If $sum|a_n|$ diverges and $sum a_n$ converges then $sum a_n^+$ diverges
$begingroup$
The problem goes as follows:
Let $a_n in mathbb{R}$, such that $sum_{n=1}^{infty} |{a_n}| = infty$ and $sum_{n=1}^{m} {a_n} to a in mathbb{R}$ as $m to infty$. Let $a_n^+= max{{a_n, 0} }$. Show that $sum_{n=1}^{infty} a_n^+= infty$.
Please do check the proof and make necessary corrections.
Proof:
We can say (not rigorously here) that the series cannot be entirely comprised of positive numbers and there must be a good amount of negative numbers in the series (informally). Essentially, we need to show, that if we replace the negative numbers by $0$, the series still diverges to $infty$.
Now, denoting the set of entirety of negative numbers by $N= {a_u}_n $ and positive numbers $P= {a_v}_n = {a_v^+}_n $ We know, $a_u= -|a_u| forall u in N$
We can write $sum_{n=1}^{infty} {a_n} =sum_{v=1}^{s} {|a_v|} -sum_{u=1}^{t} {|a_u|} = a ... (1)$. Now, both of the sets $N$ and $P$ must contain infinite number of elements, otherwise, by $sum_{n=1}^{infty} |{a_n}| = infty ... (2)$, the series will diverge, a contradiction. Now adding (1) and (2) we get $2sum_{v=1}^{s} {|a_v|} = infty + a =infty implies$ $sum_{n=1}^{infty} a_n^+= infty $ [ The negative terms can be considered to be vanished by $a_n^+= max{{a_n, 0} }$] .
sequences-and-series proof-verification
edited Dec 2 '18 at 9:15
Did
247k23223459
asked Dec 2 '18 at 8:57
Subhasis BiswasSubhasis Biswas
488311
$endgroup$
add a comment |
$begingroup$
The problem goes as follows:
Let $a_n in mathbb{R}$, such that $sum_{n=1}^{infty} |{a_n}| = infty$ and $sum_{n=1}^{m} {a_n} to a in mathbb{R}$ as $m to infty$. Let $a_n^+= max{{a_n, 0} }$. Show that $sum_{n=1}^{infty} a_n^+= infty$.
Please do check the proof and make necessary corrections.
Proof:
We can say (not rigorously here) that the series cannot be entirely comprised of positive numbers and there must be a good amount of negative numbers in the series (informally). Essentially, we need to show, that if we replace the negative numbers by $0$, the series still diverges to $infty$.
Now, denoting the set of entirety of negative numbers by $N= {a_u}_n $ and positive numbers $P= {a_v}_n = {a_v^+}_n $ We know, $a_u= -|a_u| forall u in N$
We can write $sum_{n=1}^{infty} {a_n} =sum_{v=1}^{s} {|a_v|} -sum_{u=1}^{t} {|a_u|} = a ... (1)$. Now, both of the sets $N$ and $P$ must contain infinite number of elements, otherwise, by $sum_{n=1}^{infty} |{a_n}| = infty ... (2)$, the series will diverge, a contradiction. Now adding (1) and (2) we get $2sum_{v=1}^{s} {|a_v|} = infty + a =infty implies$ $sum_{n=1}^{infty} a_n^+= infty $ [ The negative terms can be considered to be vanished by $a_n^+= max{{a_n, 0} }$] .
sequences-and-series proof-verification
edited Dec 2 '18 at 9:15
Did
247k23223459
asked Dec 2 '18 at 8:57
Subhasis BiswasSubhasis Biswas
488311
$endgroup$
1
$begingroup$
More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
$endgroup$
– Did
Dec 2 '18 at 9:17
add a comment |
$begingroup$
The problem goes as follows:
Let $a_n in mathbb{R}$, such that $sum_{n=1}^{infty} |{a_n}| = infty$ and $sum_{n=1}^{m} {a_n} to a in mathbb{R}$ as $m to infty$. Let $a_n^+= max{{a_n, 0} }$. Show that $sum_{n=1}^{infty} a_n^+= infty$.
