+x^n),mathrm dx,$$since $ln$ is increasing.
answered Nov 20 at 20:49
José Carlos Santos
141k19111207
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 at 21:19
add a comment |
up vote
1
down vote
What you did is correct. In order to show that $(I_n)_{ninmathbb N}$ is monotonic, you do$$bigl(forall xin[0,1]bigr):1+x^{n+1}leqslant1+x^nimpliesint_0^1ln(1+x^{n+1}),mathrm dxleqslantint_0^1ln(1+x^n),mathrm dx,$$since $ln$ is increasing.
answered Nov 20 at 20:49
José Carlos Santos
141k19111207
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 at 21:19
add a comment |
up vote
1
down vote
up vote
1
down vote
What you did is correct. In order to show that $(I_n)_{ninmathbb N}$ is monotonic, you do$$bigl(forall xin[0,1]bigr):1+x^{n+1}leqslant1+x^nimpliesint_0^1ln(1+x^{n+1}),mathrm dxleqslantint_0^1ln(1+x^n),mathrm dx,$$since $ln$ is increasing.
answered Nov 20 at 20:49
José Carlos Santos
141k19111207
What you did is correct. In order to show that $(I_n)_{ninmathbb N}$ is monotonic, you do$$bigl(forall xin[0,1]bigr):1+x^{n+1}leqslant1+x^nimpliesint_0^1ln(1+x^{n+1}),mathrm dxleqslantint_0^1ln(1+x^n),mathrm dx,$$since $ln$ is increasing.
answered Nov 20 at 20:49
José Carlos Santos
141k19111207
answered Nov 20 at 20:49
José Carlos Santos
141k19111207
answered Nov 20 at 20:49
José Carlos Santos
141k19111207
answered Nov 20 at 20:49
José Carlos Santos
141k19111207
141k19111207
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 at 21:19
add a comment |
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 at 21:19
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 at 21:19
Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 at 21:19
add a comment |
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006865%2fintegral-sequence%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
For a fixed $xin(0,1)$, $x^{n+1}< x^n$ implies $log(1+x^{n+1})<log(1+x^n)$ and $I_n$ is obviously decreasing.
– Jack D'Aurizio
Nov 20 at 20:48