Boundary for Series That Converge “Fast Enough”
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I would first like to clarify that I understand the P-Series convergence and am not interested in the boundary case for convergence of those.
With that said, I am curious as to whether there is some series that forms a sort of boundary case determining whether a series will converge or diverge in a general sense. Is there a series, call it $A$, for which:
$B_n < A_n$ for all n $rightarrow B_n$ converges
AND
$B_n > A_n$ for all n $rightarrow B_n$ does not converge?
Yes, the P-series with $p=1$ acts like this (sorta, since of course $0.25n^{-1} < n^{-1}$), but only applies to P-series. Does anything like this exist for a general series? If not, why not? Can we prove that no such series exists?
I would also be interested in any examples of other classes of series where there is a "boundary" case.
This question comes from the often quoted statement that a sequence goes to zero "fast enough" for its series to converge. My thought is that if there are sequences that are "fast enough" and others that are "too slow", then there should be some transition point where any sequence that goes to zero faster than some series would converge and those that are slower would not converge. This is something along the lines of the Intermediate Value Theorem for sequences and series.
sequences-and-series
asked Dec 3 '18 at 5:03
Jon WittmerJon Wittmer
12
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|
show 1 more comment
$begingroup$
I would first like to clarify that I understand the P-Series convergence and am not interested in the boundary case for convergence of those.
With that said, I am curious as to whether there is some series that forms a sort of boundary case determining whether a series will converge or diverge in a general sense. Is there a series, call it $A$, for which:
$B_n < A_n$ for all n $rightarrow B_n$ converges
AND
$B_n > A_n$ for all n $rightarrow B_n$ does not converge?
Yes, the P-series with $p=1$ acts like this (sorta, since of course $0.25n^{-1} < n^{-1}$), but only applies to P-series. Does anything like this exist for a general series? If not, why not? Can we prove that no such series exists?
I would also be interested in any examples of other classes of series where there is a "boundary" case.
This question comes from the often quoted statement that a sequence goes to zero "fast enough" for its series to converge. My thought is that if there are sequences that are "fast enough" and others that are "too slow", then there should be some transition point where any sequence that goes to zero faster than some series would converge and those that are slower would not converge. This is something along the lines of the Intermediate Value Theorem for sequences and series.
sequences-and-series
asked Dec 3 '18 at 5:03
Jon WittmerJon Wittmer
12
$endgroup$
3
$begingroup$
No. Given any convergent series, there are bigger series that converge; given any divergent series, there are smaller series that diverge. Exercise for the reader.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 5:22
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If An and Bn converge, An + Bn converge and An-Bn converge ...
$endgroup$
– Damien
Dec 3 '18 at 6:20
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@GerryMyerson I don't disagree with you, as you could always multiply any sequence by any constant and achieve result you claim. I guess I'm trying to understand what exactly "fast enough" means. We could certainly spend the rest of our lives coming up with examples that converge/diverge, but is there some sort of formal definition for what constitutes "fast enough"?
$endgroup$
– Jon Wittmer
Dec 3 '18 at 21:10
$begingroup$
You can do more than multiply by a constant. Given any convergent series $sum a_n$ there is a convergent series $sum b_n$ with $b_n/a_ntoinfty$; given any divergent series $sum a_n$ there is a divergent series $sum b_n$ with $b_n/a_nto0$. I think the only definition of "fast enough" is the definition of convergence.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:20
$begingroup$
It may be worth having a look at math.stackexchange.com/questions/1234661/… and the links given there.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:23
|
show 1 more comment
$begingroup$
I would first like to clarify that I understand the P-Series convergence and am not interested in the boundary case for convergence of those.
With that said, I am curious as to whether there is some series that forms a sort of boundary case determining whether a series will converge or diverge in a general sense. Is there a series, call it $A$, for which:
$B_n < A_n$ for all n $rightarrow B_n$ converges
AND
$B_n > A_n$ for all n $rightarrow B_n$ does not converge?
