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A convergent regrouping of a series which tends to 0 implies convergence
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Absolute convergence of a real seriesConvergence of “alternating” harmonic series where sign is +, --, +++, ----, etc.Leibniz series convergence citeria valid for monotonic increasing zero-sequence?Uniform convergence of the given series of function.If I can't use Alt. Series Conv. test, how do I determine if it converges or diverges?Verifying if series is convergentQuestions about Divergence and Convergence in Seriesexample of a convergent series whose alternating series diverges?Necessary and sufficient conditions for the convergence of a seriesSeries convergence for even and odd terms
$begingroup$
A while ago I was trying to solve the problem of whether the following series converges:
$ 1 -1/2 -1/3 +1/4+1/5-1/6-1/7+1/8+1/9...$
My solution was to repeat a similar argument to the common proof od the Alternating series test where I show that the sequence of partial sums of positive terms converges and the sequence of partial sums of the negative terms converges, then using that show that the partial sums converge.
However, the above series can easily be grouped into an alternating series. My question is, is the convergence of a single regrouping, combined with the fact that the terms of the original series individually go to 0 enough to imply convergence?
real-analysis sequences-and-series alternative-proof
asked Apr 8 at 19:42
mmmmommmmo
1347
$endgroup$
add a comment |
$begingroup$
A while ago I was trying to solve the problem of whether the following series converges:
$ 1 -1/2 -1/3 +1/4+1/5-1/6-1/7+1/8+1/9...$
My solution was to repeat a similar argument to the common proof od the Alternating series test where I show that the sequence of partial sums of positive terms converges and the sequence of partial sums of the negative terms converges, then using that show that the partial sums converge.
However, the above series can easily be grouped into an alternating series. My question is, is the convergence of a single regrouping, combined with the fact that the terms of the original series individually go to 0 enough to imply convergence?
real-analysis sequences-and-series alternative-proof
asked Apr 8 at 19:42
mmmmommmmo
1347
$endgroup$
$begingroup$
If I understand your question correctly, the answer is no. I think you can rephrase the question: does the convergence of a series imply the convergence of every rearrangement of the series? You can easily construct counterexamples for this latter statement, though, so it isn’t true.
$endgroup$
– Clayton
Apr 8 at 19:58
add a comment |
$begingroup$
A while ago I was trying to solve the problem of whether the following series converges:
$ 1 -1/2 -1/3 +1/4+1/5-1/6-1/7+1/8+1/9...$
My solution was to repeat a similar argument to the common proof od the Alternating series test where I show that the sequence of partial sums of positive terms converges and the sequence of partial sums of the negative terms converges, then using that show that the partial sums converge.
However, the above series can easily be grouped into an alternating series. My question is, is the convergence of a single regrouping, combined with the fact that the terms of the original series individually go to 0 enough to imply convergence?
real-analysis sequences-and-series alternative-proof
asked Apr 8 at 19:42
mmmmommmmo
1347
$endgroup$
A while ago I was trying to solve the problem of whether the following series converges:
$ 1 -1/2 -1/3 +1/4+1/5-1/6-1/7+1/8+1/9...$
My solution was to repeat a similar argument to the common proof od the Alternating series test where I show that the sequence of partial sums of positive terms converges and the sequence of partial sums of the negative terms converges, then using that show that the partial sums converge.
However, the above series can easily be grouped into an alternating series. My question is, is the convergence of a single regrouping, combined with the fact that the terms of the original series individually go to 0 enough to imply convergence?
real-analysis sequences-and-series alternative-proof
real-analysis sequences-and-series alternative-proof
asked Apr 8 at 19:42
mmmmommmmo
1347
asked Apr 8 at 19:42
mmmmommmmo
1347
asked Apr 8 at 19:42
mmmmommmmo
1347
asked Apr 8 at 19:42
mmmmommmmo
1347
asked Apr 8 at 19:42
mmmmommmmo
1347
1347
$begingroup$
If I understand your question correctly, the answer is no. I think you can rephrase the question: does the convergence of a series imply the convergence of every rearrangement of the series? You can easily construct counterexamples for this latter statement, though, so it isn’t true.
$endgroup$
– Clayton
Apr 8 at 19:58
add a comment |
$begingroup$
If I understand your question correctly, the answer is no. I think you can rephrase the question: does the convergence of a series imply the convergence of every rearrangement of the series? You can easily construct counterexamples for this latter statement, though, so it isn’t true.
