Indefinite Sum Extension of a Finite Sum Equality
$begingroup$
The other night I was considering the way in which we can split a finite sum of any arithmetic function into two finite sums, one for it's odd and another for even index terms :
$$sum _{k=1}^{n} left( -1 right) ^{k}a_{{k}}=sum _{k=1}^{ lfloor frac{n}{2} rfloor }a_{{2,k}}-sum _{k=0}^{lfloor frac{n}{2} rfloor -delta left( frac{n}{2},lfloor frac{n}{2} rfloor
right) +1}a_{{2,k+1}}
$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(FS001)}$$
where:
$delta left( x,y right) =cases{1&$x=y$cr 0&$xneq y$cr}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(KD001)}$$
And then I decided to be a little brave ask myself if this equality would hold for an infinite number of natural numbers:
$$lim _{nrightarrow infty}Bigl(sum _{k=1}^{n} left( -1 right) ^{k}a_{{k}}Bigr)=lim _{nrightarrow infty}Bigl(sum _{k=1}^{ lfloor frac{n}{2} rfloor }a_{{2,k}}Bigr)-lim _{nrightarrow infty}Bigl(sum _{k=0}^{lfloor frac{n}{2} rfloor -delta left( frac{n}{2},lfloor frac{n}{2} rfloor
right) +1}a_{{2,k+1}}
Bigr)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(IL001)}$$
Therefore the equality for the indefinite sum as follows:
$$sum _{k=1}^{infty} left( -1 right) ^{k}a_{{k}}=sum _{k=1}^{infty}a_{{2,k}}-sum _{k=0}^{infty}a_{{2,k+1}}
$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(IL002)}$$
A preliminary investigation has lead to me holding a belief that this will be true in the event that $sum_{k=1}^{infty}a_{{k}}$ is a convergent sum, (therefore, as too are the sums on the right hand side of our equality above) however I have this unsettling feeling that even if for other arithmetic functions I use finite sum approximations to their divergent sums, and these results do indeed appear to confirm that the equality holds in the infinite limit, this will not always be the case and there will exist pathalogical residuals that are dependant on the specific arithmetic function in which they occur.
So my questions are:
1)Does this bad feeling I have about the indefinite extension of the equality have any merit to warrant further study?
2) If these non zero residual functions exist, would they serve as a good foundation for me to construct predicates a partition of the set of all arithmetic functions based on the nature of their from this equality?
Be nice please I don't like this subject very much.
real-analysis convergence special-functions divergent-series
$endgroup$
add a comment |
$begingroup$
The other night I was considering the way in which we can split a finite sum of any arithmetic function into two finite sums, one for it's odd and another for even index terms :
$$sum _{k=1}^{n} left( -1 right) ^{k}a_{{k}}=sum _{k=1}^{ lfloor frac{n}{2} rfloor }a_{{2,k}}-sum _{k=0}^{lfloor frac{n}{2} rfloor -delta left( frac{n}{2},lfloor frac{n}{2} rfloor
right) +1}a_{{2,k+1}}
$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(FS001)}$$
where:
$delta left( x,y right) =cases{1&$x=y$cr 0&$xneq y$cr}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(KD001)}$$
And then I decided to be a little brave ask myself if this equality would hold for an infinite number of natural numbers:
$$lim _{nrightarrow infty}Bigl(sum _{k=1}^{n} left( -1 right) ^{k}a_{{k}}Bigr)=lim _{nrightarrow infty}Bigl(sum _{k=1}^{ lfloor frac{n}{2} rfloor }a_{{2,k}}Bigr)-lim _{nrightarrow infty}Bigl(sum _{k=0}^{lfloor frac{n}{2} rfloor -delta left( frac{n}{2},lfloor frac{n}{2} rfloor
right) +1}a_{{2,k+1}}
Bigr)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(IL001)}$$
Therefore the equality for the indefinite sum as follows:
$$sum _{k=1}^{infty} left( -1 right) ^{k}a_{{k}}=sum _{k=1}^{infty}a_{{2,k}}-sum _{k=0}^{infty}a_{{2,k+1}}
$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(IL002)}$$
A preliminary investigation has lead to me holding a belief that this will be true in the event that $sum_{k=1}^{infty}a_{{k}}$ is a convergent sum, (therefore, as too are the sums on the right hand side of our equality above) however I have this unsettling feeling that even if for other arithmetic functions I use finite sum approximations to their divergent sums, and these results do indeed appear to confirm that the equality holds in the infinite limit, this will not always be the case and there will exist pathalogical residuals that are dependant on the specific arithmetic function in which they occur.
