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Proof verification for associative property of convergent series
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I know this question from Abbott's book has been referenced: Prove that if an infinite series converges, the associative property holds . However, I was wondering if the following is an alternative proof for such a statement.
From a theorem in Abbott's book we know any subsequence of a convergent sequence converges to the same limit as the original sequence. Define the following subsequence of the sequence of partial sums on $a_n$:
fix $ k$, and $forall m in mathbbN$, $ s_km_1$, $s_km_2,... $ (i.e) $s_3, s_6, s_9,...$ . By the theorem above, this converges to the same limit as the original sequence. Now define another subsequence, $s_k(m_1+1), s_k(m_2+1), ...$, i.e the previous sequence shifted by k.
Using these two subsequences define the following sequence:
$p_m = s_k(m_1+1) - s_k(m_1) = a_km_1+1 + a_km_1+2, ... +
a_k(m_1+1)$ where the $a_n$ are in the original sequence. These $p_m$ represent a grouping of $k$ terms in the original sequence.
Define the sequence of partial sums on the $p_m$ including $p_0$, and define $p_0 = s_k$, as $t_1 = p_0 + p_1 = s_2k , t_2 = p_0 + p_1 + p_2 = s_3k$ thus this sequence of partial sums is a subsequence of the original sequence of partial sums on the $a_n$ and thus converges to the same limit.
I feel like something is definitely missing.
real-analysis proof-verification
asked Jul 29 at 19:05
dylan7
5371310
add a comment |Â
up vote
0
down vote
favorite
I know this question from Abbott's book has been referenced: Prove that if an infinite series converges, the associative property holds . However, I was wondering if the following is an alternative proof for such a statement.
From a theorem in Abbott's book we know any subsequence of a convergent sequence converges to the same limit as the original sequence. Define the following subsequence of the sequence of partial sums on $a_n$:
fix $ k$, and $forall m in mathbbN$, $ s_km_1$, $s_km_2,... $ (i.e) $s_3, s_6, s_9,...$ . By the theorem above, this converges to the same limit as the original sequence. Now define another subsequence, $s_k(m_1+1), s_k(m_2+1), ...$, i.e the previous sequence shifted by k.
Using these two subsequences define the following sequence:
$p_m = s_k(m_1+1) - s_k(m_1) = a_km_1+1 + a_km_1+2, ... +
a_k(m_1+1)$ where the $a_n$ are in the original sequence. These $p_m$ represent a grouping of $k$ terms in the original sequence.
Define the sequence of partial sums on the $p_m$ including $p_0$, and define $p_0 = s_k$, as $t_1 = p_0 + p_1 = s_2k , t_2 = p_0 + p_1 + p_2 = s_3k$ thus this sequence of partial sums is a subsequence of the original sequence of partial sums on the $a_n$ and thus converges to the same limit.
I feel like something is definitely missing.
real-analysis proof-verification
asked Jul 29 at 19:05
dylan7
5371310
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know this question from Abbott's book has been referenced: Prove that if an infinite series converges, the associative property holds . However, I was wondering if the following is an alternative proof for such a statement.
From a theorem in Abbott's book we know any subsequence of a convergent sequence converges to the same limit as the original sequence. Define the following subsequence of the sequence of partial sums on $a_n$:
fix $ k$, and $forall m in mathbbN$, $ s_km_1$, $s_km_2,... $ (i.e) $s_3, s_6, s_9,...$ . By the theorem above, this converges to the same limit as the original sequence. Now define another subsequence, $s_k(m_1+1), s_k(m_2+1), ...$, i.e the previous sequence shifted by k.
Using these two subsequences define the following sequence:
$p_m = s_k(m_1+1) - s_k(m_1) = a_km_1+1 + a_km_1+2, ... +
a_k(m_1+1)$ where the $a_n$ are in the original sequence. These $p_m$ represent a grouping of $k$ terms in the original sequence.
Define the sequence of partial sums on the $p_m$ including $p_0$, and define $p_0 = s_k$, as $t_1 = p_0 + p_1 = s_2k , t_2 = p_0 + p_1 + p_2 = s_3k$ thus this sequence of partial sums is a subsequence of the original sequence of partial sums on the $a_n$ and thus converges to the same limit.
I feel like something is definitely missing.
real-analysis proof-verification
asked Jul 29 at 19:05
dylan7
5371310
I know this question from Abbott's book has been referenced: Prove that if an infinite series converges, the associative property holds . However, I was wondering if the following is an alternative proof for such a statement.
