Mixing
Density argument
$begingroup$
This is a problem that I met when studying density things:
Let ${ a_n }$ be a real-valued sequence. Then the following things are equivalent:
(1) $lim_{n rightarrow infty } frac{1}{n}sum_{i=1}^{n}a_n=0$
(2) There exists some $J subset mathbb{N}$ with natural density $0$, such that $lim_{n rightarrow infty, n notin J} a_n=0$
I'm a beginner on Density arguments, so can somebody help?
density-function
asked Jan 28 at 11:34
lminsllminsl
886
$endgroup$
add a comment |
$begingroup$
This is a problem that I met when studying density things:
Let ${ a_n }$ be a real-valued sequence. Then the following things are equivalent:
(1) $lim_{n rightarrow infty } frac{1}{n}sum_{i=1}^{n}a_n=0$
(2) There exists some $J subset mathbb{N}$ with natural density $0$, such that $lim_{n rightarrow infty, n notin J} a_n=0$
I'm a beginner on Density arguments, so can somebody help?
density-function
asked Jan 28 at 11:34
lminsllminsl
886
$endgroup$
$begingroup$
$(a_n)$ is just a real sequence? It fails when $a_n =(-1)^n$.
$endgroup$
– Song
Jan 28 at 11:41
add a comment |
$begingroup$
This is a problem that I met when studying density things:
Let ${ a_n }$ be a real-valued sequence. Then the following things are equivalent:
(1) $lim_{n rightarrow infty } frac{1}{n}sum_{i=1}^{n}a_n=0$
(2) There exists some $J subset mathbb{N}$ with natural density $0$, such that $lim_{n rightarrow infty, n notin J} a_n=0$
I'm a beginner on Density arguments, so can somebody help?
density-function
asked Jan 28 at 11:34
lminsllminsl
886
$endgroup$
This is a problem that I met when studying density things:
Let ${ a_n }$ be a real-valued sequence. Then the following things are equivalent:
(1) $lim_{n rightarrow infty } frac{1}{n}sum_{i=1}^{n}a_n=0$
(2) There exists some $J subset mathbb{N}$ with natural density $0$, such that $lim_{n rightarrow infty, n notin J} a_n=0$
I'm a beginner on Density arguments, so can somebody help?
density-function
density-function
asked Jan 28 at 11:34
lminsllminsl
886
asked Jan 28 at 11:34
lminsllminsl
886
asked Jan 28 at 11:34
lminsllminsl
886
asked Jan 28 at 11:34
lminsllminsl
886
asked Jan 28 at 11:34
lminsllminsl
886
886
$begingroup$
$(a_n)$ is just a real sequence? It fails when $a_n =(-1)^n$.
$endgroup$
– Song
Jan 28 at 11:41
add a comment |
$begingroup$
$(a_n)$ is just a real sequence? It fails when $a_n =(-1)^n$.
$endgroup$
– Song
Jan 28 at 11:41
$begingroup$
$(a_n)$ is just a real sequence? It fails when $a_n =(-1)^n$.
$endgroup$
– Song
Jan 28 at 11:41
$begingroup$
$(a_n)$ is just a real sequence? It fails when $a_n =(-1)^n$.
$endgroup$
– Song
Jan 28 at 11:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The statement is wrong. If $a_n=(-1)^{n}$ then 1) holds but 2) doesn't.
