Mixing
Interchanging Sums
$begingroup$
I'm looking to prove that if $a_{ij} geq 0$ then
$$ sum_{i=1}^infty sum_{j=1}^infty a_{ij} = sum_{j=1}^infty sum_{i=1}^infty a_{ij}$$.
For my proof I let $S= sum_{j=1}^infty sum_{i=1}^infty a_{ij}$ and I establish the inequality $sum_{i=1}^infty sum_{j=1}^infty a_{ij} leq S$.
I'm trying to prove the reverse inequality in the case that $S$ is infinite. i.e.,
$$ S leq sum_{i=1}^infty sum_{j=1}^infty a_{ij}$$
which means that the double sum $sum_{i=1}^infty sum_{j=1}^infty a_{ij}$ is also infinite. Any ideas how to show this? I'm stumped.
real-analysis sequences-and-series
asked Jan 23 at 21:40
RdrrRdrr
208111
$endgroup$
add a comment |
$begingroup$
I'm looking to prove that if $a_{ij} geq 0$ then
$$ sum_{i=1}^infty sum_{j=1}^infty a_{ij} = sum_{j=1}^infty sum_{i=1}^infty a_{ij}$$.
For my proof I let $S= sum_{j=1}^infty sum_{i=1}^infty a_{ij}$ and I establish the inequality $sum_{i=1}^infty sum_{j=1}^infty a_{ij} leq S$.
I'm trying to prove the reverse inequality in the case that $S$ is infinite. i.e.,
$$ S leq sum_{i=1}^infty sum_{j=1}^infty a_{ij}$$
which means that the double sum $sum_{i=1}^infty sum_{j=1}^infty a_{ij}$ is also infinite. Any ideas how to show this? I'm stumped.
real-analysis sequences-and-series
asked Jan 23 at 21:40
RdrrRdrr
208111
$endgroup$
add a comment |
$begingroup$
I'm looking to prove that if $a_{ij} geq 0$ then
$$ sum_{i=1}^infty sum_{j=1}^infty a_{ij} = sum_{j=1}^infty sum_{i=1}^infty a_{ij}$$.
For my proof I let $S= sum_{j=1}^infty sum_{i=1}^infty a_{ij}$ and I establish the inequality $sum_{i=1}^infty sum_{j=1}^infty a_{ij} leq S$.
I'm trying to prove the reverse inequality in the case that $S$ is infinite. i.e.,
$$ S leq sum_{i=1}^infty sum_{j=1}^infty a_{ij}$$
which means that the double sum $sum_{i=1}^infty sum_{j=1}^infty a_{ij}$ is also infinite. Any ideas how to show this? I'm stumped.
real-analysis sequences-and-series
asked Jan 23 at 21:40
RdrrRdrr
208111
$endgroup$
I'm looking to prove that if $a_{ij} geq 0$ then
$$ sum_{i=1}^infty sum_{j=1}^infty a_{ij} = sum_{j=1}^infty sum_{i=1}^infty a_{ij}$$.
For my proof I let $S= sum_{j=1}^infty sum_{i=1}^infty a_{ij}$ and I establish the inequality $sum_{i=1}^infty sum_{j=1}^infty a_{ij} leq S$.
I'm trying to prove the reverse inequality in the case that $S$ is infinite. i.e.,
$$ S leq sum_{i=1}^infty sum_{j=1}^infty a_{ij}$$
which means that the double sum $sum_{i=1}^infty sum_{j=1}^infty a_{ij}$ is also infinite. Any ideas how to show this? I'm stumped.
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Jan 23 at 21:40
RdrrRdrr
208111
asked Jan 23 at 21:40
RdrrRdrr
208111
asked Jan 23 at 21:40
RdrrRdrr
208111
asked Jan 23 at 21:40
RdrrRdrr
208111
asked Jan 23 at 21:40
RdrrRdrr
208111
208111
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $b_{ij}:=a_{ji}$ and apply your first result to $b_{ij}$. Then you get the reverse result.
answered Jan 23 at 21:43
NotoriousJuanGNotoriousJuanG
843
$endgroup$
$begingroup$
How? Applying the first inequality to $b_{ij}$ gives $sum_{i=1}^infty sum_{j=1}^infty b_{ij} leq sum_{j=1}^infty sum_{i=1}^infty b_{ij}$, but then this means $sum_{i=1}^infty sum_{j=1}^infty a_{ji} leq sum_{j=1}^infty sum_{i=1}^infty a_{ji}$ which is not what I want. Swapping the order of the indices doesn't swap the order of the sums.