$endgroup$
– Clayton
Apr 8 at 19:58
$begingroup$
If I understand your question correctly, the answer is no. I think you can rephrase the question: does the convergence of a series imply the convergence of every rearrangement of the series? You can easily construct counterexamples for this latter statement, though, so it isn’t true.
$endgroup$
– Clayton
Apr 8 at 19:58
$begingroup$
If I understand your question correctly, the answer is no. I think you can rephrase the question: does the convergence of a series imply the convergence of every rearrangement of the series? You can easily construct counterexamples for this latter statement, though, so it isn’t true.
$endgroup$
– Clayton
Apr 8 at 19:58
add a comment |
1 Answer
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$begingroup$
It is as long as there is some $k$ such that each group is of size at most $k$. Lets say the $k$-th group starts at position $a_k$.
Then partial sum of the original series is $sumlimits_i=1^n x_i = sumlimits_k = 1^m sumlimits_i = a_k^a_k + 1 - 1 x_i + sumlimits_i=a_m + 1^n x_i$, where $m$ is such that $a_m + 1 < n leqslant a_m + 2$.
The first sum is a partial sum of regrouped series, and the second goes to zero, as it has at most $k$ addends and $x_i to 0$.
answered Apr 8 at 19:59
mihaildmihaild
85210
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mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
It is as long as there is some $k$ such that each group is of size at most $k$. Lets say the $k$-th group starts at position $a_k$.
Then partial sum of the original series is $sumlimits_i=1^n x_i = sumlimits_k = 1^m sumlimits_i = a_k^a_k + 1 - 1 x_i + sumlimits_i=a_m + 1^n x_i$, where $m$ is such that $a_m + 1 < n leqslant a_m + 2$.
The first sum is a partial sum of regrouped series, and the second goes to zero, as it has at most $k$ addends and $x_i to 0$.
answered Apr 8 at 19:59
mihaildmihaild
85210
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
It is as long as there is some $k$ such that each group is of size at most $k$. Lets say the $k$-th group starts at position $a_k$.
Then partial sum of the original series is $sumlimits_i=1^n x_i = sumlimits_k = 1^m sumlimits_i = a_k^a_k + 1 - 1 x_i + sumlimits_i=a_m + 1^n x_i$, where $m$ is such that $a_m + 1 < n leqslant a_m + 2$.
The first sum is a partial sum of regrouped series, and the second goes to zero, as it has at most $k$ addends and $x_i to 0$.
answered Apr 8 at 19:59
mihaildmihaild
85210
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
It is as long as there is some $k$ such that each group is of size at most $k$. Lets say the $k$-th group starts at position $a_k$.
Then partial sum of the original series is $sumlimits_i=1^n x_i = sumlimits_k = 1^m sumlimits_i = a_k^a_k + 1 - 1 x_i + sumlimits_i=a_m + 1^n x_i$, where $m$ is such that $a_m + 1 < n leqslant a_m + 2$.
The first sum is a partial sum of regrouped series, and the second goes to zero, as it has at most $k$ addends and $x_i to 0$.
answered Apr 8 at 19:59
mihaildmihaild
85210
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
It is as long as there is some $k$ such that each group is of size at most $k$. Lets say the $k$-th group starts at position $a_k$.
Then partial sum of the original series is $sumlimits_i=1^n x_i = sumlimits_k = 1^m sumlimits_i = a_k^a_k + 1 - 1 x_i + sumlimits_i=a_m + 1^n x_i$, where $m$ is such that $a_m + 1 < n leqslant a_m + 2$.
The first sum is a partial sum of regrouped series, and the second goes to zero, as it has at most $k$ addends and $x_i to 0$.
answered Apr 8 at 19:59
mihaildmihaild
85210
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 8 at 19:59
mihaildmihaild
85210
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 8 at 19:59
mihaildmihaild
85210
answered Apr 8 at 19:59
mihaildmihaild
85210
85210
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
If I understand your question correctly, the answer is no. I think you can rephrase the question: does the convergence of a series imply the convergence of every rearrangement of the series? You can easily construct counterexamples for this latter statement, though, so it isn’t true.
$endgroup$
– Clayton
Apr 8 at 19:58