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Proof on positive sequence with $limsup_n a_n^1/n=1$ and $liminf_na_n^1/n <1$
Clash Royale CLAN TAG#URR8PPP
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Does a positive sequence $a_n$ with $limsup_n a_n^1/n=1$ and $liminf_na_n^1/n <1$ must have a subsequence $a_n_i$ satisfying $lim_i a_n_i^1/n_i=1$ and $lim_i |a_n_i^2-a_n_i-1a_n_i+1|^1/n_i=1$.
So Here are the hypothesis:
beginequation limsup_n a_n^1/n=1 tag1 endequation
and
beginequation liminf_n a_n^1/n<1 tag2 endequation
How to derive the conclusion: there is a subsequence $a_n_i$ fulfilling both of
beginequation lim_i a_n_i^1/n_i=1 tag3 endequation
and
beginequation lim_i |a_n_i^2-a_n_i-1a_n_i+1|^1/n_i=1 tag4 endequation
My understanding to this problem
It is obvious that $(1)$ implies $(3)$, but how $(1)$ together with $(2)$ imply $(3)$ and $(4)$ is not so clear, although trivial examples such as
$$1,delta^2,1,delta^4,cdots,1,delta^2n,cdots$$
(where $0<delta<1$) strongly support this proposition.
Given $(3)$ holds, $(4)$ may be replaced with the equivalent
beginequationlim_i left|1-fraca_n_i+1a_n_i fraca_n_i-1a_n_iright|^1/n_i =1 endequation
So this suggests that $dfraca_n_i+1a_n_i dfraca_n_i-1a_n_i$ must be "small" enough. It seems somehow related to the ratio test vs the root test in convergence of series.
What I expect
A proof (or a counter-example) is of course appreciated, but I also want to know the origin of this problem (in what literature did it emerge). Discussions giving an insight to this problem is welcomed as well.
sequences-and-series
asked Jul 24 at 17:08
Y.Lin
586
 |Â
show 2 more comments
up vote
6
down vote
favorite
Does a positive sequence $a_n$ with $limsup_n a_n^1/n=1$ and $liminf_na_n^1/n <1$ must have a subsequence $a_n_i$ satisfying $lim_i a_n_i^1/n_i=1$ and $lim_i |a_n_i^2-a_n_i-1a_n_i+1|^1/n_i=1$.
So Here are the hypothesis:
beginequation limsup_n a_n^1/n=1 tag1 endequation
and
beginequation liminf_n a_n^1/n<1 tag2 endequation
How to derive the conclusion: there is a subsequence $a_n_i$ fulfilling both of
beginequation lim_i a_n_i^1/n_i=1 tag3 endequation
and
beginequation lim_i |a_n_i^2-a_n_i-1a_n_i+1|^1/n_i=1 tag4 endequation
My understanding to this problem
It is obvious that $(1)$ implies $(3)$, but how $(1)$ together with $(2)$ imply $(3)$ and $(4)$ is not so clear, although trivial examples such as
$$1,delta^2,1,delta^4,cdots,1,delta^2n,cdots$$
(where $0<delta<1$) strongly support this proposition.
Given $(3)$ holds, $(4)$ may be replaced with the equivalent
beginequationlim_i left|1-fraca_n_i+1a_n_i fraca_n_i-1a_n_iright|^1/n_i =1 endequation
So this suggests that $dfraca_n_i+1a_n_i dfraca_n_i-1a_n_i$ must be "small" enough. It seems somehow related to the ratio test vs the root test in convergence of series.
What I expect
A proof (or a counter-example) is of course appreciated, but I also want to know the origin of this problem (in what literature did it emerge). Discussions giving an insight to this problem is welcomed as well.
sequences-and-series
asked Jul 24 at 17:08
Y.Lin
586
I think your statement of the problem is unnecessarily complicated. You could just as well have a sequence $b_n (=a_n^frac1n)$ with $limsup_n b_n=1$ and $liminf_n b_nlt 1$. Simple example $b_n=1$ for even $n$ and $b_n=0.5$ for odd $n$.
– herb steinberg
Aug 2 at 21:40
@herb-steinberg The essense of this problem lies in $(4)$. If we let $b_n=(a_n)^1/n~$, $~(4)$ would become $lim_i |b_n_i^2n_i-b_n_i-1^n_i -1b_n_i+1^n_i +1|^1/n_i=1$, which seems just complicating the main point.
– Y.Lin
Aug 2 at 23:28
For even $n, |b_n^2n-b_n-1^n-1b_n+1^n+1|^frac1n=|1-0.5^2n|^frac1napprox 1-frac1n0.5^2nto 1$. Odd $n$ gives the same result.
– herb steinberg
Aug 2 at 23:58
I am a little confused as to what you want to show. You want a series where the limit is not 1?