However we have the following:
Let ${a_{n}}⊂ℝ$ be bounded. Then the following are equivalent:
a) $(1/n)sumlimits_{k=1}^{n}|a_{k}|→0$
b) there exists $Asubset mathbb N$ such that $lim_{nin A,n to infty} a_{n}=0$ and $card(A∩{1,2,...,n})/n)→0$ as n→∞
Here is a stronger result:
Let ${a_{n}}subset
%TCIMACRO{U{211d} }%
%BeginExpansion
mathbb{R}
%EndExpansion
$ be bounded and $1leq p<infty $. Then the following are equivalent:
a) $frac{1}{n}sumlimits_{k=1}^{n}leftvert a_{k}rightvert rightarrow
0$
b) there exists $Asubset
%TCIMACRO{U{2115} }%
%BeginExpansion
mathbb{N}
%EndExpansion
$ such that $underset{nnotin A,nrightarrow infty }{lim }a_{n}=0$ and $%
frac{#{Acap {1,2,...,n})}{n}rightarrow 0$ as $nrightarrow infty $
c) $frac{1}{n}sumlimits_{k=1}^{n}leftvert a_{k}rightvert
^{p}rightarrow 0$
It suffices to show that a) and b) are equivalent. Suppose $frac{1}{n}%
sumlimits_{k=1}^{n}leftvert a_{k}rightvert rightarrow 0$. For $%
k=1,2,...$ let $I_{k}={ngeq 1:leftvert a_{n}rightvert geq frac{1}{k}%
}$. Claim: $frac{#{I_{k}cap lbrack 1,n]}}{n}rightarrow 0$ as $%
nrightarrow infty $ for each $k$. Indeed, this follows from the inequality
$frac{1}{n}sumlimits_{j=1}^{n}leftvert a_{j}rightvert geq frac{%
#{I_{k}cap lbrack 1,n]}}{nk}$. There exist integers $n_{0}<n_{1}<...$
such that $ngeq n_{k}$ implies $frac{#{I_{k+1}cap lbrack 1,n]}}{n}<%
frac{1}{k+1}$. Let $I=bigcuplimits_{k=0}^{infty }{I_{k+1}cap lbrack
n_{k},n_{k+1})}$. Let $n_{k}leq n<n_{k+1}$. Then $Icap lbrack
1,n]subset {I_{k}cap lbrack 1,n_{k}]}cup {I_{k+1}cap lbrack 1,n]}$%
. Hence $frac{#{Icap lbrack 1,n]}}{n}leq frac{#{I_{k}cap lbrack
1,n_{k}]}}{n}+frac{#{I_{k+1}cap lbrack 1,n]}}{n}<frac{n_{k}}{n}frac{%
1}{k}+frac{1}{k+1}leq frac{1}{k}+frac{1}{k+1}$ if $ngeq n_{k}$. We have
proved that $frac{#{Icap lbrack 0,n)}}{n}rightarrow 0$ as $%
nrightarrow infty $. If $n>n_{k}$ and $nnotin I$ then $nnotin I_{k+1}$
( for, otherwise, there exists $rho geq k$ such that $n_{rho }leq
n<n_{rho +1}$ and $nin I_{k+1}subset I_{rho +1}$ so $nin I_{rho
+1}cap lbrack n_{rho },n_{rho +1)}subset I$ which is a contradiction).
Thus $leftvert a_{n}rightvert <frac{1}{k+1}$ for $n>n_{k}$, $nnotin I$
completing the proof of a) implies b). For the converse part let $leftvert
a_{n}rightvert leq C$ and let $epsilon >0$. There exists $n_{epsilon }$
such that $leftvert a_{n}rightvert <epsilon $ if $n>n_{epsilon }$ and $%
nnotin I$. Also there exists $m_{epsilon }$ such that $frac{#{Icap
lbrack 0,n)}}{n}<epsilon $ if $n>m_{epsilon }$. For $n>max
{n_{epsilon },m_{epsilon }}$ we have $frac{1}{n}sum%
limits_{k=0}^{n-1}leftvert a_{k}rightvert <epsilon +epsilon C$.
answered Jan 28 at 11:48
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
$endgroup$
$begingroup$
Oh; I forgot to type the "bounded" thing.. Could you provide a proof?
$endgroup$
– lminsl
Jan 28 at 14:45
$begingroup$
@lminsl I have provided a detailed proof.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:51
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The statement is wrong. If $a_n=(-1)^{n}$ then 1) holds but 2) doesn't.
