Integral sequence











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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







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  • 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















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







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  • 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













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







share|cite|improve this question















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






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edited Nov 20 at 21:35

























asked Nov 20 at 20:44









Numbers

1116




1116












  • 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




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










2 Answers
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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)






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  • actually, this is what I did !! thx
    – Numbers
    Nov 20 at 22:13


















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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.






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  • Aaah, sure... easy and nice!! thx a lot!
    – Numbers
    Nov 20 at 21:19











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2 Answers
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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)






share|cite|improve this answer























  • actually, this is what I did !! thx
    – Numbers
    Nov 20 at 22:13















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)






share|cite|improve this answer























  • actually, this is what I did !! thx
    – Numbers
    Nov 20 at 22:13













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)






share|cite|improve this answer














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)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 20 at 21:55

























answered Nov 20 at 21:38









Amin Sassi

162




162












  • 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




actually, this is what I did !! thx
– Numbers
Nov 20 at 22:13










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.






share|cite|improve this answer





















  • Aaah, sure... easy and nice!! thx a lot!
    – Numbers
    Nov 20 at 21:19















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.






share|cite|improve this answer





















  • Aaah, sure... easy and nice!! thx a lot!
    – Numbers
    Nov 20 at 21:19













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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 at 20:49









José Carlos Santos

147k22117217




147k22117217












  • 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




Aaah, sure... easy and nice!! thx a lot!
– Numbers
Nov 20 at 21:19


















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