– herb steinberg
Aug 3 at 0:03
How about a series where $b_n=1$, except when n is a multiple of 4 (or higher),where $b_n=0$? Most of the time the terms in (4) will be $0$
– herb steinberg
Aug 3 at 0:51
 |Â
show 2 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Does a positive sequence $a_n$ with $limsup_n a_n^1/n=1$ and $liminf_na_n^1/n <1$ must have a subsequence $a_n_i$ satisfying $lim_i a_n_i^1/n_i=1$ and $lim_i |a_n_i^2-a_n_i-1a_n_i+1|^1/n_i=1$.
So Here are the hypothesis:
beginequation limsup_n a_n^1/n=1 tag1 endequation
and
beginequation liminf_n a_n^1/n<1 tag2 endequation
How to derive the conclusion: there is a subsequence $a_n_i$ fulfilling both of
beginequation lim_i a_n_i^1/n_i=1 tag3 endequation
and
beginequation lim_i |a_n_i^2-a_n_i-1a_n_i+1|^1/n_i=1 tag4 endequation
My understanding to this problem
It is obvious that $(1)$ implies $(3)$, but how $(1)$ together with $(2)$ imply $(3)$ and $(4)$ is not so clear, although trivial examples such as
$$1,delta^2,1,delta^4,cdots,1,delta^2n,cdots$$
(where $0<delta<1$) strongly support this proposition.
Given $(3)$ holds, $(4)$ may be replaced with the equivalent
beginequationlim_i left|1-fraca_n_i+1a_n_i fraca_n_i-1a_n_iright|^1/n_i =1 endequation
So this suggests that $dfraca_n_i+1a_n_i dfraca_n_i-1a_n_i$ must be "small" enough. It seems somehow related to the ratio test vs the root test in convergence of series.
What I expect
A proof (or a counter-example) is of course appreciated, but I also want to know the origin of this problem (in what literature did it emerge). Discussions giving an insight to this problem is welcomed as well.
sequences-and-series
asked Jul 24 at 17:08
Y.Lin
586
Does a positive sequence $a_n$ with $limsup_n a_n^1/n=1$ and $liminf_na_n^1/n <1$ must have a subsequence $a_n_i$ satisfying $lim_i a_n_i^1/n_i=1$ and $lim_i |a_n_i^2-a_n_i-1a_n_i+1|^1/n_i=1$.
So Here are the hypothesis:
beginequation limsup_n a_n^1/n=1 tag1 endequation
and
beginequation liminf_n a_n^1/n<1 tag2 endequation
How to derive the conclusion: there is a subsequence $a_n_i$ fulfilling both of
beginequation lim_i a_n_i^1/n_i=1 tag3 endequation
and
beginequation lim_i |a_n_i^2-a_n_i-1a_n_i+1|^1/n_i=1 tag4 endequation
My understanding to this problem
It is obvious that $(1)$ implies $(3)$, but how $(1)$ together with $(2)$ imply $(3)$ and $(4)$ is not so clear, although trivial examples such as
$$1,delta^2,1,delta^4,cdots,1,delta^2n,cdots$$
(where $0<delta<1$) strongly support this proposition.
Given $(3)$ holds, $(4)$ may be replaced with the equivalent
beginequationlim_i left|1-fraca_n_i+1a_n_i fraca_n_i-1a_n_iright|^1/n_i =1 endequation
So this suggests that $dfraca_n_i+1a_n_i dfraca_n_i-1a_n_i$ must be "small" enough. It seems somehow related to the ratio test vs the root test in convergence of series.
What I expect
A proof (or a counter-example) is of course appreciated, but I also want to know the origin of this problem (in what literature did it emerge). Discussions giving an insight to this problem is welcomed as well.
sequences-and-series
asked Jul 24 at 17:08
Y.Lin
586
edited Aug 3 at 6:16
asked Jul 24 at 17:08
Y.Lin
586
asked Jul 24 at 17:08
Y.Lin
586
asked Jul 24 at 17:08
Y.Lin
586
586
I think your statement of the problem is unnecessarily complicated. You could just as well have a sequence $b_n (=a_n^frac1n)$ with $limsup_n b_n=1$ and $liminf_n b_nlt 1$. Simple example $b_n=1$ for even $n$ and $b_n=0.5$ for odd $n$.
– herb steinberg
Aug 2 at 21:40
@herb-steinberg The essense of this problem lies in $(4)$. If we let $b_n=(a_n)^1/n~$, $~(4)$ would become $lim_i |b_n_i^2n_i-b_n_i-1^n_i -1b_n_i+1^n_i +1|^1/n_i=1$, which seems just complicating the main point.
– Y.Lin
Aug 2 at 23:28
For even $n, |b_n^2n-b_n-1^n-1b_n+1^n+1|^frac1n=|1-0.5^2n|^frac1napprox 1-frac1n0.5^2nto 1$. Odd $n$ gives the same result.
– herb steinberg
Aug 2 at 23:58
I am a little confused as to what you want to show. You want a series where the limit is not 1?
– herb steinberg
Aug 3 at 0:03
How about a series where $b_n=1$, except when n is a multiple of 4 (or higher),where $b_n=0$? Most of the time the terms in (4) will be $0$
– herb steinberg
Aug 3 at 0:51
 |Â
show 2 more comments
I think your statement of the problem is unnecessarily complicated. You could just as well have a sequence $b_n (=a_n^frac1n)$ with $limsup_n b_n=1$ and $liminf_n b_nlt 1$. Simple example $b_n=1$ for even $n$ and $b_n=0.5$ for odd $n$.
