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A question about limsup of the sequence of the average
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Let $a_n$ be a sequence of real non-negative numbers
Define $S_n = fracsum_i=1^n a_in$
Prove that
$$liminf(a_n) leq liminf(S_n) leq limsup(S_n) leq limsup(a_n)$$
I wanted to show that $infa_n:n geq t leq infS_n:ngeq t$ for all $t$
However this doesn't seem to be true
The inequality seems to hold only at n goes to infinity but not every n individually
Can anyone help me?
real-analysis limsup-and-liminf
edited 4 hours ago
GNUSupporter 8964民主女神 地下教會
12.1k72243
asked 4 hours ago
Fluffy Skye
688
add a comment |Â
up vote
1
down vote
favorite
Let $a_n$ be a sequence of real non-negative numbers
Define $S_n = fracsum_i=1^n a_in$
Prove that
$$liminf(a_n) leq liminf(S_n) leq limsup(S_n) leq limsup(a_n)$$
I wanted to show that $infa_n:n geq t leq infS_n:ngeq t$ for all $t$
However this doesn't seem to be true
The inequality seems to hold only at n goes to infinity but not every n individually
Can anyone help me?
real-analysis limsup-and-liminf
edited 4 hours ago
GNUSupporter 8964民主女神 地下教會
12.1k72243
asked 4 hours ago
Fluffy Skye
688
To see why your proposed inequality is false, you may take the first term to be strictly smaller than the rest of the sequence.
– GNUSupporter 8964民主女神 地下教會
4 hours ago
@GNUSupporter 8964民主女神 地下教會 Yes that is what I found and therefore I'm stuck.
– Fluffy Skye
4 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $a_n$ be a sequence of real non-negative numbers
Define $S_n = fracsum_i=1^n a_in$
Prove that
$$liminf(a_n) leq liminf(S_n) leq limsup(S_n) leq limsup(a_n)$$
I wanted to show that $infa_n:n geq t leq infS_n:ngeq t$ for all $t$
However this doesn't seem to be true
The inequality seems to hold only at n goes to infinity but not every n individually
Can anyone help me?
real-analysis limsup-and-liminf
edited 4 hours ago
GNUSupporter 8964民主女神 地下教會
12.1k72243
asked 4 hours ago
Fluffy Skye
688
Let $a_n$ be a sequence of real non-negative numbers
Define $S_n = fracsum_i=1^n a_in$
Prove that
$$liminf(a_n) leq liminf(S_n) leq limsup(S_n) leq limsup(a_n)$$
I wanted to show that $infa_n:n geq t leq infS_n:ngeq t$ for all $t$
However this doesn't seem to be true
The inequality seems to hold only at n goes to infinity but not every n individually
Can anyone help me?
real-analysis limsup-and-liminf
real-analysis limsup-and-liminf
edited 4 hours ago
GNUSupporter 8964民主女神 地下教會
12.1k72243
asked 4 hours ago
Fluffy Skye
688
edited 4 hours ago
GNUSupporter 8964民主女神 地下教會
12.1k72243
asked 4 hours ago
Fluffy Skye
688
edited 4 hours ago
GNUSupporter 8964民主女神 地下教會
12.1k72243
edited 4 hours ago
GNUSupporter 8964民主女神 地下教會
12.1k72243
edited 4 hours ago
GNUSupporter 8964民主女神 地下教會
12.1k72243
12.1k72243
asked 4 hours ago
Fluffy Skye
688
asked 4 hours ago
Fluffy Skye
688
asked 4 hours ago
Fluffy Skye
688
688
To see why your proposed inequality is false, you may take the first term to be strictly smaller than the rest of the sequence.
– GNUSupporter 8964民主女神 地下教會
4 hours ago
@GNUSupporter 8964民主女神 地下教會 Yes that is what I found and therefore I'm stuck.
– Fluffy Skye
4 hours ago
add a comment |Â
To see why your proposed inequality is false, you may take the first term to be strictly smaller than the rest of the sequence.
– GNUSupporter 8964民主女神 地下教會
4 hours ago
@GNUSupporter 8964民主女神 地下教會 Yes that is what I found and therefore I'm stuck.