So my questions are:
1)Does this bad feeling I have about the indefinite extension of the equality have any merit to warrant further study?
2) If these non zero residual functions exist, would they serve as a good foundation for me to construct predicates a partition of the set of all arithmetic functions based on the nature of their from this equality?
Be nice please I don't like this subject very much.
real-analysis convergence special-functions divergent-series
$endgroup$
add a comment |
$begingroup$
The other night I was considering the way in which we can split a finite sum of any arithmetic function into two finite sums, one for it's odd and another for even index terms :
$$sum _{k=1}^{n} left( -1 right) ^{k}a_{{k}}=sum _{k=1}^{ lfloor frac{n}{2} rfloor }a_{{2,k}}-sum _{k=0}^{lfloor frac{n}{2} rfloor -delta left( frac{n}{2},lfloor frac{n}{2} rfloor
right) +1}a_{{2,k+1}}
$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(FS001)}$$
where:
$delta left( x,y right) =cases{1&$x=y$cr 0&$xneq y$cr}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(KD001)}$$
And then I decided to be a little brave ask myself if this equality would hold for an infinite number of natural numbers:
$$lim _{nrightarrow infty}Bigl(sum _{k=1}^{n} left( -1 right) ^{k}a_{{k}}Bigr)=lim _{nrightarrow infty}Bigl(sum _{k=1}^{ lfloor frac{n}{2} rfloor }a_{{2,k}}Bigr)-lim _{nrightarrow infty}Bigl(sum _{k=0}^{lfloor frac{n}{2} rfloor -delta left( frac{n}{2},lfloor frac{n}{2} rfloor
right) +1}a_{{2,k+1}}
Bigr)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(IL001)}$$
Therefore the equality for the indefinite sum as follows:
$$sum _{k=1}^{infty} left( -1 right) ^{k}a_{{k}}=sum _{k=1}^{infty}a_{{2,k}}-sum _{k=0}^{infty}a_{{2,k+1}}
$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(IL002)}$$
A preliminary investigation has lead to me holding a belief that this will be true in the event that $sum_{k=1}^{infty}a_{{k}}$ is a convergent sum, (therefore, as too are the sums on the right hand side of our equality above) however I have this unsettling feeling that even if for other arithmetic functions I use finite sum approximations to their divergent sums, and these results do indeed appear to confirm that the equality holds in the infinite limit, this will not always be the case and there will exist pathalogical residuals that are dependant on the specific arithmetic function in which they occur.
So my questions are:
1)Does this bad feeling I have about the indefinite extension of the equality have any merit to warrant further study?
2) If these non zero residual functions exist, would they serve as a good foundation for me to construct predicates a partition of the set of all arithmetic functions based on the nature of their from this equality?