However we have the following:
Let ${a_{n}}⊂ℝ$ be bounded. Then the following are equivalent:
a) $(1/n)sumlimits_{k=1}^{n}|a_{k}|→0$
b) there exists $Asubset mathbb N$ such that $lim_{nin A,n to infty} a_{n}=0$ and $card(A∩{1,2,...,n})/n)→0$ as n→∞
Here is a stronger result:
Let ${a_{n}}subset
%TCIMACRO{U{211d} }%
%BeginExpansion
mathbb{R}
%EndExpansion
$ be bounded and $1leq p<infty $. Then the following are equivalent:
a) $frac{1}{n}sumlimits_{k=1}^{n}leftvert a_{k}rightvert rightarrow
0$
b) there exists $Asubset
%TCIMACRO{U{2115} }%
%BeginExpansion
mathbb{N}
%EndExpansion
$ such that $underset{nnotin A,nrightarrow infty }{lim }a_{n}=0$ and $%
frac{#{Acap {1,2,...,n})}{n}rightarrow 0$ as $nrightarrow infty $
c) $frac{1}{n}sumlimits_{k=1}^{n}leftvert a_{k}rightvert
^{p}rightarrow 0$
It suffices to show that a) and b) are equivalent. Suppose $frac{1}{n}%
sumlimits_{k=1}^{n}leftvert a_{k}rightvert rightarrow 0$. For $%
k=1,2,...$ let $I_{k}={ngeq 1:leftvert a_{n}rightvert geq frac{1}{k}%
}$. Claim: $frac{#{I_{k}cap lbrack 1,n]}}{n}rightarrow 0$ as $%
nrightarrow infty $ for each $k$. Indeed, this follows from the inequality
$frac{1}{n}sumlimits_{j=1}^{n}leftvert a_{j}rightvert geq frac{%
#{I_{k}cap lbrack 1,n]}}{nk}$. There exist integers $n_{0}<n_{1}<...$
such that $ngeq n_{k}$ implies $frac{#{I_{k+1}cap lbrack 1,n]}}{n}<%
frac{1}{k+1}$. Let $I=bigcuplimits_{k=0}^{infty }{I_{k+1}cap lbrack
n_{k},n_{k+1})}$. Let $n_{k}leq n<n_{k+1}$. Then $Icap lbrack
1,n]subset {I_{k}cap lbrack 1,n_{k}]}cup {I_{k+1}cap lbrack 1,n]}$%
. Hence $frac{#{Icap lbrack 1,n]}}{n}leq frac{#{I_{k}cap lbrack
1,n_{k}]}}{n}+frac{#{I_{k+1}cap lbrack 1,n]}}{n}<frac{n_{k}}{n}frac{%
1}{k}+frac{1}{k+1}leq frac{1}{k}+frac{1}{k+1}$ if $ngeq n_{k}$. We have
proved that $frac{#{Icap lbrack 0,n)}}{n}rightarrow 0$ as $%
nrightarrow infty $. If $n>n_{k}$ and $nnotin I$ then $nnotin I_{k+1}$
( for, otherwise, there exists $rho geq k$ such that $n_{rho }leq
n<n_{rho +1}$ and $nin I_{k+1}subset I_{rho +1}$ so $nin I_{rho
+1}cap lbrack n_{rho },n_{rho +1)}subset I$ which is a contradiction).
Thus $leftvert a_{n}rightvert <frac{1}{k+1}$ for $n>n_{k}$, $nnotin I$
completing the proof of a) implies b). For the converse part let $leftvert
a_{n}rightvert leq C$ and let $epsilon >0$. There exists $n_{epsilon }$
such that $leftvert a_{n}rightvert <epsilon $ if $n>n_{epsilon }$ and $%
nnotin I$. Also there exists $m_{epsilon }$ such that $frac{#{Icap
lbrack 0,n)}}{n}<epsilon $ if $n>m_{epsilon }$. For $n>max
{n_{epsilon },m_{epsilon }}$ we have $frac{1}{n}sum%
limits_{k=0}^{n-1}leftvert a_{k}rightvert <epsilon +epsilon C$.
answered Jan 28 at 11:48
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
$endgroup$
$begingroup$
Oh; I forgot to type the "bounded" thing.. Could you provide a proof?
$endgroup$
– lminsl
Jan 28 at 14:45
$begingroup$
@lminsl I have provided a detailed proof.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:51
add a comment |
$begingroup$
The statement is wrong. If $a_n=(-1)^{n}$ then 1) holds but 2) doesn't.