– Fluffy Skye
4 hours ago
To see why your proposed inequality is false, you may take the first term to be strictly smaller than the rest of the sequence.
– GNUSupporter 8964民主女神 地下教會
4 hours ago
To see why your proposed inequality is false, you may take the first term to be strictly smaller than the rest of the sequence.
– GNUSupporter 8964民主女神 地下教會
4 hours ago
@GNUSupporter 8964民主女神 地下教會 Yes that is what I found and therefore I'm stuck.
– Fluffy Skye
4 hours ago
@GNUSupporter 8964民主女神 地下教會 Yes that is what I found and therefore I'm stuck.
– Fluffy Skye
4 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The statement follows directly by Stolz–Cesàro theorem
$$liminf fracA_n+1-A_nB_n+1-B_n leq liminf fracA_nB_n leq limsup fracA_nB_n leq limsup fracA_n+1-A_nB_n+1-B_n$$
by $A_n=S_n$ and $B_n=n$.
answered 4 hours ago
gimusi
80.8k74090
I am new to real analysis and haven't learnt this theorem yet. Do you have any more elementary method?
– Fluffy Skye
4 hours ago
This gave me some trouble, until I realised that the $a_n$ from the theorem is supposed to be $sum_k=1^n a_k$, not $a_n$.
– Theo Bendit
4 hours ago
Also you meant $a_n = S_n$ Right?
– Fluffy Skye
4 hours ago
@TheoBendit Opsss yes of course I fix that!
– gimusi
4 hours ago
@FluffySkye A way is to prove Stolz-Cesaro theorem and then use that.
– gimusi
4 hours ago
 |Â
show 2 more comments
up vote
2
down vote
Let $l$ be the $liminf$ of $a_n$. Then for all $varepsilon > 0$, there exists an $N in mathbbN$ such that
$$n ge N implies a_n ge l - varepsilon.$$
Let $M = sum_k=1^N-1 a_k.$ Then,
$$n ge N implies sum_k=1^n a_n = M + sum_k=N^n a_n ge M + (n - N + 1)(l - varepsilon).$$
Therefore, when $n ge N$, we have
$$fracsum_k=1^n a_nn ge fracMn + left(1 - fracN - 1nright)(l - varepsilon) to l - varepsilon.$$
Hence, every subsequence of $fracsum_k=1^n a_nn$ must converge to at least $l - varepsilon$, where $varepsilon > 0$ is arbitrary. Thus, every subsequence must converge to at least $l$, proving the inequality of the limits inferior.
answered 4 hours ago
Theo Bendit
15.6k12147
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The statement follows directly by Stolz–Cesàro theorem
$$liminf fracA_n+1-A_nB_n+1-B_n leq liminf fracA_nB_n leq limsup fracA_nB_n leq limsup fracA_n+1-A_nB_n+1-B_n$$
by $A_n=S_n$ and $B_n=n$.
answered 4 hours ago
gimusi
80.8k74090
I am new to real analysis and haven't learnt this theorem yet. Do you have any more elementary method?
– Fluffy Skye
4 hours ago
This gave me some trouble, until I realised that the $a_n$ from the theorem is supposed to be $sum_k=1^n a_k$, not $a_n$.
– Theo Bendit
4 hours ago
Also you meant $a_n = S_n$ Right?
– Fluffy Skye
4 hours ago
@TheoBendit Opsss yes of course I fix that!
– gimusi
4 hours ago
@FluffySkye A way is to prove Stolz-Cesaro theorem and then use that.
– gimusi
4 hours ago
 |Â
show 2 more comments
up vote
3
down vote
accepted
The statement follows directly by Stolz–Cesàro theorem
$$liminf fracA_n+1-A_nB_n+1-B_n leq liminf fracA_nB_n leq limsup fracA_nB_n leq limsup fracA_n+1-A_nB_n+1-B_n$$
by $A_n=S_n$ and $B_n=n$.
answered 4 hours ago
gimusi
80.8k74090
I am new to real analysis and haven't learnt this theorem yet. Do you have any more elementary method?