Please do check the proof and make necessary corrections.
Proof:
We can say (not rigorously here) that the series cannot be entirely comprised of positive numbers and there must be a good amount of negative numbers in the series (informally). Essentially, we need to show, that if we replace the negative numbers by $0$, the series still diverges to $infty$.
Now, denoting the set of entirety of negative numbers by $N= {a_u}_n $ and positive numbers $P= {a_v}_n = {a_v^+}_n $ We know, $a_u= -|a_u| forall u in N$
We can write $sum_{n=1}^{infty} {a_n} =sum_{v=1}^{s} {|a_v|} -sum_{u=1}^{t} {|a_u|} = a ... (1)$. Now, both of the sets $N$ and $P$ must contain infinite number of elements, otherwise, by $sum_{n=1}^{infty} |{a_n}| = infty ... (2)$, the series will diverge, a contradiction. Now adding (1) and (2) we get $2sum_{v=1}^{s} {|a_v|} = infty + a =infty implies$ $sum_{n=1}^{infty} a_n^+= infty $ [ The negative terms can be considered to be vanished by $a_n^+= max{{a_n, 0} }$] .
sequences-and-series proof-verification
edited Dec 2 '18 at 9:15
Did
247k23223459
asked Dec 2 '18 at 8:57
Subhasis BiswasSubhasis Biswas
488311
$endgroup$
The problem goes as follows:
Let $a_n in mathbb{R}$, such that $sum_{n=1}^{infty} |{a_n}| = infty$ and $sum_{n=1}^{m} {a_n} to a in mathbb{R}$ as $m to infty$. Let $a_n^+= max{{a_n, 0} }$. Show that $sum_{n=1}^{infty} a_n^+= infty$.
Please do check the proof and make necessary corrections.
Proof:
We can say (not rigorously here) that the series cannot be entirely comprised of positive numbers and there must be a good amount of negative numbers in the series (informally). Essentially, we need to show, that if we replace the negative numbers by $0$, the series still diverges to $infty$.
Now, denoting the set of entirety of negative numbers by $N= {a_u}_n $ and positive numbers $P= {a_v}_n = {a_v^+}_n $ We know, $a_u= -|a_u| forall u in N$
We can write $sum_{n=1}^{infty} {a_n} =sum_{v=1}^{s} {|a_v|} -sum_{u=1}^{t} {|a_u|} = a ... (1)$. Now, both of the sets $N$ and $P$ must contain infinite number of elements, otherwise, by $sum_{n=1}^{infty} |{a_n}| = infty ... (2)$, the series will diverge, a contradiction. Now adding (1) and (2) we get $2sum_{v=1}^{s} {|a_v|} = infty + a =infty implies$ $sum_{n=1}^{infty} a_n^+= infty $ [ The negative terms can be considered to be vanished by $a_n^+= max{{a_n, 0} }$] .
sequences-and-series proof-verification
sequences-and-series proof-verification
edited Dec 2 '18 at 9:15
Did
247k23223459
asked Dec 2 '18 at 8:57
Subhasis BiswasSubhasis Biswas
488311
edited Dec 2 '18 at 9:15
Did
247k23223459
asked Dec 2 '18 at 8:57
Subhasis BiswasSubhasis Biswas
488311
edited Dec 2 '18 at 9:15
Did
247k23223459
edited Dec 2 '18 at 9:15
Did
247k23223459
edited Dec 2 '18 at 9:15
Did
247k23223459
247k23223459
asked Dec 2 '18 at 8:57
Subhasis BiswasSubhasis Biswas
488311
asked Dec 2 '18 at 8:57
Subhasis BiswasSubhasis Biswas
488311
asked Dec 2 '18 at 8:57
Subhasis BiswasSubhasis Biswas
488311
488311
1
$begingroup$
More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
$endgroup$
– Did
Dec 2 '18 at 9:17
add a comment |
1
$begingroup$
More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
$endgroup$
– Did
Dec 2 '18 at 9:17
1
1
$begingroup$
More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
$endgroup$
– Did
Dec 2 '18 at 9:17
$begingroup$
More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
$endgroup$
– Did
Dec 2 '18 at 9:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For all $n in mathbb{N}$,
$$ a_n^+ = max(a_n, 0) = frac{a_n + vert a_n vert}{2}. $$
Therefore, for $N in mathbb{N}$ :
$$ sum_{n=0}^{N} a_n^+ =frac{1}{2} sum_{n=0}^{N} a_n + frac{1}{2} sum_{n=0}^{N} vert a_n vert $$
where $displaystyle sum_{n=0}^{N} a_n ; to ; a$ as $N to +infty$ since $displaystyle sum_{n=0}^{+infty} a_n = a$ and $displaystyle sum_{n=0}^{N} vert a_n vert to +infty$ since $displaystyle sum_{n=0}^{+infty} vert a_n vert = +infty$.
