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Is it true that $2frac(2n+1)^2n+1(2n+2)^2n+2 le fracn^n(n+1)^n+1 le 2frac(2n)^2n(2n+1)^2n+1$
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I conjectured the following for positive integers $n$
$$2frac(2n+1)^2n+1(2n+2)^2n+2 le fracn^n(n+1)^n+1 le 2frac(2n)^2n(2n+1)^2n+1$$
It seems to be true for the (few) values I have tried and I was wondering if it is true for all $n in mathbb N_>0$? If so how might I see this?
inequality
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I conjectured the following for positive integers $n$
$$2frac(2n+1)^2n+1(2n+2)^2n+2 le fracn^n(n+1)^n+1 le 2frac(2n)^2n(2n+1)^2n+1$$
It seems to be true for the (few) values I have tried and I was wondering if it is true for all $n in mathbb N_>0$? If so how might I see this?
inequality
How about by induction? How many values (and which values) have you tested?
– amWhy
Jul 15 at 23:16
I tried $n=1$ and $n=24$. I tried induction but the induction step doesn't come out
– user544680
Jul 15 at 23:26
1
The first test: Plot them.
– David G. Stork
Jul 15 at 23:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I conjectured the following for positive integers $n$
$$2frac(2n+1)^2n+1(2n+2)^2n+2 le fracn^n(n+1)^n+1 le 2frac(2n)^2n(2n+1)^2n+1$$
It seems to be true for the (few) values I have tried and I was wondering if it is true for all $n in mathbb N_>0$? If so how might I see this?
inequality
I conjectured the following for positive integers $n$
$$2frac(2n+1)^2n+1(2n+2)^2n+2 le fracn^n(n+1)^n+1 le 2frac(2n)^2n(2n+1)^2n+1$$
It seems to be true for the (few) values I have tried and I was wondering if it is true for all $n in mathbb N_>0$? If so how might I see this?
inequality
asked Jul 15 at 23:12
user544680
How about by induction? How many values (and which values) have you tested?
– amWhy
Jul 15 at 23:16
I tried $n=1$ and $n=24$. I tried induction but the induction step doesn't come out
– user544680
Jul 15 at 23:26
1
The first test: Plot them.
– David G. Stork
Jul 15 at 23:37
add a comment |Â
How about by induction? How many values (and which values) have you tested?
– amWhy
Jul 15 at 23:16
I tried $n=1$ and $n=24$. I tried induction but the induction step doesn't come out
– user544680
Jul 15 at 23:26
1
The first test: Plot them.
– David G. Stork
Jul 15 at 23:37
How about by induction? How many values (and which values) have you tested?
– amWhy
Jul 15 at 23:16
How about by induction? How many values (and which values) have you tested?
– amWhy
Jul 15 at 23:16
I tried $n=1$ and $n=24$. I tried induction but the induction step doesn't come out
– user544680
Jul 15 at 23:26
I tried $n=1$ and $n=24$. I tried induction but the induction step doesn't come out
– user544680
Jul 15 at 23:26
1
1
The first test: Plot them.
– David G. Stork
Jul 15 at 23:37
The first test: Plot them.
– David G. Stork
Jul 15 at 23:37
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
To prove
$$
2frac(2n+1)^2n+1(2n+2)^2n+2lefracn^n(n+1)^n+1le2frac(2n)^2n(2n+1)^2n+1
$$
note that the left-hand inequality is equivalent to
$$
left(1+frac12n+1right)^2n+1geleft(1+frac1nright)^n
$$
which follows from the monotonic increase of $left(1+frac1nright)^n$ proven in this answer using only Bernoulli's Inequality.