– Fluffy Skye
4 hours ago
This gave me some trouble, until I realised that the $a_n$ from the theorem is supposed to be $sum_k=1^n a_k$, not $a_n$.
– Theo Bendit
4 hours ago
Also you meant $a_n = S_n$ Right?
– Fluffy Skye
4 hours ago
@TheoBendit Opsss yes of course I fix that!
– gimusi
4 hours ago
@FluffySkye A way is to prove Stolz-Cesaro theorem and then use that.
– gimusi
4 hours ago
 |Â
show 2 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The statement follows directly by Stolz–Cesàro theorem
$$liminf fracA_n+1-A_nB_n+1-B_n leq liminf fracA_nB_n leq limsup fracA_nB_n leq limsup fracA_n+1-A_nB_n+1-B_n$$
by $A_n=S_n$ and $B_n=n$.
answered 4 hours ago
gimusi
80.8k74090
The statement follows directly by Stolz–Cesàro theorem
$$liminf fracA_n+1-A_nB_n+1-B_n leq liminf fracA_nB_n leq limsup fracA_nB_n leq limsup fracA_n+1-A_nB_n+1-B_n$$
by $A_n=S_n$ and $B_n=n$.
answered 4 hours ago
gimusi
80.8k74090
edited 4 hours ago
answered 4 hours ago
gimusi
80.8k74090
answered 4 hours ago
gimusi
80.8k74090
answered 4 hours ago
gimusi
80.8k74090
80.8k74090
I am new to real analysis and haven't learnt this theorem yet. Do you have any more elementary method?
– Fluffy Skye
4 hours ago
This gave me some trouble, until I realised that the $a_n$ from the theorem is supposed to be $sum_k=1^n a_k$, not $a_n$.
– Theo Bendit
4 hours ago
Also you meant $a_n = S_n$ Right?
– Fluffy Skye
4 hours ago
@TheoBendit Opsss yes of course I fix that!
– gimusi
4 hours ago
@FluffySkye A way is to prove Stolz-Cesaro theorem and then use that.
– gimusi
4 hours ago
 |Â
show 2 more comments
I am new to real analysis and haven't learnt this theorem yet. Do you have any more elementary method?
– Fluffy Skye
4 hours ago
This gave me some trouble, until I realised that the $a_n$ from the theorem is supposed to be $sum_k=1^n a_k$, not $a_n$.
– Theo Bendit
4 hours ago
Also you meant $a_n = S_n$ Right?
– Fluffy Skye
4 hours ago
@TheoBendit Opsss yes of course I fix that!
– gimusi
4 hours ago
@FluffySkye A way is to prove Stolz-Cesaro theorem and then use that.
– gimusi
4 hours ago
I am new to real analysis and haven't learnt this theorem yet. Do you have any more elementary method?
– Fluffy Skye
4 hours ago
I am new to real analysis and haven't learnt this theorem yet. Do you have any more elementary method?
– Fluffy Skye
4 hours ago
This gave me some trouble, until I realised that the $a_n$ from the theorem is supposed to be $sum_k=1^n a_k$, not $a_n$.
– Theo Bendit
4 hours ago
This gave me some trouble, until I realised that the $a_n$ from the theorem is supposed to be $sum_k=1^n a_k$, not $a_n$.
– Theo Bendit
4 hours ago
Also you meant $a_n = S_n$ Right?
– Fluffy Skye
4 hours ago
Also you meant $a_n = S_n$ Right?
– Fluffy Skye
4 hours ago
@TheoBendit Opsss yes of course I fix that!
– gimusi
4 hours ago
@TheoBendit Opsss yes of course I fix that!
– gimusi
4 hours ago
@FluffySkye A way is to prove Stolz-Cesaro theorem and then use that.
– gimusi
4 hours ago
@FluffySkye A way is to prove Stolz-Cesaro theorem and then use that.
