Inequality involving $limsup$ and $liminf$: $ liminf(a_n+1/a_n) le liminf((a_n)^(1/n)) le limsup((a_n)^(1/n)) le limsup(a_n+1/a_n)$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Root test is stronger than ratio test?Given that $a_n > 0$, prove: $liminf left(fraca_n+1a_nright)leq liminf; sqrt[n]a_nleqlimsupleft(fraca_n+1a_nright)$Do the sequences from the ratio and root tests converge to the same limit?Ratio test and the Root testShow root test is stronger than ratio testFinding limit using inequalities: $liminf fraca_n+1a_n le liminf (a_n)^ 1/nlelimsup (a_n)^ 1/nle limsup fraca_n+1a_n$Proof of limit inequalityShow that $limsup|s_n|^1over nle limsup|s_n+1over s_n|$Inequality between limits inferiorThe inequality $limsup fracc_n+1c_n geqslant limsup sqrt[n]c_n$How to work out the limit of this bounded sequence?Rudin Series ratio and root test.Inequalities involving liminf and limsupFind $limsup$ and $liminf$ of a sequence and prove $liminf a_n leq limsup a_n$.Trying to prove $liminf (A_n) subseteq limsup (A_n)$For $s_n$ a sequence in $Bbb R$, if $lim s_n$ defined as a real number, then $liminf s_n = lim s_n = limsup s_n$.Convergene of a sequence if and only if limsup and liminf agree$liminf A_n$ and $limsup B_n$$liminf a_nleq lim a_n_k leq limsup a_n$Proofs involving limsup and liminf
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Inequality involving $limsup$ and $liminf$: $ liminf(a_n+1/a_n) le liminf((a_n)^(1/n)) le limsup((a_n)^(1/n)) le limsup(a_n+1/a_n)$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Root test is stronger than ratio test?Given that $a_n > 0$, prove: $liminf left(fraca_n+1a_nright)leq liminf; sqrt[n]a_nleqlimsupleft(fraca_n+1a_nright)$Do the sequences from the ratio and root tests converge to the same limit?Ratio test and the Root testShow root test is stronger than ratio testFinding limit using inequalities: $liminf fraca_n+1a_n le liminf (a_n)^ 1/nlelimsup (a_n)^ 1/nle limsup fraca_n+1a_n$Proof of limit inequalityShow that $limsup|s_n|^1over nle limsup|s_n+1over s_n|$Inequality between limits inferiorThe inequality $limsup fracc_n+1c_n geqslant limsup sqrt[n]c_n$How to work out the limit of this bounded sequence?Rudin Series ratio and root test.Inequalities involving liminf and limsupFind $limsup$ and $liminf$ of a sequence and prove $liminf a_n leq limsup a_n$.Trying to prove $liminf (A_n) subseteq limsup (A_n)$For $s_n$ a sequence in $Bbb R$, if $lim s_n$ defined as a real number, then $liminf s_n = lim s_n = limsup s_n$.Convergene of a sequence if and only if limsup and liminf agree$liminf A_n$ and $limsup B_n$$liminf a_nleq lim a_n_k leq limsup a_n$Proofs involving limsup and liminf
$begingroup$
This may have been asked before, however I was unable to find any duplicate.
This comes from pg. 52 of "Mathematical Analysis: An Introduction" by Browder. Problem 14:
If $(a_n)$ is a sequence in $mathbb R$ and $a_n > 0$ for every $n$. Then show:
$$
liminf(a_n+1/a_n) le liminf((a_n)^(1/n)) le limsup((a_n)^(1/n)) le limsup(a_n+1/a_n)$$
The middle inequality is clear. However I am having a hard time showing the ones on the left and right. (It seems like the approach should be similar for each). This is homework, so it'd be great if someone could give me a hint to get started on at least one of the inequalities.
Thanks.
real-analysis analysis inequality limsup-and-liminf
edited May 7 '16 at 7:42
Martin Sleziak
45k10123277
asked Oct 2 '11 at 23:36
RelsiarkRelsiark
9515
$endgroup$
add a comment |
$begingroup$
This may have been asked before, however I was unable to find any duplicate.