From a theorem in Abbott's book we know any subsequence of a convergent sequence converges to the same limit as the original sequence. Define the following subsequence of the sequence of partial sums on $a_n$:
fix $ k$, and $forall m in mathbbN$, $ s_km_1$, $s_km_2,... $ (i.e) $s_3, s_6, s_9,...$ . By the theorem above, this converges to the same limit as the original sequence. Now define another subsequence, $s_k(m_1+1), s_k(m_2+1), ...$, i.e the previous sequence shifted by k.
Using these two subsequences define the following sequence:
$p_m = s_k(m_1+1) - s_k(m_1) = a_km_1+1 + a_km_1+2, ... +
a_k(m_1+1)$ where the $a_n$ are in the original sequence. These $p_m$ represent a grouping of $k$ terms in the original sequence.
Define the sequence of partial sums on the $p_m$ including $p_0$, and define $p_0 = s_k$, as $t_1 = p_0 + p_1 = s_2k , t_2 = p_0 + p_1 + p_2 = s_3k$ thus this sequence of partial sums is a subsequence of the original sequence of partial sums on the $a_n$ and thus converges to the same limit.
I feel like something is definitely missing.
real-analysis proof-verification
asked Jul 29 at 19:05
dylan7
5371310
asked Jul 29 at 19:05
dylan7
5371310
asked Jul 29 at 19:05
dylan7
5371310
asked Jul 29 at 19:05
dylan7
5371310
5371310
add a comment |Â
add a comment |Â
1 Answer
1
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0
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What appears to be missing here is consideration for an arbitrary grouping of sum terms. The associative property would, for example, demand that
$$sum_n=1^infty frac(-1)^n+1n = left( frac11 right) + left( -frac12 + frac13 right) + left( -frac14 + frac15 - frac16 right) + ldots,$$
which your proof doesn't seem to cover.
answered Jul 29 at 19:48
Theo Bendit
11.8k1841
1
The proof proposed in the Question is for a convergent sequence, and you give one example (albeit a conditionally convergent sequence). It remains true with your grouping of terms that the new partial sums are a subsequence of the original partial sums, so it seems that the idea proposed in the Question still applies. Is your objection to the mechanics of producing "an arbitrary grouping of sum terms"? A little elaboration on how you would prefer to set up the "bookkeeping" for that would be helpful. The notation in the Question seemed perfectly general (if a bit brusque) for the situation.
– hardmath
Jul 29 at 21:00
1
This is a good note, and I do have to clarify my answer, but I'm on my way out of the door. :-) For now, let me say that I don't think the asker did assume a general grouping, only groupings that group $k$ elements together, where $k$ is a fixed integer. The grouping above does not fit this mould.
– Theo Bendit
Jul 29 at 21:05
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
What appears to be missing here is consideration for an arbitrary grouping of sum terms. The associative property would, for example, demand that
$$sum_n=1^infty frac(-1)^n+1n = left( frac11 right) + left( -frac12 + frac13 right) + left( -frac14 + frac15 - frac16 right) + ldots,$$
which your proof doesn't seem to cover.
answered Jul 29 at 19:48
Theo Bendit
11.8k1841
1
The proof proposed in the Question is for a convergent sequence, and you give one example (albeit a conditionally convergent sequence). It remains true with your grouping of terms that the new partial sums are a subsequence of the original partial sums, so it seems that the idea proposed in the Question still applies. Is your objection to the mechanics of producing "an arbitrary grouping of sum terms"? A little elaboration on how you would prefer to set up the "bookkeeping" for that would be helpful. The notation in the Question seemed perfectly general (if a bit brusque) for the situation.
– hardmath
Jul 29 at 21:00
1
This is a good note, and I do have to clarify my answer, but I'm on my way out of the door. :-) For now, let me say that I don't think the asker did assume a general grouping, only groupings that group $k$ elements together, where $k$ is a fixed integer. The grouping above does not fit this mould.
– Theo Bendit
Jul 29 at 21:05
add a comment |Â
up vote
0
down vote
What appears to be missing here is consideration for an arbitrary grouping of sum terms. The associative property would, for example, demand that
$$sum_n=1^infty frac(-1)^n+1n = left( frac11 right) + left( -frac12 + frac13 right) + left( -frac14 + frac15 - frac16 right) + ldots,$$
which your proof doesn't seem to cover.
answered Jul 29 at 19:48
Theo Bendit
11.8k1841
1
The proof proposed in the Question is for a convergent sequence, and you give one example (albeit a conditionally convergent sequence). It remains true with your grouping of terms that the new partial sums are a subsequence of the original partial sums, so it seems that the idea proposed in the Question still applies. Is your objection to the mechanics of producing "an arbitrary grouping of sum terms"? A little elaboration on how you would prefer to set up the "bookkeeping" for that would be helpful. The notation in the Question seemed perfectly general (if a bit brusque) for the situation.