The right-hand inequality is equivalent to
$$
left(1+frac1nright)^n+1geleft(1+frac12nright)^2n+1
$$
which follows from the monotonic decrease of $left(1+frac1nright)^n+1$ proven in the same answer, again using only Bernoulli's Inequality.
answered Jul 16 at 5:22
robjohn♦
258k26297612
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up vote
0
down vote
The function
begineqnarray*
left( 1+ frac1x right)^x
endeqnarray*
is increasing so
begineqnarray*
left( 1+ frac12n right)^2n leq left( 1+ frac12n+1 right)^2n+1 leq left( 1+ frac12n+1 right)^2n+2.
endeqnarray*
Square root this to get
begineqnarray*
left( 1+ frac12n right)^n leq left( 1+ frac12n+1 right)^n+1.
endeqnarray*
This can be rearranged to give the upper bound ...
begineqnarray*
n ln(1+frac12n) leq (n+1) ln (1+frac12n+1) \
endeqnarray*
begineqnarray*
n( ln(n+frac12) -ln(n)) leq (n+1) (ln(n+1)- ln(n+frac12)) \
endeqnarray*
begineqnarray*(2n+1) ln(n+frac12) leq n ln (n) +(n+1) ln(n+1) \
endeqnarray*
begineqnarray*n ln n +(2n+1) ln(2n+1) leq ln 2 +2n ln(2n) +(n+1) ln(n+1) \
endeqnarray*
begineqnarray*
fracn^n(n+1)^n+1 leq 2 frac(2n)^2n(2n+1)^2n+1.
endeqnarray*
The lower bound can probably be shown similarly.
answered Jul 16 at 0:45
Donald Splutterwit
21.3k21243
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up vote
0
down vote
At least, for large values of $n$, take the logarithms and expand as series to get
$$logleft(2frac(2n+1)^2n+1(2n+2)^2n+2right)=-log left(nright)-1-frac34 n+frac724
n^2+Oleft(frac1n^3right)$$
$$logleft(fracn^n(n+1)^n+1right)=-log left(nright)-1-frac12 n+frac16
n^2+Oleft(frac1n^3right)$$
$$logleft(2frac(2n)^2n(2n+1)^2n+1right)=-log left(nright)-1-frac14 n+frac124
n^2+Oleft(frac1n^3right)$$ Now, using $A=e^log(A)$, we have, for large $n$,
$$2frac(2n+1)^2n+1(2n+2)^2n+2=frac 1e left(frac1n-frac34 n^2+frac5596
n^3right)+Oleft(frac1n^4right)$$
$$fracn^n(n+1)^n+1=frac 1e left(frac1n-frac12 n^2+frac724
n^3right)+Oleft(frac1n^4right)$$
$$2frac(2n)^2n(2n+1)^2n+1=frac 1e left(frac1n-frac14 n^2+frac796
n^3right)+Oleft(frac1n^4right)$$ Now, you can conclude since
$$fracn^n(n+1)^n+1-2frac(2n+1)^2n+1(2n+2)^2n+2=frac14 e n^2-frac932 e n^3+Oleft(frac1n^4right)$$
$$2frac(2n)^2n(2n+1)^2n+1-fracn^n(n+1)^n+1=frac14 e n^2-frac732 e n^3+Oleft(frac1n^4right)$$
answered Jul 16 at 4:53
Claude Leibovici
112k1055126
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
To prove
$$
2frac(2n+1)^2n+1(2n+2)^2n+2lefracn^n(n+1)^n+1le2frac(2n)^2n(2n+1)^2n+1
$$
note that the left-hand inequality is equivalent to
$$
left(1+frac12n+1right)^2n+1geleft(1+frac1nright)^n
$$
which follows from the monotonic increase of $left(1+frac1nright)^n$ proven in this answer using only Bernoulli's Inequality.