– herb steinberg
Aug 2 at 21:40
@herb-steinberg The essense of this problem lies in $(4)$. If we let $b_n=(a_n)^1/n~$, $~(4)$ would become $lim_i |b_n_i^2n_i-b_n_i-1^n_i -1b_n_i+1^n_i +1|^1/n_i=1$, which seems just complicating the main point.
– Y.Lin
Aug 2 at 23:28
For even $n, |b_n^2n-b_n-1^n-1b_n+1^n+1|^frac1n=|1-0.5^2n|^frac1napprox 1-frac1n0.5^2nto 1$. Odd $n$ gives the same result.
– herb steinberg
Aug 2 at 23:58
I am a little confused as to what you want to show. You want a series where the limit is not 1?
– herb steinberg
Aug 3 at 0:03
How about a series where $b_n=1$, except when n is a multiple of 4 (or higher),where $b_n=0$? Most of the time the terms in (4) will be $0$
– herb steinberg
Aug 3 at 0:51
I think your statement of the problem is unnecessarily complicated. You could just as well have a sequence $b_n (=a_n^frac1n)$ with $limsup_n b_n=1$ and $liminf_n b_nlt 1$. Simple example $b_n=1$ for even $n$ and $b_n=0.5$ for odd $n$.
– herb steinberg
Aug 2 at 21:40
I think your statement of the problem is unnecessarily complicated. You could just as well have a sequence $b_n (=a_n^frac1n)$ with $limsup_n b_n=1$ and $liminf_n b_nlt 1$. Simple example $b_n=1$ for even $n$ and $b_n=0.5$ for odd $n$.
– herb steinberg
Aug 2 at 21:40
@herb-steinberg The essense of this problem lies in $(4)$. If we let $b_n=(a_n)^1/n~$, $~(4)$ would become $lim_i |b_n_i^2n_i-b_n_i-1^n_i -1b_n_i+1^n_i +1|^1/n_i=1$, which seems just complicating the main point.
– Y.Lin
Aug 2 at 23:28
@herb-steinberg The essense of this problem lies in $(4)$. If we let $b_n=(a_n)^1/n~$, $~(4)$ would become $lim_i |b_n_i^2n_i-b_n_i-1^n_i -1b_n_i+1^n_i +1|^1/n_i=1$, which seems just complicating the main point.
– Y.Lin
Aug 2 at 23:28
For even $n, |b_n^2n-b_n-1^n-1b_n+1^n+1|^frac1n=|1-0.5^2n|^frac1napprox 1-frac1n0.5^2nto 1$. Odd $n$ gives the same result.
– herb steinberg
Aug 2 at 23:58
For even $n, |b_n^2n-b_n-1^n-1b_n+1^n+1|^frac1n=|1-0.5^2n|^frac1napprox 1-frac1n0.5^2nto 1$. Odd $n$ gives the same result.
– herb steinberg
Aug 2 at 23:58
I am a little confused as to what you want to show. You want a series where the limit is not 1?
– herb steinberg
Aug 3 at 0:03
I am a little confused as to what you want to show. You want a series where the limit is not 1?
– herb steinberg
Aug 3 at 0:03
How about a series where $b_n=1$, except when n is a multiple of 4 (or higher),where $b_n=0$? Most of the time the terms in (4) will be $0$
– herb steinberg
Aug 3 at 0:51
How about a series where $b_n=1$, except when n is a multiple of 4 (or higher),where $b_n=0$? Most of the time the terms in (4) will be $0$
– herb steinberg
Aug 3 at 0:51
 |Â
show 2 more comments
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I think your statement of the problem is unnecessarily complicated. You could just as well have a sequence $b_n (=a_n^frac1n)$ with $limsup_n b_n=1$ and $liminf_n b_nlt 1$. Simple example $b_n=1$ for even $n$ and $b_n=0.5$ for odd $n$.
– herb steinberg
Aug 2 at 21:40
@herb-steinberg The essense of this problem lies in $(4)$. If we let $b_n=(a_n)^1/n~$, $~(4)$ would become $lim_i |b_n_i^2n_i-b_n_i-1^n_i -1b_n_i+1^n_i +1|^1/n_i=1$, which seems just complicating the main point.
– Y.Lin
Aug 2 at 23:28
For even $n, |b_n^2n-b_n-1^n-1b_n+1^n+1|^frac1n=|1-0.5^2n|^frac1napprox 1-frac1n0.5^2nto 1$. Odd $n$ gives the same result.
– herb steinberg
Aug 2 at 23:58
I am a little confused as to what you want to show. You want a series where the limit is not 1?
– herb steinberg
Aug 3 at 0:03
How about a series where $b_n=1$, except when n is a multiple of 4 (or higher),where $b_n=0$? Most of the time the terms in (4) will be $0$
– herb steinberg
Aug 3 at 0:51