This comes from pg. 52 of "Mathematical Analysis: An Introduction" by Browder. Problem 14:
If $(a_n)$ is a sequence in $mathbb R$ and $a_n > 0$ for every $n$. Then show:
$$
liminf(a_n+1/a_n) le liminf((a_n)^(1/n)) le limsup((a_n)^(1/n)) le limsup(a_n+1/a_n)$$
The middle inequality is clear. However I am having a hard time showing the ones on the left and right. (It seems like the approach should be similar for each). This is homework, so it'd be great if someone could give me a hint to get started on at least one of the inequalities.
Thanks.
real-analysis analysis inequality limsup-and-liminf
edited May 7 '16 at 7:42
Martin Sleziak
45k10123277
asked Oct 2 '11 at 23:36
RelsiarkRelsiark
9515
$endgroup$
1
$begingroup$
It may sound somewhat unhelpful, but the best technique is usually to ditch the tricks (at least at first) and go one step at a time with the definitions for every term. After a while that you have become comfortable with the definitions, it gets a lot easier and a lot clearer how "tricks" and other techniques work.
$endgroup$
– Asaf Karagila♦
Oct 2 '11 at 23:56
$begingroup$
See also these two answers: math.stackexchange.com/questions/76743/limit-of-fraca-n1a-n/… and math.stackexchange.com/questions/28476/…
$endgroup$
– Martin Sleziak
Nov 1 '11 at 9:53
1
$begingroup$
This is also given as Problem 2.4.26 in the book Kaczor, Nowak: Problems in Mathematical Analysis. The problem is stated on p.46 and a solution is given on p.205.
$endgroup$
– Martin Sleziak
Jul 23 '12 at 5:55
1
$begingroup$
Also in Ross, Elementary Analysis 2ed p.79-80 with solution to third inequality.
$endgroup$
– user203509
May 1 '16 at 19:18
$begingroup$
Actualy you can take $ln$ on each term and use Stolz-Cesaro
$endgroup$
– Tony Ma
May 16 '18 at 11:45
add a comment |
$begingroup$
This may have been asked before, however I was unable to find any duplicate.
This comes from pg. 52 of "Mathematical Analysis: An Introduction" by Browder. Problem 14:
If $(a_n)$ is a sequence in $mathbb R$ and $a_n > 0$ for every $n$. Then show:
$$
liminf(a_n+1/a_n) le liminf((a_n)^(1/n)) le limsup((a_n)^(1/n)) le limsup(a_n+1/a_n)$$
The middle inequality is clear. However I am having a hard time showing the ones on the left and right. (It seems like the approach should be similar for each). This is homework, so it'd be great if someone could give me a hint to get started on at least one of the inequalities.
Thanks.
real-analysis analysis inequality limsup-and-liminf
edited May 7 '16 at 7:42
Martin Sleziak
45k10123277
asked Oct 2 '11 at 23:36
RelsiarkRelsiark
9515
$endgroup$
This may have been asked before, however I was unable to find any duplicate.
This comes from pg. 52 of "Mathematical Analysis: An Introduction" by Browder. Problem 14:
If $(a_n)$ is a sequence in $mathbb R$ and $a_n > 0$ for every $n$. Then show:
$$
liminf(a_n+1/a_n) le liminf((a_n)^(1/n)) le limsup((a_n)^(1/n)) le limsup(a_n+1/a_n)$$
The middle inequality is clear. However I am having a hard time showing the ones on the left and right. (It seems like the approach should be similar for each). This is homework, so it'd be great if someone could give me a hint to get started on at least one of the inequalities.