The right-hand inequality is equivalent to
$$
left(1+frac1nright)^n+1geleft(1+frac12nright)^2n+1
$$
which follows from the monotonic decrease of $left(1+frac1nright)^n+1$ proven in the same answer, again using only Bernoulli's Inequality.
answered Jul 16 at 5:22
robjohn♦
258k26297612
add a comment |Â
up vote
1
down vote
accepted
To prove
$$
2frac(2n+1)^2n+1(2n+2)^2n+2lefracn^n(n+1)^n+1le2frac(2n)^2n(2n+1)^2n+1
$$
note that the left-hand inequality is equivalent to
$$
left(1+frac12n+1right)^2n+1geleft(1+frac1nright)^n
$$
which follows from the monotonic increase of $left(1+frac1nright)^n$ proven in this answer using only Bernoulli's Inequality.
The right-hand inequality is equivalent to
$$
left(1+frac1nright)^n+1geleft(1+frac12nright)^2n+1
$$
which follows from the monotonic decrease of $left(1+frac1nright)^n+1$ proven in the same answer, again using only Bernoulli's Inequality.
answered Jul 16 at 5:22
robjohn♦
258k26297612
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
To prove
$$
2frac(2n+1)^2n+1(2n+2)^2n+2lefracn^n(n+1)^n+1le2frac(2n)^2n(2n+1)^2n+1
$$
note that the left-hand inequality is equivalent to
$$
left(1+frac12n+1right)^2n+1geleft(1+frac1nright)^n
$$
which follows from the monotonic increase of $left(1+frac1nright)^n$ proven in this answer using only Bernoulli's Inequality.
The right-hand inequality is equivalent to
$$
left(1+frac1nright)^n+1geleft(1+frac12nright)^2n+1
$$
which follows from the monotonic decrease of $left(1+frac1nright)^n+1$ proven in the same answer, again using only Bernoulli's Inequality.
answered Jul 16 at 5:22
robjohn♦
258k26297612
To prove
$$
2frac(2n+1)^2n+1(2n+2)^2n+2lefracn^n(n+1)^n+1le2frac(2n)^2n(2n+1)^2n+1
$$
note that the left-hand inequality is equivalent to
$$
left(1+frac12n+1right)^2n+1geleft(1+frac1nright)^n
$$
which follows from the monotonic increase of $left(1+frac1nright)^n$ proven in this answer using only Bernoulli's Inequality.
The right-hand inequality is equivalent to
$$
left(1+frac1nright)^n+1geleft(1+frac12nright)^2n+1
$$
which follows from the monotonic decrease of $left(1+frac1nright)^n+1$ proven in the same answer, again using only Bernoulli's Inequality.
answered Jul 16 at 5:22
robjohn♦
258k26297612
answered Jul 16 at 5:22
robjohn♦
258k26297612
answered Jul 16 at 5:22
robjohn♦
258k26297612
answered Jul 16 at 5:22
robjohn♦
258k26297612
258k26297612
add a comment |Â
add a comment |Â
up vote
0
down vote
The function
begineqnarray*
left( 1+ frac1x right)^x
endeqnarray*
is increasing so
begineqnarray*
left( 1+ frac12n right)^2n leq left( 1+ frac12n+1 right)^2n+1 leq left( 1+ frac12n+1 right)^2n+2.
endeqnarray*
Square root this to get
begineqnarray*
left( 1+ frac12n right)^n leq left( 1+ frac12n+1 right)^n+1.
endeqnarray*
This can be rearranged to give the upper bound ...
begineqnarray*
n ln(1+frac12n) leq (n+1) ln (1+frac12n+1) \
endeqnarray*
begineqnarray*
n( ln(n+frac12) -ln(n)) leq (n+1) (ln(n+1)- ln(n+frac12)) \
endeqnarray*
begineqnarray*(2n+1) ln(n+frac12) leq n ln (n) +(n+1) ln(n+1) \
endeqnarray*
begineqnarray*n ln n +(2n+1) ln(2n+1) leq ln 2 +2n ln(2n) +(n+1) ln(n+1) \
endeqnarray*
begineqnarray*
fracn^n(n+1)^n+1 leq 2 frac(2n)^2n(2n+1)^2n+1.
endeqnarray*
The lower bound can probably be shown similarly.