However we have the following:
Let ${a_{n}}⊂ℝ$ be bounded. Then the following are equivalent:
a) $(1/n)sumlimits_{k=1}^{n}|a_{k}|→0$
b) there exists $Asubset mathbb N$ such that $lim_{nin A,n to infty} a_{n}=0$ and $card(A∩{1,2,...,n})/n)→0$ as n→∞
Here is a stronger result:
Let ${a_{n}}subset
%TCIMACRO{U{211d} }%
%BeginExpansion
mathbb{R}
%EndExpansion
$ be bounded and $1leq p<infty $. Then the following are equivalent:
a) $frac{1}{n}sumlimits_{k=1}^{n}leftvert a_{k}rightvert rightarrow
0$
b) there exists $Asubset
%TCIMACRO{U{2115} }%
%BeginExpansion
mathbb{N}
%EndExpansion
$ such that $underset{nnotin A,nrightarrow infty }{lim }a_{n}=0$ and $%
frac{#{Acap {1,2,...,n})}{n}rightarrow 0$ as $nrightarrow infty $
c) $frac{1}{n}sumlimits_{k=1}^{n}leftvert a_{k}rightvert
^{p}rightarrow 0$
It suffices to show that a) and b) are equivalent. Suppose $frac{1}{n}%
sumlimits_{k=1}^{n}leftvert a_{k}rightvert rightarrow 0$. For $%
k=1,2,...$ let $I_{k}={ngeq 1:leftvert a_{n}rightvert geq frac{1}{k}%
}$. Claim: $frac{#{I_{k}cap lbrack 1,n]}}{n}rightarrow 0$ as $%
nrightarrow infty $ for each $k$. Indeed, this follows from the inequality
$frac{1}{n}sumlimits_{j=1}^{n}leftvert a_{j}rightvert geq frac{%
#{I_{k}cap lbrack 1,n]}}{nk}$. There exist integers $n_{0}<n_{1}<...$
such that $ngeq n_{k}$ implies $frac{#{I_{k+1}cap lbrack 1,n]}}{n}<%
frac{1}{k+1}$. Let $I=bigcuplimits_{k=0}^{infty }{I_{k+1}cap lbrack
n_{k},n_{k+1})}$. Let $n_{k}leq n<n_{k+1}$. Then $Icap lbrack
1,n]subset {I_{k}cap lbrack 1,n_{k}]}cup {I_{k+1}cap lbrack 1,n]}$%
. Hence $frac{#{Icap lbrack 1,n]}}{n}leq frac{#{I_{k}cap lbrack
1,n_{k}]}}{n}+frac{#{I_{k+1}cap lbrack 1,n]}}{n}<frac{n_{k}}{n}frac{%
1}{k}+frac{1}{k+1}leq frac{1}{k}+frac{1}{k+1}$ if $ngeq n_{k}$. We have
proved that $frac{#{Icap lbrack 0,n)}}{n}rightarrow 0$ as $%
nrightarrow infty $. If $n>n_{k}$ and $nnotin I$ then $nnotin I_{k+1}$
( for, otherwise, there exists $rho geq k$ such that $n_{rho }leq
n<n_{rho +1}$ and $nin I_{k+1}subset I_{rho +1}$ so $nin I_{rho
+1}cap lbrack n_{rho },n_{rho +1)}subset I$ which is a contradiction).
Thus $leftvert a_{n}rightvert <frac{1}{k+1}$ for $n>n_{k}$, $nnotin I$
completing the proof of a) implies b). For the converse part let $leftvert
a_{n}rightvert leq C$ and let $epsilon >0$. There exists $n_{epsilon }$
such that $leftvert a_{n}rightvert <epsilon $ if $n>n_{epsilon }$ and $%
nnotin I$. Also there exists $m_{epsilon }$ such that $frac{#{Icap
lbrack 0,n)}}{n}<epsilon $ if $n>m_{epsilon }$. For $n>max
{n_{epsilon },m_{epsilon }}$ we have $frac{1}{n}sum%
limits_{k=0}^{n-1}leftvert a_{k}rightvert <epsilon +epsilon C$.
answered Jan 28 at 11:48
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
$endgroup$
$begingroup$
Oh; I forgot to type the "bounded" thing.. Could you provide a proof?
$endgroup$
– lminsl
Jan 28 at 14:45
$begingroup$
@lminsl I have provided a detailed proof.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:51
add a comment |
$begingroup$
The statement is wrong. If $a_n=(-1)^{n}$ then 1) holds but 2) doesn't.