answered Dec 2 '18 at 9:16
jibounetjibounet
5,6481032
$endgroup$
3
$begingroup$
Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
$endgroup$
– Martin R
Dec 2 '18 at 9:18
$begingroup$
can you please check the validity of the solution
$endgroup$
– Subhasis Biswas
Dec 2 '18 at 9:35
add a comment |
$begingroup$
(1) is rearranging the terms of a conditionally convergent series, which isn't guaranteed to equal the same series. Thus, this step is not justified.
answered Dec 2 '18 at 9:38
modest_mildewmodest_mildew
112
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
For all $n in mathbb{N}$,
$$ a_n^+ = max(a_n, 0) = frac{a_n + vert a_n vert}{2}. $$
Therefore, for $N in mathbb{N}$ :
$$ sum_{n=0}^{N} a_n^+ =frac{1}{2} sum_{n=0}^{N} a_n + frac{1}{2} sum_{n=0}^{N} vert a_n vert $$
where $displaystyle sum_{n=0}^{N} a_n ; to ; a$ as $N to +infty$ since $displaystyle sum_{n=0}^{+infty} a_n = a$ and $displaystyle sum_{n=0}^{N} vert a_n vert to +infty$ since $displaystyle sum_{n=0}^{+infty} vert a_n vert = +infty$.
answered Dec 2 '18 at 9:16
jibounetjibounet
5,6481032
$endgroup$
3
$begingroup$
Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
$endgroup$
– Martin R
Dec 2 '18 at 9:18
$begingroup$
can you please check the validity of the solution
$endgroup$
– Subhasis Biswas
Dec 2 '18 at 9:35
add a comment |
$begingroup$
For all $n in mathbb{N}$,
$$ a_n^+ = max(a_n, 0) = frac{a_n + vert a_n vert}{2}. $$
Therefore, for $N in mathbb{N}$ :
$$ sum_{n=0}^{N} a_n^+ =frac{1}{2} sum_{n=0}^{N} a_n + frac{1}{2} sum_{n=0}^{N} vert a_n vert $$
where $displaystyle sum_{n=0}^{N} a_n ; to ; a$ as $N to +infty$ since $displaystyle sum_{n=0}^{+infty} a_n = a$ and $displaystyle sum_{n=0}^{N} vert a_n vert to +infty$ since $displaystyle sum_{n=0}^{+infty} vert a_n vert = +infty$.
answered Dec 2 '18 at 9:16
jibounetjibounet
5,6481032
$endgroup$
3
$begingroup$
Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
$endgroup$
– Martin R
Dec 2 '18 at 9:18
$begingroup$
can you please check the validity of the solution
$endgroup$
– Subhasis Biswas
Dec 2 '18 at 9:35
add a comment |
$begingroup$
For all $n in mathbb{N}$,
$$ a_n^+ = max(a_n, 0) = frac{a_n + vert a_n vert}{2}. $$
Therefore, for $N in mathbb{N}$ :
$$ sum_{n=0}^{N} a_n^+ =frac{1}{2} sum_{n=0}^{N} a_n + frac{1}{2} sum_{n=0}^{N} vert a_n vert $$
where $displaystyle sum_{n=0}^{N} a_n ; to ; a$ as $N to +infty$ since $displaystyle sum_{n=0}^{+infty} a_n = a$ and $displaystyle sum_{n=0}^{N} vert a_n vert to +infty$ since $displaystyle sum_{n=0}^{+infty} vert a_n vert = +infty$.