answered Jul 16 at 0:45
Donald Splutterwit
21.3k21243
add a comment |Â
up vote
0
down vote
The function
begineqnarray*
left( 1+ frac1x right)^x
endeqnarray*
is increasing so
begineqnarray*
left( 1+ frac12n right)^2n leq left( 1+ frac12n+1 right)^2n+1 leq left( 1+ frac12n+1 right)^2n+2.
endeqnarray*
Square root this to get
begineqnarray*
left( 1+ frac12n right)^n leq left( 1+ frac12n+1 right)^n+1.
endeqnarray*
This can be rearranged to give the upper bound ...
begineqnarray*
n ln(1+frac12n) leq (n+1) ln (1+frac12n+1) \
endeqnarray*
begineqnarray*
n( ln(n+frac12) -ln(n)) leq (n+1) (ln(n+1)- ln(n+frac12)) \
endeqnarray*
begineqnarray*(2n+1) ln(n+frac12) leq n ln (n) +(n+1) ln(n+1) \
endeqnarray*
begineqnarray*n ln n +(2n+1) ln(2n+1) leq ln 2 +2n ln(2n) +(n+1) ln(n+1) \
endeqnarray*
begineqnarray*
fracn^n(n+1)^n+1 leq 2 frac(2n)^2n(2n+1)^2n+1.
endeqnarray*
The lower bound can probably be shown similarly.
answered Jul 16 at 0:45
Donald Splutterwit
21.3k21243
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The function
begineqnarray*
left( 1+ frac1x right)^x
endeqnarray*
is increasing so
begineqnarray*
left( 1+ frac12n right)^2n leq left( 1+ frac12n+1 right)^2n+1 leq left( 1+ frac12n+1 right)^2n+2.
endeqnarray*
Square root this to get
begineqnarray*
left( 1+ frac12n right)^n leq left( 1+ frac12n+1 right)^n+1.
endeqnarray*
This can be rearranged to give the upper bound ...
begineqnarray*
n ln(1+frac12n) leq (n+1) ln (1+frac12n+1) \
endeqnarray*
begineqnarray*
n( ln(n+frac12) -ln(n)) leq (n+1) (ln(n+1)- ln(n+frac12)) \
endeqnarray*
begineqnarray*(2n+1) ln(n+frac12) leq n ln (n) +(n+1) ln(n+1) \
endeqnarray*
begineqnarray*n ln n +(2n+1) ln(2n+1) leq ln 2 +2n ln(2n) +(n+1) ln(n+1) \
endeqnarray*
begineqnarray*
fracn^n(n+1)^n+1 leq 2 frac(2n)^2n(2n+1)^2n+1.
endeqnarray*
The lower bound can probably be shown similarly.
answered Jul 16 at 0:45
Donald Splutterwit
21.3k21243
The function
begineqnarray*
left( 1+ frac1x right)^x
endeqnarray*
is increasing so
begineqnarray*
left( 1+ frac12n right)^2n leq left( 1+ frac12n+1 right)^2n+1 leq left( 1+ frac12n+1 right)^2n+2.
endeqnarray*
Square root this to get
begineqnarray*
left( 1+ frac12n right)^n leq left( 1+ frac12n+1 right)^n+1.
endeqnarray*
This can be rearranged to give the upper bound ...
begineqnarray*
n ln(1+frac12n) leq (n+1) ln (1+frac12n+1) \
endeqnarray*
begineqnarray*
n( ln(n+frac12) -ln(n)) leq (n+1) (ln(n+1)- ln(n+frac12)) \
endeqnarray*
begineqnarray*(2n+1) ln(n+frac12) leq n ln (n) +(n+1) ln(n+1) \
endeqnarray*
begineqnarray*n ln n +(2n+1) ln(2n+1) leq ln 2 +2n ln(2n) +(n+1) ln(n+1) \
endeqnarray*
begineqnarray*
fracn^n(n+1)^n+1 leq 2 frac(2n)^2n(2n+1)^2n+1.