However we have the following:
Let ${a_{n}}⊂ℝ$ be bounded. Then the following are equivalent:
a) $(1/n)sumlimits_{k=1}^{n}|a_{k}|→0$
b) there exists $Asubset mathbb N$ such that $lim_{nin A,n to infty} a_{n}=0$ and $card(A∩{1,2,...,n})/n)→0$ as n→∞
Here is a stronger result:
Let ${a_{n}}subset
%TCIMACRO{U{211d} }%
%BeginExpansion
mathbb{R}
%EndExpansion
$ be bounded and $1leq p<infty $. Then the following are equivalent:
a) $frac{1}{n}sumlimits_{k=1}^{n}leftvert a_{k}rightvert rightarrow
0$
b) there exists $Asubset
%TCIMACRO{U{2115} }%
%BeginExpansion
mathbb{N}
%EndExpansion
$ such that $underset{nnotin A,nrightarrow infty }{lim }a_{n}=0$ and $%
frac{#{Acap {1,2,...,n})}{n}rightarrow 0$ as $nrightarrow infty $
c) $frac{1}{n}sumlimits_{k=1}^{n}leftvert a_{k}rightvert
^{p}rightarrow 0$
It suffices to show that a) and b) are equivalent. Suppose $frac{1}{n}%
sumlimits_{k=1}^{n}leftvert a_{k}rightvert rightarrow 0$. For $%
k=1,2,...$ let $I_{k}={ngeq 1:leftvert a_{n}rightvert geq frac{1}{k}%
}$. Claim: $frac{#{I_{k}cap lbrack 1,n]}}{n}rightarrow 0$ as $%
nrightarrow infty $ for each $k$. Indeed, this follows from the inequality
$frac{1}{n}sumlimits_{j=1}^{n}leftvert a_{j}rightvert geq frac{%
#{I_{k}cap lbrack 1,n]}}{nk}$. There exist integers $n_{0}<n_{1}<...$
such that $ngeq n_{k}$ implies $frac{#{I_{k+1}cap lbrack 1,n]}}{n}<%
frac{1}{k+1}$. Let $I=bigcuplimits_{k=0}^{infty }{I_{k+1}cap lbrack
n_{k},n_{k+1})}$. Let $n_{k}leq n<n_{k+1}$. Then $Icap lbrack
1,n]subset {I_{k}cap lbrack 1,n_{k}]}cup {I_{k+1}cap lbrack 1,n]}$%
. Hence $frac{#{Icap lbrack 1,n]}}{n}leq frac{#{I_{k}cap lbrack
1,n_{k}]}}{n}+frac{#{I_{k+1}cap lbrack 1,n]}}{n}<frac{n_{k}}{n}frac{%
1}{k}+frac{1}{k+1}leq frac{1}{k}+frac{1}{k+1}$ if $ngeq n_{k}$. We have
proved that $frac{#{Icap lbrack 0,n)}}{n}rightarrow 0$ as $%
nrightarrow infty $. If $n>n_{k}$ and $nnotin I$ then $nnotin I_{k+1}$
( for, otherwise, there exists $rho geq k$ such that $n_{rho }leq
n<n_{rho +1}$ and $nin I_{k+1}subset I_{rho +1}$ so $nin I_{rho
+1}cap lbrack n_{rho },n_{rho +1)}subset I$ which is a contradiction).
Thus $leftvert a_{n}rightvert <frac{1}{k+1}$ for $n>n_{k}$, $nnotin I$
completing the proof of a) implies b). For the converse part let $leftvert
a_{n}rightvert leq C$ and let $epsilon >0$. There exists $n_{epsilon }$
such that $leftvert a_{n}rightvert <epsilon $ if $n>n_{epsilon }$ and $%
nnotin I$. Also there exists $m_{epsilon }$ such that $frac{#{Icap
lbrack 0,n)}}{n}<epsilon $ if $n>m_{epsilon }$. For $n>max
{n_{epsilon },m_{epsilon }}$ we have $frac{1}{n}sum%
limits_{k=0}^{n-1}leftvert a_{k}rightvert <epsilon +epsilon C$.
answered Jan 28 at 11:48
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
$endgroup$
The statement is wrong. If $a_n=(-1)^{n}$ then 1) holds but 2) doesn't.
