Proving $left( 1-frac{2}{n} right )^{frac {nln n}{4}}-left( 1-frac{1}{n} right )^{frac {2nln n}{4}}<0$
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As a part of a solution I'm writing I need to prove: $$left( 1-frac{2}{n} right )^{frac {nln n}{4}}-left( 1-frac{1}{n} right )^{frac {2nln n}{4}}<0$$ for large enough $n$. I checked in Wolfram-Alpha and it looks like it's true.. I've tried using:
$$1-xleq e^{-x}$$ and
$$1-xgeq e^{-2x}$$
but I get $$n^{-frac {1}{2}}-n^{-1}>0$$
Thanks.
sequences-and-series inequality asymptotics
asked Nov 13 at 19:07
Amihai Zivan
1,85811524
add a comment |
up vote
0
down vote
favorite
As a part of a solution I'm writing I need to prove: $$left( 1-frac{2}{n} right )^{frac {nln n}{4}}-left( 1-frac{1}{n} right )^{frac {2nln n}{4}}<0$$ for large enough $n$. I checked in Wolfram-Alpha and it looks like it's true.. I've tried using:
$$1-xleq e^{-x}$$ and
$$1-xgeq e^{-2x}$$
but I get $$n^{-frac {1}{2}}-n^{-1}>0$$
Thanks.
sequences-and-series inequality asymptotics
asked Nov 13 at 19:07
Amihai Zivan
1,85811524
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
As a part of a solution I'm writing I need to prove: $$left( 1-frac{2}{n} right )^{frac {nln n}{4}}-left( 1-frac{1}{n} right )^{frac {2nln n}{4}}<0$$ for large enough $n$. I checked in Wolfram-Alpha and it looks like it's true.. I've tried using:
$$1-xleq e^{-x}$$ and
$$1-xgeq e^{-2x}$$
but I get $$n^{-frac {1}{2}}-n^{-1}>0$$
Thanks.
sequences-and-series inequality asymptotics
asked Nov 13 at 19:07
Amihai Zivan
1,85811524
As a part of a solution I'm writing I need to prove: $$left( 1-frac{2}{n} right )^{frac {nln n}{4}}-left( 1-frac{1}{n} right )^{frac {2nln n}{4}}<0$$ for large enough $n$. I checked in Wolfram-Alpha and it looks like it's true.. I've tried using:
$$1-xleq e^{-x}$$ and
$$1-xgeq e^{-2x}$$
but I get $$n^{-frac {1}{2}}-n^{-1}>0$$
Thanks.
sequences-and-series inequality asymptotics
sequences-and-series inequality asymptotics
asked Nov 13 at 19:07
Amihai Zivan
1,85811524
asked Nov 13 at 19:07
Amihai Zivan
1,85811524
asked Nov 13 at 19:07
Amihai Zivan
1,85811524
asked Nov 13 at 19:07
Amihai Zivan
1,85811524
asked Nov 13 at 19:07
Amihai Zivan
1,85811524
1,85811524
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
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accepted
The hint:
$$frac{left(1-frac{1}{n}right)^2}{1-frac{2}{n}}>1.$$
Thus, $$left(frac{left(1-frac{1}{n}right)^2}{1-frac{2}{n}}right)^{frac{nln{n}}{4}}>1.$$
answered Nov 13 at 19:15
Michael Rozenberg
94.2k1588183
I've been hinted to use the above inequalities but they doesn't seem to work and your's nicer!
– Amihai Zivan
Nov 13 at 19:22
Small nit: if $n < e$, the $>$ would get inverted, correct? So we need $n >= e$ for this, correct?
– user1952500
Nov 13 at 19:29
1
@user1952500 I think it's true for all real $n>2$.
– Michael Rozenberg
Nov 13 at 19:33
Sorry, completely my mistake, it should be as you say.
– user1952500
Nov 13 at 19:42
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The hint:
$$frac{left(1-frac{1}{n}right)^2}{1-frac{2}{n}}>1.$$
Thus, $$left(frac{left(1-frac{1}{n}right)^2}{1-frac{2}{n}}right)^{frac{nln{n}}{4}}>1.$$
answered Nov 13 at 19:15
Michael Rozenberg
94.2k1588183
I've been hinted to use the above inequalities but they doesn't seem to work and your's nicer!