Yes, the P-series with $p=1$ acts like this (sorta, since of course $0.25n^{-1} < n^{-1}$), but only applies to P-series. Does anything like this exist for a general series? If not, why not? Can we prove that no such series exists?
I would also be interested in any examples of other classes of series where there is a "boundary" case.
This question comes from the often quoted statement that a sequence goes to zero "fast enough" for its series to converge. My thought is that if there are sequences that are "fast enough" and others that are "too slow", then there should be some transition point where any sequence that goes to zero faster than some series would converge and those that are slower would not converge. This is something along the lines of the Intermediate Value Theorem for sequences and series.
sequences-and-series
asked Dec 3 '18 at 5:03
Jon WittmerJon Wittmer
12
$endgroup$
I would first like to clarify that I understand the P-Series convergence and am not interested in the boundary case for convergence of those.
With that said, I am curious as to whether there is some series that forms a sort of boundary case determining whether a series will converge or diverge in a general sense. Is there a series, call it $A$, for which:
$B_n < A_n$ for all n $rightarrow B_n$ converges
AND
$B_n > A_n$ for all n $rightarrow B_n$ does not converge?
Yes, the P-series with $p=1$ acts like this (sorta, since of course $0.25n^{-1} < n^{-1}$), but only applies to P-series. Does anything like this exist for a general series? If not, why not? Can we prove that no such series exists?
I would also be interested in any examples of other classes of series where there is a "boundary" case.
This question comes from the often quoted statement that a sequence goes to zero "fast enough" for its series to converge. My thought is that if there are sequences that are "fast enough" and others that are "too slow", then there should be some transition point where any sequence that goes to zero faster than some series would converge and those that are slower would not converge. This is something along the lines of the Intermediate Value Theorem for sequences and series.
sequences-and-series
sequences-and-series
asked Dec 3 '18 at 5:03
Jon WittmerJon Wittmer
12
asked Dec 3 '18 at 5:03
Jon WittmerJon Wittmer
12
asked Dec 3 '18 at 5:03
Jon WittmerJon Wittmer
12
asked Dec 3 '18 at 5:03
Jon WittmerJon Wittmer
12
asked Dec 3 '18 at 5:03
Jon WittmerJon Wittmer
12
12
3
$begingroup$
No. Given any convergent series, there are bigger series that converge; given any divergent series, there are smaller series that diverge. Exercise for the reader.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 5:22
$begingroup$
If An and Bn converge, An + Bn converge and An-Bn converge ...
$endgroup$
– Damien
Dec 3 '18 at 6:20
$begingroup$
@GerryMyerson I don't disagree with you, as you could always multiply any sequence by any constant and achieve result you claim. I guess I'm trying to understand what exactly "fast enough" means. We could certainly spend the rest of our lives coming up with examples that converge/diverge, but is there some sort of formal definition for what constitutes "fast enough"?
$endgroup$
– Jon Wittmer
Dec 3 '18 at 21:10
$begingroup$
You can do more than multiply by a constant. Given any convergent series $sum a_n$ there is a convergent series $sum b_n$ with $b_n/a_ntoinfty$; given any divergent series $sum a_n$ there is a divergent series $sum b_n$ with $b_n/a_nto0$. I think the only definition of "fast enough" is the definition of convergence.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:20
$begingroup$
It may be worth having a look at math.stackexchange.com/questions/1234661/… and the links given there.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:23
|
show 1 more comment
3
$begingroup$
No. Given any convergent series, there are bigger series that converge; given any divergent series, there are smaller series that diverge. Exercise for the reader.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 5:22
$begingroup$
If An and Bn converge, An + Bn converge and An-Bn converge ...
$endgroup$
– Damien
Dec 3 '18 at 6:20
$begingroup$
@GerryMyerson I don't disagree with you, as you could always multiply any sequence by any constant and achieve result you claim. I guess I'm trying to understand what exactly "fast enough" means. We could certainly spend the rest of our lives coming up with examples that converge/diverge, but is there some sort of formal definition for what constitutes "fast enough"?