$endgroup$
– Rdrr
Jan 23 at 21:47
$begingroup$
Related:math.stackexchange.com/questions/3079002/…
$endgroup$
– Peter Szilas
Jan 23 at 22:09
$begingroup$
That doesn't appear to answer my specific question. Basicall I want to show if the RHS sum diverges to infinity, then so does LHS. The hypothesis include the convergence of the sums.
$endgroup$
– Rdrr
Jan 23 at 23:11
$begingroup$
Note that $sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ji}=sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ij}$ and $sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ji}=sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ij}$ by swapping the names of the indices i and j. This leads to the conclusion you want.
$endgroup$
– NotoriousJuanG
Jan 24 at 0:42
add a comment |
$begingroup$
Note that since $a_{ij} geqslant 0$,
$$S' = sum_{i=1}^infty sum_{j=1}^infty a_{ij} geqslant sum_{i=1}^infty sum_{j=1}^N a_{ij} = sum_{j=1}^N sum_{i=1}^infty a_{ij}, $$
where the last equality follows because $sum_{j=1}^N lim_{M to infty} S_M = lim_{M to infty} sum_{j=1}^N S_M$.
(The limit of a finite sum of sequences equals the sum of the individual limits, and this is true if we have extended nonnegative real-valued limits).
Thus,
$$S' geqslant limsup_{N to infty} sum_{j=1}^N sum_{i=1}^infty a_{ij} = S=+infty$$
answered Jan 23 at 23:17
RRLRRL
52.6k42573
$endgroup$
add a comment |
$begingroup$
Do not bother about convergence at all. We have $ sumlimits_{i=1}^{N} sumlimits_{j=1}^{M} a_{ij} =sumlimits_{j=1}^{M} sumlimits_{i=1}^{N} a_{ij} leq sumlimits_{j=1}^{infty} sumlimits_{i=1}^{infty} a_{ij}$. Let $M to infty$ and then $N to infty $ to get $sumlimits_{i=1}^{infty} sumlimits_{j=1}^{infty} a_{ij} leq sumlimits_{j=1}^{infty} sumlimits_{i=1}^{infty} a_{ij}$. By symmtery the reverse inequality also holds.
answered Jan 23 at 23:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $b_{ij}:=a_{ji}$ and apply your first result to $b_{ij}$. Then you get the reverse result.
answered Jan 23 at 21:43
NotoriousJuanGNotoriousJuanG
843
$endgroup$
$begingroup$
How? Applying the first inequality to $b_{ij}$ gives $sum_{i=1}^infty sum_{j=1}^infty b_{ij} leq sum_{j=1}^infty sum_{i=1}^infty b_{ij}$, but then this means $sum_{i=1}^infty sum_{j=1}^infty a_{ji} leq sum_{j=1}^infty sum_{i=1}^infty a_{ji}$ which is not what I want. Swapping the order of the indices doesn't swap the order of the sums.
$endgroup$
– Rdrr
Jan 23 at 21:47
$begingroup$
Related:math.stackexchange.com/questions/3079002/…
$endgroup$
– Peter Szilas
Jan 23 at 22:09
$begingroup$
That doesn't appear to answer my specific question. Basicall I want to show if the RHS sum diverges to infinity, then so does LHS. The hypothesis include the convergence of the sums.
$endgroup$
– Rdrr
Jan 23 at 23:11
$begingroup$
Note that $sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ji}=sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ij}$ and $sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ji}=sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ij}$ by swapping the names of the indices i and j. This leads to the conclusion you want.
$endgroup$
– NotoriousJuanG
Jan 24 at 0:42
add a comment |
$begingroup$
Let $b_{ij}:=a_{ji}$ and apply your first result to $b_{ij}$. Then you get the reverse result.
answered Jan 23 at 21:43
NotoriousJuanGNotoriousJuanG
843
$endgroup$
$begingroup$
How? Applying the first inequality to $b_{ij}$ gives $sum_{i=1}^infty sum_{j=1}^infty b_{ij} leq sum_{j=1}^infty sum_{i=1}^infty b_{ij}$, but then this means $sum_{i=1}^infty sum_{j=1}^infty a_{ji} leq sum_{j=1}^infty sum_{i=1}^infty a_{ji}$ which is not what I want. Swapping the order of the indices doesn't swap the order of the sums.