endeqnarray*
The lower bound can probably be shown similarly.
answered Jul 16 at 0:45
Donald Splutterwit
21.3k21243
answered Jul 16 at 0:45
Donald Splutterwit
21.3k21243
answered Jul 16 at 0:45
Donald Splutterwit
21.3k21243
answered Jul 16 at 0:45
Donald Splutterwit
21.3k21243
21.3k21243
add a comment |Â
add a comment |Â
up vote
0
down vote
At least, for large values of $n$, take the logarithms and expand as series to get
$$logleft(2frac(2n+1)^2n+1(2n+2)^2n+2right)=-log left(nright)-1-frac34 n+frac724
n^2+Oleft(frac1n^3right)$$
$$logleft(fracn^n(n+1)^n+1right)=-log left(nright)-1-frac12 n+frac16
n^2+Oleft(frac1n^3right)$$
$$logleft(2frac(2n)^2n(2n+1)^2n+1right)=-log left(nright)-1-frac14 n+frac124
n^2+Oleft(frac1n^3right)$$ Now, using $A=e^log(A)$, we have, for large $n$,
$$2frac(2n+1)^2n+1(2n+2)^2n+2=frac 1e left(frac1n-frac34 n^2+frac5596
n^3right)+Oleft(frac1n^4right)$$
$$fracn^n(n+1)^n+1=frac 1e left(frac1n-frac12 n^2+frac724
n^3right)+Oleft(frac1n^4right)$$
$$2frac(2n)^2n(2n+1)^2n+1=frac 1e left(frac1n-frac14 n^2+frac796
n^3right)+Oleft(frac1n^4right)$$ Now, you can conclude since
$$fracn^n(n+1)^n+1-2frac(2n+1)^2n+1(2n+2)^2n+2=frac14 e n^2-frac932 e n^3+Oleft(frac1n^4right)$$
$$2frac(2n)^2n(2n+1)^2n+1-fracn^n(n+1)^n+1=frac14 e n^2-frac732 e n^3+Oleft(frac1n^4right)$$
answered Jul 16 at 4:53
Claude Leibovici
112k1055126
add a comment |Â
up vote
0
down vote
At least, for large values of $n$, take the logarithms and expand as series to get
$$logleft(2frac(2n+1)^2n+1(2n+2)^2n+2right)=-log left(nright)-1-frac34 n+frac724
n^2+Oleft(frac1n^3right)$$
$$logleft(fracn^n(n+1)^n+1right)=-log left(nright)-1-frac12 n+frac16
n^2+Oleft(frac1n^3right)$$
$$logleft(2frac(2n)^2n(2n+1)^2n+1right)=-log left(nright)-1-frac14 n+frac124
n^2+Oleft(frac1n^3right)$$ Now, using $A=e^log(A)$, we have, for large $n$,
$$2frac(2n+1)^2n+1(2n+2)^2n+2=frac 1e left(frac1n-frac34 n^2+frac5596
n^3right)+Oleft(frac1n^4right)$$
$$fracn^n(n+1)^n+1=frac 1e left(frac1n-frac12 n^2+frac724
n^3right)+Oleft(frac1n^4right)$$
$$2frac(2n)^2n(2n+1)^2n+1=frac 1e left(frac1n-frac14 n^2+frac796
n^3right)+Oleft(frac1n^4right)$$ Now, you can conclude since
$$fracn^n(n+1)^n+1-2frac(2n+1)^2n+1(2n+2)^2n+2=frac14 e n^2-frac932 e n^3+Oleft(frac1n^4right)$$
$$2frac(2n)^2n(2n+1)^2n+1-fracn^n(n+1)^n+1=frac14 e n^2-frac732 e n^3+Oleft(frac1n^4right)$$
answered Jul 16 at 4:53
Claude Leibovici
112k1055126
add a comment |Â
up vote
0
down vote
up vote
0
down vote
At least, for large values of $n$, take the logarithms and expand as series to get