Be nice please I don't like this subject very much.
real-analysis convergence special-functions divergent-series
$endgroup$
The other night I was considering the way in which we can split a finite sum of any arithmetic function into two finite sums, one for it's odd and another for even index terms :
$$sum _{k=1}^{n} left( -1 right) ^{k}a_{{k}}=sum _{k=1}^{ lfloor frac{n}{2} rfloor }a_{{2,k}}-sum _{k=0}^{lfloor frac{n}{2} rfloor -delta left( frac{n}{2},lfloor frac{n}{2} rfloor
right) +1}a_{{2,k+1}}
$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(FS001)}$$
where:
$delta left( x,y right) =cases{1&$x=y$cr 0&$xneq y$cr}$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(KD001)}$$
And then I decided to be a little brave ask myself if this equality would hold for an infinite number of natural numbers:
$$lim _{nrightarrow infty}Bigl(sum _{k=1}^{n} left( -1 right) ^{k}a_{{k}}Bigr)=lim _{nrightarrow infty}Bigl(sum _{k=1}^{ lfloor frac{n}{2} rfloor }a_{{2,k}}Bigr)-lim _{nrightarrow infty}Bigl(sum _{k=0}^{lfloor frac{n}{2} rfloor -delta left( frac{n}{2},lfloor frac{n}{2} rfloor
right) +1}a_{{2,k+1}}
Bigr)$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(IL001)}$$
Therefore the equality for the indefinite sum as follows:
$$sum _{k=1}^{infty} left( -1 right) ^{k}a_{{k}}=sum _{k=1}^{infty}a_{{2,k}}-sum _{k=0}^{infty}a_{{2,k+1}}
$$
$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(IL002)}$$
A preliminary investigation has lead to me holding a belief that this will be true in the event that $sum_{k=1}^{infty}a_{{k}}$ is a convergent sum, (therefore, as too are the sums on the right hand side of our equality above) however I have this unsettling feeling that even if for other arithmetic functions I use finite sum approximations to their divergent sums, and these results do indeed appear to confirm that the equality holds in the infinite limit, this will not always be the case and there will exist pathalogical residuals that are dependant on the specific arithmetic function in which they occur.
So my questions are:
1)Does this bad feeling I have about the indefinite extension of the equality have any merit to warrant further study?
2) If these non zero residual functions exist, would they serve as a good foundation for me to construct predicates a partition of the set of all arithmetic functions based on the nature of their from this equality?
Be nice please I don't like this subject very much.
real-analysis convergence special-functions divergent-series
real-analysis convergence special-functions divergent-series
edited Dec 20 '18 at 11:14
Adam
asked Dec 20 '18 at 10:15
AdamAdam
54114
54114
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $a_k=(-1)^{k}frac 1 k$ then all the three series under consideration are divergent so the equation is not valid. Howoever, if $sum a_k$ and $sum (-1)^{k}a_k$ are both assumed to be convergent then everything is fine.
$endgroup$
$begingroup$
isn't $a_k$ divergent there? ah right the alternating does, ok you need to elaborate further are there any other examples that do not hold?
$endgroup$
– Adam
Dec 20 '18 at 10:21
$begingroup$
ok so that answers my first question, my concern was valid.
$endgroup$
– Adam
Dec 20 '18 at 10:24
$begingroup$
ok, but for the harmonic I need to understand why the finite sum equality fails in the infinite limit
$endgroup$
– Adam
Dec 20 '18 at 10:26
$begingroup$
Thankyou for your help Sir.
$endgroup$
– Adam
Dec 20 '18 at 10:28
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
If $a_k=(-1)^{k}frac 1 k$ then all the three series under consideration are divergent so the equation is not valid. Howoever, if $sum a_k$ and $sum (-1)^{k}a_k$ are both assumed to be convergent then everything is fine.
$endgroup$
$begingroup$
isn't $a_k$ divergent there? ah right the alternating does, ok you need to elaborate further are there any other examples that do not hold?
$endgroup$
– Adam
Dec 20 '18 at 10:21
$begingroup$
ok so that answers my first question, my concern was valid.
$endgroup$
– Adam
Dec 20 '18 at 10:24
$begingroup$
ok, but for the harmonic I need to understand why the finite sum equality fails in the infinite limit
$endgroup$
– Adam
Dec 20 '18 at 10:26
$begingroup$
Thankyou for your help Sir.