$endgroup$
– Rdrr
Jan 23 at 21:47
$begingroup$
Related:math.stackexchange.com/questions/3079002/…
$endgroup$
– Peter Szilas
Jan 23 at 22:09
$begingroup$
That doesn't appear to answer my specific question. Basicall I want to show if the RHS sum diverges to infinity, then so does LHS. The hypothesis include the convergence of the sums.
$endgroup$
– Rdrr
Jan 23 at 23:11
$begingroup$
Note that $sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ji}=sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ij}$ and $sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ji}=sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ij}$ by swapping the names of the indices i and j. This leads to the conclusion you want.
$endgroup$
– NotoriousJuanG
Jan 24 at 0:42
add a comment |
$begingroup$
Let $b_{ij}:=a_{ji}$ and apply your first result to $b_{ij}$. Then you get the reverse result.
answered Jan 23 at 21:43
NotoriousJuanGNotoriousJuanG
843
$endgroup$
Let $b_{ij}:=a_{ji}$ and apply your first result to $b_{ij}$. Then you get the reverse result.
answered Jan 23 at 21:43
NotoriousJuanGNotoriousJuanG
843
answered Jan 23 at 21:43
NotoriousJuanGNotoriousJuanG
843
answered Jan 23 at 21:43
NotoriousJuanGNotoriousJuanG
843
answered Jan 23 at 21:43
NotoriousJuanGNotoriousJuanG
843
843
$begingroup$
How? Applying the first inequality to $b_{ij}$ gives $sum_{i=1}^infty sum_{j=1}^infty b_{ij} leq sum_{j=1}^infty sum_{i=1}^infty b_{ij}$, but then this means $sum_{i=1}^infty sum_{j=1}^infty a_{ji} leq sum_{j=1}^infty sum_{i=1}^infty a_{ji}$ which is not what I want. Swapping the order of the indices doesn't swap the order of the sums.
$endgroup$
– Rdrr
Jan 23 at 21:47
$begingroup$
Related:math.stackexchange.com/questions/3079002/…
$endgroup$
– Peter Szilas
Jan 23 at 22:09
$begingroup$
That doesn't appear to answer my specific question. Basicall I want to show if the RHS sum diverges to infinity, then so does LHS. The hypothesis include the convergence of the sums.
$endgroup$
– Rdrr
Jan 23 at 23:11
$begingroup$
Note that $sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ji}=sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ij}$ and $sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ji}=sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ij}$ by swapping the names of the indices i and j. This leads to the conclusion you want.
$endgroup$
– NotoriousJuanG
Jan 24 at 0:42
add a comment |
$begingroup$
How? Applying the first inequality to $b_{ij}$ gives $sum_{i=1}^infty sum_{j=1}^infty b_{ij} leq sum_{j=1}^infty sum_{i=1}^infty b_{ij}$, but then this means $sum_{i=1}^infty sum_{j=1}^infty a_{ji} leq sum_{j=1}^infty sum_{i=1}^infty a_{ji}$ which is not what I want. Swapping the order of the indices doesn't swap the order of the sums.
$endgroup$
– Rdrr
Jan 23 at 21:47
$begingroup$
Related:math.stackexchange.com/questions/3079002/…
$endgroup$
– Peter Szilas
Jan 23 at 22:09
$begingroup$
That doesn't appear to answer my specific question. Basicall I want to show if the RHS sum diverges to infinity, then so does LHS. The hypothesis include the convergence of the sums.
$endgroup$
– Rdrr
Jan 23 at 23:11
$begingroup$
Note that $sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ji}=sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ij}$ and $sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ji}=sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ij}$ by swapping the names of the indices i and j. This leads to the conclusion you want.
$endgroup$
– NotoriousJuanG
Jan 24 at 0:42
$begingroup$
How? Applying the first inequality to $b_{ij}$ gives $sum_{i=1}^infty sum_{j=1}^infty b_{ij} leq sum_{j=1}^infty sum_{i=1}^infty b_{ij}$, but then this means $sum_{i=1}^infty sum_{j=1}^infty a_{ji} leq sum_{j=1}^infty sum_{i=1}^infty a_{ji}$ which is not what I want. Swapping the order of the indices doesn't swap the order of the sums.