Thanks.
real-analysis analysis inequality limsup-and-liminf
real-analysis analysis inequality limsup-and-liminf
edited May 7 '16 at 7:42
Martin Sleziak
45k10123277
asked Oct 2 '11 at 23:36
RelsiarkRelsiark
9515
edited May 7 '16 at 7:42
Martin Sleziak
45k10123277
asked Oct 2 '11 at 23:36
RelsiarkRelsiark
9515
edited May 7 '16 at 7:42
Martin Sleziak
45k10123277
edited May 7 '16 at 7:42
Martin Sleziak
45k10123277
edited May 7 '16 at 7:42
Martin Sleziak
45k10123277
45k10123277
asked Oct 2 '11 at 23:36
RelsiarkRelsiark
9515
asked Oct 2 '11 at 23:36
RelsiarkRelsiark
9515
asked Oct 2 '11 at 23:36
RelsiarkRelsiark
9515
9515
1
$begingroup$
It may sound somewhat unhelpful, but the best technique is usually to ditch the tricks (at least at first) and go one step at a time with the definitions for every term. After a while that you have become comfortable with the definitions, it gets a lot easier and a lot clearer how "tricks" and other techniques work.
$endgroup$
– Asaf Karagila♦
Oct 2 '11 at 23:56
$begingroup$
See also these two answers: math.stackexchange.com/questions/76743/limit-of-fraca-n1a-n/… and math.stackexchange.com/questions/28476/…
$endgroup$
– Martin Sleziak
Nov 1 '11 at 9:53
1
$begingroup$
This is also given as Problem 2.4.26 in the book Kaczor, Nowak: Problems in Mathematical Analysis. The problem is stated on p.46 and a solution is given on p.205.
$endgroup$
– Martin Sleziak
Jul 23 '12 at 5:55
1
$begingroup$
Also in Ross, Elementary Analysis 2ed p.79-80 with solution to third inequality.
$endgroup$
– user203509
May 1 '16 at 19:18
$begingroup$
Actualy you can take $ln$ on each term and use Stolz-Cesaro
$endgroup$
– Tony Ma
May 16 '18 at 11:45
add a comment |
1
$begingroup$
It may sound somewhat unhelpful, but the best technique is usually to ditch the tricks (at least at first) and go one step at a time with the definitions for every term. After a while that you have become comfortable with the definitions, it gets a lot easier and a lot clearer how "tricks" and other techniques work.
$endgroup$
– Asaf Karagila♦
Oct 2 '11 at 23:56
$begingroup$
See also these two answers: math.stackexchange.com/questions/76743/limit-of-fraca-n1a-n/… and math.stackexchange.com/questions/28476/…
$endgroup$
– Martin Sleziak
Nov 1 '11 at 9:53
1
$begingroup$
This is also given as Problem 2.4.26 in the book Kaczor, Nowak: Problems in Mathematical Analysis. The problem is stated on p.46 and a solution is given on p.205.
$endgroup$
– Martin Sleziak
Jul 23 '12 at 5:55
1
$begingroup$
Also in Ross, Elementary Analysis 2ed p.79-80 with solution to third inequality.
$endgroup$
– user203509
May 1 '16 at 19:18
$begingroup$
Actualy you can take $ln$ on each term and use Stolz-Cesaro
$endgroup$
– Tony Ma
May 16 '18 at 11:45
1
1
$begingroup$
It may sound somewhat unhelpful, but the best technique is usually to ditch the tricks (at least at first) and go one step at a time with the definitions for every term. After a while that you have become comfortable with the definitions, it gets a lot easier and a lot clearer how "tricks" and other techniques work.
$endgroup$
– Asaf Karagila♦
Oct 2 '11 at 23:56
$begingroup$
It may sound somewhat unhelpful, but the best technique is usually to ditch the tricks (at least at first) and go one step at a time with the definitions for every term. After a while that you have become comfortable with the definitions, it gets a lot easier and a lot clearer how "tricks" and other techniques work.
$endgroup$
– Asaf Karagila♦
Oct 2 '11 at 23:56
$begingroup$
See also these two answers: math.stackexchange.com/questions/76743/limit-of-fraca-n1a-n/… and math.stackexchange.com/questions/28476/…
$endgroup$
– Martin Sleziak
Nov 1 '11 at 9:53
$begingroup$
See also these two answers: math.stackexchange.com/questions/76743/limit-of-fraca-n1a-n/… and math.stackexchange.com/questions/28476/…
$endgroup$
– Martin Sleziak
Nov 1 '11 at 9:53
1
1
$begingroup$
This is also given as Problem 2.4.26 in the book Kaczor, Nowak: Problems in Mathematical Analysis. The problem is stated on p.46 and a solution is given on p.205.