$endgroup$
– Jon Wittmer
Dec 3 '18 at 21:10
$begingroup$
You can do more than multiply by a constant. Given any convergent series $sum a_n$ there is a convergent series $sum b_n$ with $b_n/a_ntoinfty$; given any divergent series $sum a_n$ there is a divergent series $sum b_n$ with $b_n/a_nto0$. I think the only definition of "fast enough" is the definition of convergence.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:20
$begingroup$
It may be worth having a look at math.stackexchange.com/questions/1234661/… and the links given there.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:23
3
3
$begingroup$
No. Given any convergent series, there are bigger series that converge; given any divergent series, there are smaller series that diverge. Exercise for the reader.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 5:22
$begingroup$
No. Given any convergent series, there are bigger series that converge; given any divergent series, there are smaller series that diverge. Exercise for the reader.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 5:22
$begingroup$
If An and Bn converge, An + Bn converge and An-Bn converge ...
$endgroup$
– Damien
Dec 3 '18 at 6:20
$begingroup$
If An and Bn converge, An + Bn converge and An-Bn converge ...
$endgroup$
– Damien
Dec 3 '18 at 6:20
$begingroup$
@GerryMyerson I don't disagree with you, as you could always multiply any sequence by any constant and achieve result you claim. I guess I'm trying to understand what exactly "fast enough" means. We could certainly spend the rest of our lives coming up with examples that converge/diverge, but is there some sort of formal definition for what constitutes "fast enough"?
$endgroup$
– Jon Wittmer
Dec 3 '18 at 21:10
$begingroup$
@GerryMyerson I don't disagree with you, as you could always multiply any sequence by any constant and achieve result you claim. I guess I'm trying to understand what exactly "fast enough" means. We could certainly spend the rest of our lives coming up with examples that converge/diverge, but is there some sort of formal definition for what constitutes "fast enough"?
$endgroup$
– Jon Wittmer
Dec 3 '18 at 21:10
$begingroup$
You can do more than multiply by a constant. Given any convergent series $sum a_n$ there is a convergent series $sum b_n$ with $b_n/a_ntoinfty$; given any divergent series $sum a_n$ there is a divergent series $sum b_n$ with $b_n/a_nto0$. I think the only definition of "fast enough" is the definition of convergence.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:20
$begingroup$
You can do more than multiply by a constant. Given any convergent series $sum a_n$ there is a convergent series $sum b_n$ with $b_n/a_ntoinfty$; given any divergent series $sum a_n$ there is a divergent series $sum b_n$ with $b_n/a_nto0$. I think the only definition of "fast enough" is the definition of convergence.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:20
$begingroup$
It may be worth having a look at math.stackexchange.com/questions/1234661/… and the links given there.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:23
$begingroup$
It may be worth having a look at math.stackexchange.com/questions/1234661/… and the links given there.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:23
|
show 1 more comment
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$begingroup$
No. Given any convergent series, there are bigger series that converge; given any divergent series, there are smaller series that diverge. Exercise for the reader.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 5:22
$begingroup$
If An and Bn converge, An + Bn converge and An-Bn converge ...
$endgroup$
– Damien
Dec 3 '18 at 6:20
$begingroup$
@GerryMyerson I don't disagree with you, as you could always multiply any sequence by any constant and achieve result you claim. I guess I'm trying to understand what exactly "fast enough" means. We could certainly spend the rest of our lives coming up with examples that converge/diverge, but is there some sort of formal definition for what constitutes "fast enough"?
$endgroup$
– Jon Wittmer
Dec 3 '18 at 21:10
$begingroup$
You can do more than multiply by a constant. Given any convergent series $sum a_n$ there is a convergent series $sum b_n$ with $b_n/a_ntoinfty$; given any divergent series $sum a_n$ there is a divergent series $sum b_n$ with $b_n/a_nto0$. I think the only definition of "fast enough" is the definition of convergence.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:20
$begingroup$
It may be worth having a look at math.stackexchange.com/questions/1234661/… and the links given there.
$endgroup$
– Gerry Myerson
Dec 3 '18 at 21:23