$endgroup$
– Rdrr
Jan 23 at 21:47
$begingroup$
How? Applying the first inequality to $b_{ij}$ gives $sum_{i=1}^infty sum_{j=1}^infty b_{ij} leq sum_{j=1}^infty sum_{i=1}^infty b_{ij}$, but then this means $sum_{i=1}^infty sum_{j=1}^infty a_{ji} leq sum_{j=1}^infty sum_{i=1}^infty a_{ji}$ which is not what I want. Swapping the order of the indices doesn't swap the order of the sums.
$endgroup$
– Rdrr
Jan 23 at 21:47
$begingroup$
Related:math.stackexchange.com/questions/3079002/…
$endgroup$
– Peter Szilas
Jan 23 at 22:09
$begingroup$
Related:math.stackexchange.com/questions/3079002/…
$endgroup$
– Peter Szilas
Jan 23 at 22:09
$begingroup$
That doesn't appear to answer my specific question. Basicall I want to show if the RHS sum diverges to infinity, then so does LHS. The hypothesis include the convergence of the sums.
$endgroup$
– Rdrr
Jan 23 at 23:11
$begingroup$
That doesn't appear to answer my specific question. Basicall I want to show if the RHS sum diverges to infinity, then so does LHS. The hypothesis include the convergence of the sums.
$endgroup$
– Rdrr
Jan 23 at 23:11
$begingroup$
Note that $sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ji}=sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ij}$ and $sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ji}=sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ij}$ by swapping the names of the indices i and j. This leads to the conclusion you want.
$endgroup$
– NotoriousJuanG
Jan 24 at 0:42
$begingroup$
Note that $sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ji}=sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ij}$ and $sum_{j=1}^{infty}sum_{i=1}^{infty}a_{ji}=sum_{i=1}^{infty}sum_{j=1}^{infty}a_{ij}$ by swapping the names of the indices i and j. This leads to the conclusion you want.
$endgroup$
– NotoriousJuanG
Jan 24 at 0:42
add a comment |
$begingroup$
Note that since $a_{ij} geqslant 0$,
$$S' = sum_{i=1}^infty sum_{j=1}^infty a_{ij} geqslant sum_{i=1}^infty sum_{j=1}^N a_{ij} = sum_{j=1}^N sum_{i=1}^infty a_{ij}, $$
where the last equality follows because $sum_{j=1}^N lim_{M to infty} S_M = lim_{M to infty} sum_{j=1}^N S_M$.
(The limit of a finite sum of sequences equals the sum of the individual limits, and this is true if we have extended nonnegative real-valued limits).
Thus,
$$S' geqslant limsup_{N to infty} sum_{j=1}^N sum_{i=1}^infty a_{ij} = S=+infty$$
answered Jan 23 at 23:17
RRLRRL
52.6k42573
$endgroup$
add a comment |
$begingroup$
Note that since $a_{ij} geqslant 0$,
$$S' = sum_{i=1}^infty sum_{j=1}^infty a_{ij} geqslant sum_{i=1}^infty sum_{j=1}^N a_{ij} = sum_{j=1}^N sum_{i=1}^infty a_{ij}, $$
where the last equality follows because $sum_{j=1}^N lim_{M to infty} S_M = lim_{M to infty} sum_{j=1}^N S_M$.
(The limit of a finite sum of sequences equals the sum of the individual limits, and this is true if we have extended nonnegative real-valued limits).
Thus,
$$S' geqslant limsup_{N to infty} sum_{j=1}^N sum_{i=1}^infty a_{ij} = S=+infty$$
answered Jan 23 at 23:17
RRLRRL
52.6k42573
$endgroup$
add a comment |
$begingroup$
Note that since $a_{ij} geqslant 0$,
$$S' = sum_{i=1}^infty sum_{j=1}^infty a_{ij} geqslant sum_{i=1}^infty sum_{j=1}^N a_{ij} = sum_{j=1}^N sum_{i=1}^infty a_{ij}, $$
where the last equality follows because $sum_{j=1}^N lim_{M to infty} S_M = lim_{M to infty} sum_{j=1}^N S_M$.