answered Dec 2 '18 at 9:16
jibounetjibounet
5,6481032
$endgroup$
For all $n in mathbb{N}$,
$$ a_n^+ = max(a_n, 0) = frac{a_n + vert a_n vert}{2}. $$
Therefore, for $N in mathbb{N}$ :
$$ sum_{n=0}^{N} a_n^+ =frac{1}{2} sum_{n=0}^{N} a_n + frac{1}{2} sum_{n=0}^{N} vert a_n vert $$
where $displaystyle sum_{n=0}^{N} a_n ; to ; a$ as $N to +infty$ since $displaystyle sum_{n=0}^{+infty} a_n = a$ and $displaystyle sum_{n=0}^{N} vert a_n vert to +infty$ since $displaystyle sum_{n=0}^{+infty} vert a_n vert = +infty$.
answered Dec 2 '18 at 9:16
jibounetjibounet
5,6481032
answered Dec 2 '18 at 9:16
jibounetjibounet
5,6481032
answered Dec 2 '18 at 9:16
jibounetjibounet
5,6481032
answered Dec 2 '18 at 9:16
jibounetjibounet
5,6481032
5,6481032
3
$begingroup$
Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
$endgroup$
– Martin R
Dec 2 '18 at 9:18
$begingroup$
can you please check the validity of the solution
$endgroup$
– Subhasis Biswas
Dec 2 '18 at 9:35
add a comment |
3
$begingroup$
Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
$endgroup$
– Martin R
Dec 2 '18 at 9:18
$begingroup$
can you please check the validity of the solution
$endgroup$
– Subhasis Biswas
Dec 2 '18 at 9:35
3
3
$begingroup$
Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
$endgroup$
– Martin R
Dec 2 '18 at 9:18
$begingroup$
Shouldn't an answer to the question “Please do check the proof” (also tagged with [proof-verification]) make some observations about the proposed proof?
$endgroup$
– Martin R
Dec 2 '18 at 9:18
$begingroup$
can you please check the validity of the solution
$endgroup$
– Subhasis Biswas
Dec 2 '18 at 9:35
$begingroup$
can you please check the validity of the solution
$endgroup$
– Subhasis Biswas
Dec 2 '18 at 9:35
add a comment |
$begingroup$
(1) is rearranging the terms of a conditionally convergent series, which isn't guaranteed to equal the same series. Thus, this step is not justified.
answered Dec 2 '18 at 9:38
modest_mildewmodest_mildew
112
$endgroup$
add a comment |
$begingroup$
(1) is rearranging the terms of a conditionally convergent series, which isn't guaranteed to equal the same series. Thus, this step is not justified.
answered Dec 2 '18 at 9:38
modest_mildewmodest_mildew
112
$endgroup$
add a comment |
$begingroup$
(1) is rearranging the terms of a conditionally convergent series, which isn't guaranteed to equal the same series. Thus, this step is not justified.
answered Dec 2 '18 at 9:38
modest_mildewmodest_mildew
112
$endgroup$
(1) is rearranging the terms of a conditionally convergent series, which isn't guaranteed to equal the same series. Thus, this step is not justified.
answered Dec 2 '18 at 9:38
modest_mildewmodest_mildew
112
answered Dec 2 '18 at 9:38
modest_mildewmodest_mildew
112
answered Dec 2 '18 at 9:38
modest_mildewmodest_mildew
112
answered Dec 2 '18 at 9:38
modest_mildewmodest_mildew
112
112
add a comment |
add a comment |
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$begingroup$
More simply, $a_n=a_n^+-a_n^-$ and $|a_n|=a_n^++a_n^-$ hence, if $sum a_n$ and $sum a_n^+$ both converge then $sum a_n^-$ converges as well hence $sum|a_n|$ converges. Now take the contraposition.
$endgroup$
– Did
Dec 2 '18 at 9:17