– hardmath
Jul 29 at 21:00
1
This is a good note, and I do have to clarify my answer, but I'm on my way out of the door. :-) For now, let me say that I don't think the asker did assume a general grouping, only groupings that group $k$ elements together, where $k$ is a fixed integer. The grouping above does not fit this mould.
– Theo Bendit
Jul 29 at 21:05
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What appears to be missing here is consideration for an arbitrary grouping of sum terms. The associative property would, for example, demand that
$$sum_n=1^infty frac(-1)^n+1n = left( frac11 right) + left( -frac12 + frac13 right) + left( -frac14 + frac15 - frac16 right) + ldots,$$
which your proof doesn't seem to cover.
answered Jul 29 at 19:48
Theo Bendit
11.8k1841
What appears to be missing here is consideration for an arbitrary grouping of sum terms. The associative property would, for example, demand that
$$sum_n=1^infty frac(-1)^n+1n = left( frac11 right) + left( -frac12 + frac13 right) + left( -frac14 + frac15 - frac16 right) + ldots,$$
which your proof doesn't seem to cover.
answered Jul 29 at 19:48
Theo Bendit
11.8k1841
answered Jul 29 at 19:48
Theo Bendit
11.8k1841
answered Jul 29 at 19:48
Theo Bendit
11.8k1841
answered Jul 29 at 19:48
Theo Bendit
11.8k1841
11.8k1841
1
The proof proposed in the Question is for a convergent sequence, and you give one example (albeit a conditionally convergent sequence). It remains true with your grouping of terms that the new partial sums are a subsequence of the original partial sums, so it seems that the idea proposed in the Question still applies. Is your objection to the mechanics of producing "an arbitrary grouping of sum terms"? A little elaboration on how you would prefer to set up the "bookkeeping" for that would be helpful. The notation in the Question seemed perfectly general (if a bit brusque) for the situation.
– hardmath
Jul 29 at 21:00
1
This is a good note, and I do have to clarify my answer, but I'm on my way out of the door. :-) For now, let me say that I don't think the asker did assume a general grouping, only groupings that group $k$ elements together, where $k$ is a fixed integer. The grouping above does not fit this mould.
– Theo Bendit
Jul 29 at 21:05
add a comment |Â
1
The proof proposed in the Question is for a convergent sequence, and you give one example (albeit a conditionally convergent sequence). It remains true with your grouping of terms that the new partial sums are a subsequence of the original partial sums, so it seems that the idea proposed in the Question still applies. Is your objection to the mechanics of producing "an arbitrary grouping of sum terms"? A little elaboration on how you would prefer to set up the "bookkeeping" for that would be helpful. The notation in the Question seemed perfectly general (if a bit brusque) for the situation.
– hardmath
Jul 29 at 21:00
1
This is a good note, and I do have to clarify my answer, but I'm on my way out of the door. :-) For now, let me say that I don't think the asker did assume a general grouping, only groupings that group $k$ elements together, where $k$ is a fixed integer. The grouping above does not fit this mould.
– Theo Bendit
Jul 29 at 21:05
1
1
The proof proposed in the Question is for a convergent sequence, and you give one example (albeit a conditionally convergent sequence). It remains true with your grouping of terms that the new partial sums are a subsequence of the original partial sums, so it seems that the idea proposed in the Question still applies. Is your objection to the mechanics of producing "an arbitrary grouping of sum terms"? A little elaboration on how you would prefer to set up the "bookkeeping" for that would be helpful. The notation in the Question seemed perfectly general (if a bit brusque) for the situation.
– hardmath
Jul 29 at 21:00
The proof proposed in the Question is for a convergent sequence, and you give one example (albeit a conditionally convergent sequence). It remains true with your grouping of terms that the new partial sums are a subsequence of the original partial sums, so it seems that the idea proposed in the Question still applies. Is your objection to the mechanics of producing "an arbitrary grouping of sum terms"? A little elaboration on how you would prefer to set up the "bookkeeping" for that would be helpful. The notation in the Question seemed perfectly general (if a bit brusque) for the situation.
– hardmath
Jul 29 at 21:00
1
1
This is a good note, and I do have to clarify my answer, but I'm on my way out of the door. :-) For now, let me say that I don't think the asker did assume a general grouping, only groupings that group $k$ elements together, where $k$ is a fixed integer. The grouping above does not fit this mould.
– Theo Bendit
Jul 29 at 21:05
This is a good note, and I do have to clarify my answer, but I'm on my way out of the door. :-) For now, let me say that I don't think the asker did assume a general grouping, only groupings that group $k$ elements together, where $k$ is a fixed integer. The grouping above does not fit this mould.
– Theo Bendit
Jul 29 at 21:05
add a comment |Â
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