$endgroup$
– Adam
Dec 20 '18 at 10:28
add a comment |
$begingroup$
If $a_k=(-1)^{k}frac 1 k$ then all the three series under consideration are divergent so the equation is not valid. Howoever, if $sum a_k$ and $sum (-1)^{k}a_k$ are both assumed to be convergent then everything is fine.
$endgroup$
$begingroup$
isn't $a_k$ divergent there? ah right the alternating does, ok you need to elaborate further are there any other examples that do not hold?
$endgroup$
– Adam
Dec 20 '18 at 10:21
$begingroup$
ok so that answers my first question, my concern was valid.
$endgroup$
– Adam
Dec 20 '18 at 10:24
$begingroup$
ok, but for the harmonic I need to understand why the finite sum equality fails in the infinite limit
$endgroup$
– Adam
Dec 20 '18 at 10:26
$begingroup$
Thankyou for your help Sir.
$endgroup$
– Adam
Dec 20 '18 at 10:28
add a comment |
$begingroup$
If $a_k=(-1)^{k}frac 1 k$ then all the three series under consideration are divergent so the equation is not valid. Howoever, if $sum a_k$ and $sum (-1)^{k}a_k$ are both assumed to be convergent then everything is fine.
$endgroup$
If $a_k=(-1)^{k}frac 1 k$ then all the three series under consideration are divergent so the equation is not valid. Howoever, if $sum a_k$ and $sum (-1)^{k}a_k$ are both assumed to be convergent then everything is fine.
edited Dec 20 '18 at 10:25
answered Dec 20 '18 at 10:20
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
60.6k42161
$begingroup$
isn't $a_k$ divergent there? ah right the alternating does, ok you need to elaborate further are there any other examples that do not hold?
$endgroup$
– Adam
Dec 20 '18 at 10:21
$begingroup$
ok so that answers my first question, my concern was valid.
$endgroup$
– Adam
Dec 20 '18 at 10:24
$begingroup$
ok, but for the harmonic I need to understand why the finite sum equality fails in the infinite limit
$endgroup$
– Adam
Dec 20 '18 at 10:26
$begingroup$
Thankyou for your help Sir.
$endgroup$
– Adam
Dec 20 '18 at 10:28
add a comment |
$begingroup$
isn't $a_k$ divergent there? ah right the alternating does, ok you need to elaborate further are there any other examples that do not hold?
$endgroup$
– Adam
Dec 20 '18 at 10:21
$begingroup$
ok so that answers my first question, my concern was valid.
$endgroup$
– Adam
Dec 20 '18 at 10:24
$begingroup$
ok, but for the harmonic I need to understand why the finite sum equality fails in the infinite limit
$endgroup$
– Adam
Dec 20 '18 at 10:26
$begingroup$
Thankyou for your help Sir.
$endgroup$
– Adam
Dec 20 '18 at 10:28
$begingroup$
isn't $a_k$ divergent there? ah right the alternating does, ok you need to elaborate further are there any other examples that do not hold?
$endgroup$
– Adam
Dec 20 '18 at 10:21
$begingroup$
isn't $a_k$ divergent there? ah right the alternating does, ok you need to elaborate further are there any other examples that do not hold?
$endgroup$
– Adam
Dec 20 '18 at 10:21
$begingroup$
ok so that answers my first question, my concern was valid.
$endgroup$
– Adam
Dec 20 '18 at 10:24
$begingroup$
ok so that answers my first question, my concern was valid.
$endgroup$
– Adam
Dec 20 '18 at 10:24
$begingroup$
ok, but for the harmonic I need to understand why the finite sum equality fails in the infinite limit
$endgroup$
– Adam
Dec 20 '18 at 10:26
$begingroup$
ok, but for the harmonic I need to understand why the finite sum equality fails in the infinite limit
$endgroup$
– Adam
Dec 20 '18 at 10:26
$begingroup$
Thankyou for your help Sir.
$endgroup$
– Adam
Dec 20 '18 at 10:28
$begingroup$
Thankyou for your help Sir.
$endgroup$
– Adam
Dec 20 '18 at 10:28
add a comment |
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