– Amihai Zivan
Nov 13 at 19:22
Small nit: if $n < e$, the $>$ would get inverted, correct? So we need $n >= e$ for this, correct?
– user1952500
Nov 13 at 19:29
1
@user1952500 I think it's true for all real $n>2$.
– Michael Rozenberg
Nov 13 at 19:33
Sorry, completely my mistake, it should be as you say.
– user1952500
Nov 13 at 19:42
add a comment |
up vote
4
down vote
accepted
The hint:
$$frac{left(1-frac{1}{n}right)^2}{1-frac{2}{n}}>1.$$
Thus, $$left(frac{left(1-frac{1}{n}right)^2}{1-frac{2}{n}}right)^{frac{nln{n}}{4}}>1.$$
answered Nov 13 at 19:15
Michael Rozenberg
94.2k1588183
I've been hinted to use the above inequalities but they doesn't seem to work and your's nicer!
– Amihai Zivan
Nov 13 at 19:22
Small nit: if $n < e$, the $>$ would get inverted, correct? So we need $n >= e$ for this, correct?
– user1952500
Nov 13 at 19:29
1
@user1952500 I think it's true for all real $n>2$.
– Michael Rozenberg
Nov 13 at 19:33
Sorry, completely my mistake, it should be as you say.
– user1952500
Nov 13 at 19:42
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The hint:
$$frac{left(1-frac{1}{n}right)^2}{1-frac{2}{n}}>1.$$
Thus, $$left(frac{left(1-frac{1}{n}right)^2}{1-frac{2}{n}}right)^{frac{nln{n}}{4}}>1.$$
answered Nov 13 at 19:15
Michael Rozenberg
94.2k1588183
The hint:
$$frac{left(1-frac{1}{n}right)^2}{1-frac{2}{n}}>1.$$
Thus, $$left(frac{left(1-frac{1}{n}right)^2}{1-frac{2}{n}}right)^{frac{nln{n}}{4}}>1.$$
answered Nov 13 at 19:15
Michael Rozenberg
94.2k1588183
answered Nov 13 at 19:15
Michael Rozenberg
94.2k1588183
answered Nov 13 at 19:15
Michael Rozenberg
94.2k1588183
answered Nov 13 at 19:15
Michael Rozenberg
94.2k1588183
94.2k1588183
I've been hinted to use the above inequalities but they doesn't seem to work and your's nicer!
– Amihai Zivan
Nov 13 at 19:22
Small nit: if $n < e$, the $>$ would get inverted, correct? So we need $n >= e$ for this, correct?
– user1952500
Nov 13 at 19:29
1
@user1952500 I think it's true for all real $n>2$.
– Michael Rozenberg
Nov 13 at 19:33
Sorry, completely my mistake, it should be as you say.
– user1952500
Nov 13 at 19:42
add a comment |
I've been hinted to use the above inequalities but they doesn't seem to work and your's nicer!
– Amihai Zivan
Nov 13 at 19:22
Small nit: if $n < e$, the $>$ would get inverted, correct? So we need $n >= e$ for this, correct?
– user1952500
Nov 13 at 19:29
1
@user1952500 I think it's true for all real $n>2$.
– Michael Rozenberg
Nov 13 at 19:33
Sorry, completely my mistake, it should be as you say.
– user1952500
Nov 13 at 19:42
I've been hinted to use the above inequalities but they doesn't seem to work and your's nicer!
– Amihai Zivan
Nov 13 at 19:22
I've been hinted to use the above inequalities but they doesn't seem to work and your's nicer!
– Amihai Zivan
Nov 13 at 19:22
Small nit: if $n < e$, the $>$ would get inverted, correct? So we need $n >= e$ for this, correct?
– user1952500
Nov 13 at 19:29
Small nit: if $n < e$, the $>$ would get inverted, correct? So we need $n >= e$ for this, correct?
– user1952500
Nov 13 at 19:29
1
1
@user1952500 I think it's true for all real $n>2$.
– Michael Rozenberg
Nov 13 at 19:33
@user1952500 I think it's true for all real $n>2$.
– Michael Rozenberg
Nov 13 at 19:33
Sorry, completely my mistake, it should be as you say.
– user1952500
Nov 13 at 19:42
Sorry, completely my mistake, it should be as you say.
– user1952500
Nov 13 at 19:42
add a comment |
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