$endgroup$
– Martin Sleziak
Jul 23 '12 at 5:55
$begingroup$
This is also given as Problem 2.4.26 in the book Kaczor, Nowak: Problems in Mathematical Analysis. The problem is stated on p.46 and a solution is given on p.205.
$endgroup$
– Martin Sleziak
Jul 23 '12 at 5:55
1
1
$begingroup$
Also in Ross, Elementary Analysis 2ed p.79-80 with solution to third inequality.
$endgroup$
– user203509
May 1 '16 at 19:18
$begingroup$
Also in Ross, Elementary Analysis 2ed p.79-80 with solution to third inequality.
$endgroup$
– user203509
May 1 '16 at 19:18
$begingroup$
Actualy you can take $ln$ on each term and use Stolz-Cesaro
$endgroup$
– Tony Ma
May 16 '18 at 11:45
$begingroup$
Actualy you can take $ln$ on each term and use Stolz-Cesaro
$endgroup$
– Tony Ma
May 16 '18 at 11:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A somewhat detailed hint for the right hand side inequality:
Suppose
$$
r := lim sup fraca_n+1a_n.
$$
(If the above expression is $infty$, then there is nothing to prove. So assume $0 leq r < infty$.) Fix any $epsilon > 0$. This means that there exists $N$ such that for $n geq N$, we have
$$
fraca_n+1a_n leq r + epsilon.
$$
From this, can you deduce that for $n geq N$, we have
$$
fraca_na_N leq (r+epsilon)^n-N?
$$
Rearranging a bit,
$$
a_n leq (r+epsilon)^n left( fraca_N(r+epsilon)^N right),
$$
so that
$$
a_n^1/n leq (r+epsilon) left( fraca_N(r+epsilon)^N right)^1/n.
$$
Can you take it from here?
You do not have to work so hard for the left hand side inequality: there is a simple way to obtain the left hand side inequality using the right hand side one in a black-box way. I will leave you to figure it out.
answered Oct 2 '11 at 23:51
SrivatsanSrivatsan
21.1k371126
$endgroup$
add a comment |
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1 Answer
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votes
$begingroup$
A somewhat detailed hint for the right hand side inequality:
Suppose
$$
r := lim sup fraca_n+1a_n.
$$
(If the above expression is $infty$, then there is nothing to prove. So assume $0 leq r < infty$.) Fix any $epsilon > 0$. This means that there exists $N$ such that for $n geq N$, we have
$$
fraca_n+1a_n leq r + epsilon.
$$
From this, can you deduce that for $n geq N$, we have
$$
fraca_na_N leq (r+epsilon)^n-N?
$$
Rearranging a bit,
$$
a_n leq (r+epsilon)^n left( fraca_N(r+epsilon)^N right),
$$
so that
$$
a_n^1/n leq (r+epsilon) left( fraca_N(r+epsilon)^N right)^1/n.
$$
Can you take it from here?
You do not have to work so hard for the left hand side inequality: there is a simple way to obtain the left hand side inequality using the right hand side one in a black-box way. I will leave you to figure it out.
answered Oct 2 '11 at 23:51
SrivatsanSrivatsan
21.1k371126
$endgroup$
add a comment |
$begingroup$
A somewhat detailed hint for the right hand side inequality:
Suppose
$$
r := lim sup fraca_n+1a_n.
$$
(If the above expression is $infty$, then there is nothing to prove. So assume $0 leq r < infty$.) Fix any $epsilon > 0$. This means that there exists $N$ such that for $n geq N$, we have
$$
fraca_n+1a_n leq r + epsilon.
$$
From this, can you deduce that for $n geq N$, we have
$$
fraca_na_N leq (r+epsilon)^n-N?
$$
Rearranging a bit,
$$
a_n leq (r+epsilon)^n left( fraca_N(r+epsilon)^N right),
$$
so that
$$
a_n^1/n leq (r+epsilon) left( fraca_N(r+epsilon)^N right)^1/n.
$$
Can you take it from here?