$$logleft(2frac(2n+1)^2n+1(2n+2)^2n+2right)=-log left(nright)-1-frac34 n+frac724
n^2+Oleft(frac1n^3right)$$
$$logleft(fracn^n(n+1)^n+1right)=-log left(nright)-1-frac12 n+frac16
n^2+Oleft(frac1n^3right)$$
$$logleft(2frac(2n)^2n(2n+1)^2n+1right)=-log left(nright)-1-frac14 n+frac124
n^2+Oleft(frac1n^3right)$$ Now, using $A=e^log(A)$, we have, for large $n$,
$$2frac(2n+1)^2n+1(2n+2)^2n+2=frac 1e left(frac1n-frac34 n^2+frac5596
n^3right)+Oleft(frac1n^4right)$$
$$fracn^n(n+1)^n+1=frac 1e left(frac1n-frac12 n^2+frac724
n^3right)+Oleft(frac1n^4right)$$
$$2frac(2n)^2n(2n+1)^2n+1=frac 1e left(frac1n-frac14 n^2+frac796
n^3right)+Oleft(frac1n^4right)$$ Now, you can conclude since
$$fracn^n(n+1)^n+1-2frac(2n+1)^2n+1(2n+2)^2n+2=frac14 e n^2-frac932 e n^3+Oleft(frac1n^4right)$$
$$2frac(2n)^2n(2n+1)^2n+1-fracn^n(n+1)^n+1=frac14 e n^2-frac732 e n^3+Oleft(frac1n^4right)$$
answered Jul 16 at 4:53
Claude Leibovici
112k1055126
At least, for large values of $n$, take the logarithms and expand as series to get
$$logleft(2frac(2n+1)^2n+1(2n+2)^2n+2right)=-log left(nright)-1-frac34 n+frac724
n^2+Oleft(frac1n^3right)$$
$$logleft(fracn^n(n+1)^n+1right)=-log left(nright)-1-frac12 n+frac16
n^2+Oleft(frac1n^3right)$$
$$logleft(2frac(2n)^2n(2n+1)^2n+1right)=-log left(nright)-1-frac14 n+frac124
n^2+Oleft(frac1n^3right)$$ Now, using $A=e^log(A)$, we have, for large $n$,
$$2frac(2n+1)^2n+1(2n+2)^2n+2=frac 1e left(frac1n-frac34 n^2+frac5596
n^3right)+Oleft(frac1n^4right)$$
$$fracn^n(n+1)^n+1=frac 1e left(frac1n-frac12 n^2+frac724
n^3right)+Oleft(frac1n^4right)$$
$$2frac(2n)^2n(2n+1)^2n+1=frac 1e left(frac1n-frac14 n^2+frac796
n^3right)+Oleft(frac1n^4right)$$ Now, you can conclude since
$$fracn^n(n+1)^n+1-2frac(2n+1)^2n+1(2n+2)^2n+2=frac14 e n^2-frac932 e n^3+Oleft(frac1n^4right)$$
$$2frac(2n)^2n(2n+1)^2n+1-fracn^n(n+1)^n+1=frac14 e n^2-frac732 e n^3+Oleft(frac1n^4right)$$
answered Jul 16 at 4:53
Claude Leibovici
112k1055126
edited Jul 16 at 5:00
answered Jul 16 at 4:53
Claude Leibovici
112k1055126
answered Jul 16 at 4:53
Claude Leibovici
112k1055126
answered Jul 16 at 4:53
Claude Leibovici
112k1055126
112k1055126
add a comment |Â
add a comment |Â
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How about by induction? How many values (and which values) have you tested?
– amWhy
Jul 15 at 23:16
I tried $n=1$ and $n=24$. I tried induction but the induction step doesn't come out
– user544680
Jul 15 at 23:26
1
The first test: Plot them.
– David G. Stork
Jul 15 at 23:37