However we have the following:
Let ${a_{n}}⊂ℝ$ be bounded. Then the following are equivalent:
a) $(1/n)sumlimits_{k=1}^{n}|a_{k}|→0$
b) there exists $Asubset mathbb N$ such that $lim_{nin A,n to infty} a_{n}=0$ and $card(A∩{1,2,...,n})/n)→0$ as n→∞
Here is a stronger result:
Let ${a_{n}}subset
%TCIMACRO{U{211d} }%
%BeginExpansion
mathbb{R}
%EndExpansion
$ be bounded and $1leq p<infty $. Then the following are equivalent:
a) $frac{1}{n}sumlimits_{k=1}^{n}leftvert a_{k}rightvert rightarrow
0$
b) there exists $Asubset
%TCIMACRO{U{2115} }%
%BeginExpansion
mathbb{N}
%EndExpansion
$ such that $underset{nnotin A,nrightarrow infty }{lim }a_{n}=0$ and $%
frac{#{Acap {1,2,...,n})}{n}rightarrow 0$ as $nrightarrow infty $
c) $frac{1}{n}sumlimits_{k=1}^{n}leftvert a_{k}rightvert
^{p}rightarrow 0$
It suffices to show that a) and b) are equivalent. Suppose $frac{1}{n}%
sumlimits_{k=1}^{n}leftvert a_{k}rightvert rightarrow 0$. For $%
k=1,2,...$ let $I_{k}={ngeq 1:leftvert a_{n}rightvert geq frac{1}{k}%
}$. Claim: $frac{#{I_{k}cap lbrack 1,n]}}{n}rightarrow 0$ as $%
nrightarrow infty $ for each $k$. Indeed, this follows from the inequality
$frac{1}{n}sumlimits_{j=1}^{n}leftvert a_{j}rightvert geq frac{%
#{I_{k}cap lbrack 1,n]}}{nk}$. There exist integers $n_{0}<n_{1}<...$
such that $ngeq n_{k}$ implies $frac{#{I_{k+1}cap lbrack 1,n]}}{n}<%
frac{1}{k+1}$. Let $I=bigcuplimits_{k=0}^{infty }{I_{k+1}cap lbrack
n_{k},n_{k+1})}$. Let $n_{k}leq n<n_{k+1}$. Then $Icap lbrack
1,n]subset {I_{k}cap lbrack 1,n_{k}]}cup {I_{k+1}cap lbrack 1,n]}$%
. Hence $frac{#{Icap lbrack 1,n]}}{n}leq frac{#{I_{k}cap lbrack
1,n_{k}]}}{n}+frac{#{I_{k+1}cap lbrack 1,n]}}{n}<frac{n_{k}}{n}frac{%
1}{k}+frac{1}{k+1}leq frac{1}{k}+frac{1}{k+1}$ if $ngeq n_{k}$. We have
proved that $frac{#{Icap lbrack 0,n)}}{n}rightarrow 0$ as $%
nrightarrow infty $. If $n>n_{k}$ and $nnotin I$ then $nnotin I_{k+1}$
( for, otherwise, there exists $rho geq k$ such that $n_{rho }leq
n<n_{rho +1}$ and $nin I_{k+1}subset I_{rho +1}$ so $nin I_{rho
+1}cap lbrack n_{rho },n_{rho +1)}subset I$ which is a contradiction).
Thus $leftvert a_{n}rightvert <frac{1}{k+1}$ for $n>n_{k}$, $nnotin I$
completing the proof of a) implies b). For the converse part let $leftvert
a_{n}rightvert leq C$ and let $epsilon >0$. There exists $n_{epsilon }$
such that $leftvert a_{n}rightvert <epsilon $ if $n>n_{epsilon }$ and $%
nnotin I$. Also there exists $m_{epsilon }$ such that $frac{#{Icap
lbrack 0,n)}}{n}<epsilon $ if $n>m_{epsilon }$. For $n>max
{n_{epsilon },m_{epsilon }}$ we have $frac{1}{n}sum%
limits_{k=0}^{n-1}leftvert a_{k}rightvert <epsilon +epsilon C$.
answered Jan 28 at 11:48
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
edited Jan 28 at 23:50
answered Jan 28 at 11:48
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
answered Jan 28 at 11:48
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
answered Jan 28 at 11:48
Kavi Rama MurthyKavi Rama Murthy
70.7k53170
70.7k53170
$begingroup$
Oh; I forgot to type the "bounded" thing.. Could you provide a proof?
$endgroup$
– lminsl
Jan 28 at 14:45
$begingroup$
@lminsl I have provided a detailed proof.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:51
add a comment |
$begingroup$
Oh; I forgot to type the "bounded" thing.. Could you provide a proof?
$endgroup$
– lminsl
Jan 28 at 14:45
$begingroup$
@lminsl I have provided a detailed proof.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:51
$begingroup$
Oh; I forgot to type the "bounded" thing.. Could you provide a proof?
$endgroup$
– lminsl
Jan 28 at 14:45
$begingroup$
Oh; I forgot to type the "bounded" thing.. Could you provide a proof?
$endgroup$
– lminsl
Jan 28 at 14:45
$begingroup$
@lminsl I have provided a detailed proof.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:51
$begingroup$
@lminsl I have provided a detailed proof.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 23:51
add a comment |
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$begingroup$
$(a_n)$ is just a real sequence? It fails when $a_n =(-1)^n$.
$endgroup$
– Song
Jan 28 at 11:41