(The limit of a finite sum of sequences equals the sum of the individual limits, and this is true if we have extended nonnegative real-valued limits).
Thus,
$$S' geqslant limsup_{N to infty} sum_{j=1}^N sum_{i=1}^infty a_{ij} = S=+infty$$
answered Jan 23 at 23:17
RRLRRL
52.6k42573
$endgroup$
Note that since $a_{ij} geqslant 0$,
$$S' = sum_{i=1}^infty sum_{j=1}^infty a_{ij} geqslant sum_{i=1}^infty sum_{j=1}^N a_{ij} = sum_{j=1}^N sum_{i=1}^infty a_{ij}, $$
where the last equality follows because $sum_{j=1}^N lim_{M to infty} S_M = lim_{M to infty} sum_{j=1}^N S_M$.
(The limit of a finite sum of sequences equals the sum of the individual limits, and this is true if we have extended nonnegative real-valued limits).
Thus,
$$S' geqslant limsup_{N to infty} sum_{j=1}^N sum_{i=1}^infty a_{ij} = S=+infty$$
answered Jan 23 at 23:17
RRLRRL
52.6k42573
edited Jan 23 at 23:23
answered Jan 23 at 23:17
RRLRRL
52.6k42573
answered Jan 23 at 23:17
RRLRRL
52.6k42573
answered Jan 23 at 23:17
RRLRRL
52.6k42573
52.6k42573
add a comment |
add a comment |
$begingroup$
Do not bother about convergence at all. We have $ sumlimits_{i=1}^{N} sumlimits_{j=1}^{M} a_{ij} =sumlimits_{j=1}^{M} sumlimits_{i=1}^{N} a_{ij} leq sumlimits_{j=1}^{infty} sumlimits_{i=1}^{infty} a_{ij}$. Let $M to infty$ and then $N to infty $ to get $sumlimits_{i=1}^{infty} sumlimits_{j=1}^{infty} a_{ij} leq sumlimits_{j=1}^{infty} sumlimits_{i=1}^{infty} a_{ij}$. By symmtery the reverse inequality also holds.
answered Jan 23 at 23:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
add a comment |
$begingroup$
Do not bother about convergence at all. We have $ sumlimits_{i=1}^{N} sumlimits_{j=1}^{M} a_{ij} =sumlimits_{j=1}^{M} sumlimits_{i=1}^{N} a_{ij} leq sumlimits_{j=1}^{infty} sumlimits_{i=1}^{infty} a_{ij}$. Let $M to infty$ and then $N to infty $ to get $sumlimits_{i=1}^{infty} sumlimits_{j=1}^{infty} a_{ij} leq sumlimits_{j=1}^{infty} sumlimits_{i=1}^{infty} a_{ij}$. By symmtery the reverse inequality also holds.
answered Jan 23 at 23:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
add a comment |
$begingroup$
Do not bother about convergence at all. We have $ sumlimits_{i=1}^{N} sumlimits_{j=1}^{M} a_{ij} =sumlimits_{j=1}^{M} sumlimits_{i=1}^{N} a_{ij} leq sumlimits_{j=1}^{infty} sumlimits_{i=1}^{infty} a_{ij}$. Let $M to infty$ and then $N to infty $ to get $sumlimits_{i=1}^{infty} sumlimits_{j=1}^{infty} a_{ij} leq sumlimits_{j=1}^{infty} sumlimits_{i=1}^{infty} a_{ij}$. By symmtery the reverse inequality also holds.
answered Jan 23 at 23:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
Do not bother about convergence at all. We have $ sumlimits_{i=1}^{N} sumlimits_{j=1}^{M} a_{ij} =sumlimits_{j=1}^{M} sumlimits_{i=1}^{N} a_{ij} leq sumlimits_{j=1}^{infty} sumlimits_{i=1}^{infty} a_{ij}$. Let $M to infty$ and then $N to infty $ to get $sumlimits_{i=1}^{infty} sumlimits_{j=1}^{infty} a_{ij} leq sumlimits_{j=1}^{infty} sumlimits_{i=1}^{infty} a_{ij}$. By symmtery the reverse inequality also holds.
answered Jan 23 at 23:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
answered Jan 23 at 23:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
answered Jan 23 at 23:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
answered Jan 23 at 23:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
67.5k53067
add a comment |
add a comment |
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