You do not have to work so hard for the left hand side inequality: there is a simple way to obtain the left hand side inequality using the right hand side one in a black-box way. I will leave you to figure it out.
answered Oct 2 '11 at 23:51
SrivatsanSrivatsan
21.1k371126
$endgroup$
add a comment |
$begingroup$
A somewhat detailed hint for the right hand side inequality:
Suppose
$$
r := lim sup fraca_n+1a_n.
$$
(If the above expression is $infty$, then there is nothing to prove. So assume $0 leq r < infty$.) Fix any $epsilon > 0$. This means that there exists $N$ such that for $n geq N$, we have
$$
fraca_n+1a_n leq r + epsilon.
$$
From this, can you deduce that for $n geq N$, we have
$$
fraca_na_N leq (r+epsilon)^n-N?
$$
Rearranging a bit,
$$
a_n leq (r+epsilon)^n left( fraca_N(r+epsilon)^N right),
$$
so that
$$
a_n^1/n leq (r+epsilon) left( fraca_N(r+epsilon)^N right)^1/n.
$$
Can you take it from here?
You do not have to work so hard for the left hand side inequality: there is a simple way to obtain the left hand side inequality using the right hand side one in a black-box way. I will leave you to figure it out.
answered Oct 2 '11 at 23:51
SrivatsanSrivatsan
21.1k371126
$endgroup$
A somewhat detailed hint for the right hand side inequality:
Suppose
$$
r := lim sup fraca_n+1a_n.
$$
(If the above expression is $infty$, then there is nothing to prove. So assume $0 leq r < infty$.) Fix any $epsilon > 0$. This means that there exists $N$ such that for $n geq N$, we have
$$
fraca_n+1a_n leq r + epsilon.
$$
From this, can you deduce that for $n geq N$, we have
$$
fraca_na_N leq (r+epsilon)^n-N?
$$
Rearranging a bit,
$$
a_n leq (r+epsilon)^n left( fraca_N(r+epsilon)^N right),
$$
so that
$$
a_n^1/n leq (r+epsilon) left( fraca_N(r+epsilon)^N right)^1/n.
$$
Can you take it from here?
You do not have to work so hard for the left hand side inequality: there is a simple way to obtain the left hand side inequality using the right hand side one in a black-box way. I will leave you to figure it out.
answered Oct 2 '11 at 23:51
SrivatsanSrivatsan
21.1k371126
answered Oct 2 '11 at 23:51
SrivatsanSrivatsan
21.1k371126
answered Oct 2 '11 at 23:51
SrivatsanSrivatsan
21.1k371126
answered Oct 2 '11 at 23:51
SrivatsanSrivatsan
21.1k371126
21.1k371126
add a comment |
add a comment |
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1
$begingroup$
It may sound somewhat unhelpful, but the best technique is usually to ditch the tricks (at least at first) and go one step at a time with the definitions for every term. After a while that you have become comfortable with the definitions, it gets a lot easier and a lot clearer how "tricks" and other techniques work.
$endgroup$
– Asaf Karagila♦
Oct 2 '11 at 23:56
$begingroup$
See also these two answers: math.stackexchange.com/questions/76743/limit-of-fraca-n1a-n/… and math.stackexchange.com/questions/28476/…
$endgroup$
– Martin Sleziak
Nov 1 '11 at 9:53
1
$begingroup$
This is also given as Problem 2.4.26 in the book Kaczor, Nowak: Problems in Mathematical Analysis. The problem is stated on p.46 and a solution is given on p.205.
$endgroup$
– Martin Sleziak
Jul 23 '12 at 5:55
1
$begingroup$
Also in Ross, Elementary Analysis 2ed p.79-80 with solution to third inequality.
$endgroup$
– user203509
May 1 '16 at 19:18
$begingroup$
Actualy you can take $ln$ on each term and use Stolz-Cesaro
$endgroup$
– Tony Ma
May 16 '18 at 11:45