– gimusi
4 hours ago
 |Â
show 2 more comments
up vote
2
down vote
Let $l$ be the $liminf$ of $a_n$. Then for all $varepsilon > 0$, there exists an $N in mathbbN$ such that
$$n ge N implies a_n ge l - varepsilon.$$
Let $M = sum_k=1^N-1 a_k.$ Then,
$$n ge N implies sum_k=1^n a_n = M + sum_k=N^n a_n ge M + (n - N + 1)(l - varepsilon).$$
Therefore, when $n ge N$, we have
$$fracsum_k=1^n a_nn ge fracMn + left(1 - fracN - 1nright)(l - varepsilon) to l - varepsilon.$$
Hence, every subsequence of $fracsum_k=1^n a_nn$ must converge to at least $l - varepsilon$, where $varepsilon > 0$ is arbitrary. Thus, every subsequence must converge to at least $l$, proving the inequality of the limits inferior.
answered 4 hours ago
Theo Bendit
15.6k12147
add a comment |Â
up vote
2
down vote
Let $l$ be the $liminf$ of $a_n$. Then for all $varepsilon > 0$, there exists an $N in mathbbN$ such that
$$n ge N implies a_n ge l - varepsilon.$$
Let $M = sum_k=1^N-1 a_k.$ Then,
$$n ge N implies sum_k=1^n a_n = M + sum_k=N^n a_n ge M + (n - N + 1)(l - varepsilon).$$
Therefore, when $n ge N$, we have
$$fracsum_k=1^n a_nn ge fracMn + left(1 - fracN - 1nright)(l - varepsilon) to l - varepsilon.$$
Hence, every subsequence of $fracsum_k=1^n a_nn$ must converge to at least $l - varepsilon$, where $varepsilon > 0$ is arbitrary. Thus, every subsequence must converge to at least $l$, proving the inequality of the limits inferior.
answered 4 hours ago
Theo Bendit
15.6k12147
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $l$ be the $liminf$ of $a_n$. Then for all $varepsilon > 0$, there exists an $N in mathbbN$ such that
$$n ge N implies a_n ge l - varepsilon.$$
Let $M = sum_k=1^N-1 a_k.$ Then,
$$n ge N implies sum_k=1^n a_n = M + sum_k=N^n a_n ge M + (n - N + 1)(l - varepsilon).$$
Therefore, when $n ge N$, we have
$$fracsum_k=1^n a_nn ge fracMn + left(1 - fracN - 1nright)(l - varepsilon) to l - varepsilon.$$
Hence, every subsequence of $fracsum_k=1^n a_nn$ must converge to at least $l - varepsilon$, where $varepsilon > 0$ is arbitrary. Thus, every subsequence must converge to at least $l$, proving the inequality of the limits inferior.
answered 4 hours ago
Theo Bendit
15.6k12147
Let $l$ be the $liminf$ of $a_n$. Then for all $varepsilon > 0$, there exists an $N in mathbbN$ such that
$$n ge N implies a_n ge l - varepsilon.$$
Let $M = sum_k=1^N-1 a_k.$ Then,
$$n ge N implies sum_k=1^n a_n = M + sum_k=N^n a_n ge M + (n - N + 1)(l - varepsilon).$$
Therefore, when $n ge N$, we have
$$fracsum_k=1^n a_nn ge fracMn + left(1 - fracN - 1nright)(l - varepsilon) to l - varepsilon.$$
Hence, every subsequence of $fracsum_k=1^n a_nn$ must converge to at least $l - varepsilon$, where $varepsilon > 0$ is arbitrary. Thus, every subsequence must converge to at least $l$, proving the inequality of the limits inferior.
answered 4 hours ago
Theo Bendit
15.6k12147
answered 4 hours ago
Theo Bendit
15.6k12147
answered 4 hours ago
Theo Bendit
15.6k12147
answered 4 hours ago
Theo Bendit
15.6k12147
15.6k12147
add a comment |Â
add a comment |Â
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To see why your proposed inequality is false, you may take the first term to be strictly smaller than the rest of the sequence.
– GNUSupporter 8964民主女神 地下教會
4 hours ago
@GNUSupporter 8964民主女神 地下教會 Yes that is what I found and therefore I'm stuck.
– Fluffy Skye
4 hours ago