Product of corresponding terms in two Infinite series
$begingroup$
Given two series (over complex numbers) that are convergent but not absolutely convergent:
$sum_{n=0}^infty a_n = K_0$
$sum_{n=0}^infty b_n = K_1$
(where $K_0$ and $K_1$ are some real numbers)
Now, let us consider the series that is formed by multiplying the $i^{th}$ term of first series with the $i^{th}$ term of second series:
$sum_{n=0}^infty a_n*b_n = Sum_0$
(where $Sum_0$ is equal the sum of infinite terms in the series)
can we comment something on the value of $Sum_0$?
I read the Cauchy Product (How to Multiply Two Infinite Series Correctly?) but that deals with all terms in first series being multiplied with all terms in second series where as we are restricted to a subset of multiplying only corresponding terms in both series. So that doesn't seem to help.
real-analysis
asked Dec 15 '18 at 16:32
TheoryQuest1TheoryQuest1
1498
$endgroup$
add a comment |
$begingroup$
Given two series (over complex numbers) that are convergent but not absolutely convergent:
$sum_{n=0}^infty a_n = K_0$
$sum_{n=0}^infty b_n = K_1$
(where $K_0$ and $K_1$ are some real numbers)
Now, let us consider the series that is formed by multiplying the $i^{th}$ term of first series with the $i^{th}$ term of second series:
$sum_{n=0}^infty a_n*b_n = Sum_0$
(where $Sum_0$ is equal the sum of infinite terms in the series)
can we comment something on the value of $Sum_0$?
I read the Cauchy Product (How to Multiply Two Infinite Series Correctly?) but that deals with all terms in first series being multiplied with all terms in second series where as we are restricted to a subset of multiplying only corresponding terms in both series. So that doesn't seem to help.
real-analysis
asked Dec 15 '18 at 16:32
TheoryQuest1TheoryQuest1
1498
$endgroup$
3
$begingroup$
Well, the product (in the sense you specify) doesn't even need to converge. Say $a_n=b_n=frac {(-1)^n}{sqrt n}$.
$endgroup$
– lulu
Dec 15 '18 at 16:35
1
$begingroup$
It could also be zero if $a_n$ is zero on odd integers and $b_n$ on even integers. I don't think you can say anything about it.
$endgroup$
– Yanko
Dec 15 '18 at 16:36
$begingroup$
Thank you. As a follow up: in case, if both $k_0$ and $K_1$ are 0, still its unanswerable?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:39
add a comment |
$begingroup$
Given two series (over complex numbers) that are convergent but not absolutely convergent:
$sum_{n=0}^infty a_n = K_0$
$sum_{n=0}^infty b_n = K_1$
(where $K_0$ and $K_1$ are some real numbers)
Now, let us consider the series that is formed by multiplying the $i^{th}$ term of first series with the $i^{th}$ term of second series:
$sum_{n=0}^infty a_n*b_n = Sum_0$
(where $Sum_0$ is equal the sum of infinite terms in the series)
can we comment something on the value of $Sum_0$?
I read the Cauchy Product (How to Multiply Two Infinite Series Correctly?) but that deals with all terms in first series being multiplied with all terms in second series where as we are restricted to a subset of multiplying only corresponding terms in both series. So that doesn't seem to help.
real-analysis
asked Dec 15 '18 at 16:32
TheoryQuest1TheoryQuest1
1498
$endgroup$
Given two series (over complex numbers) that are convergent but not absolutely convergent:
$sum_{n=0}^infty a_n = K_0$
$sum_{n=0}^infty b_n = K_1$
(where $K_0$ and $K_1$ are some real numbers)
Now, let us consider the series that is formed by multiplying the $i^{th}$ term of first series with the $i^{th}$ term of second series:
$sum_{n=0}^infty a_n*b_n = Sum_0$
(where $Sum_0$ is equal the sum of infinite terms in the series)
can we comment something on the value of $Sum_0$?
I read the Cauchy Product (How to Multiply Two Infinite Series Correctly?) but that deals with all terms in first series being multiplied with all terms in second series where as we are restricted to a subset of multiplying only corresponding terms in both series. So that doesn't seem to help.
real-analysis
real-analysis
asked Dec 15 '18 at 16:32
TheoryQuest1TheoryQuest1
1498
asked Dec 15 '18 at 16:32
TheoryQuest1TheoryQuest1
1498
asked Dec 15 '18 at 16:32
TheoryQuest1TheoryQuest1
1498
asked Dec 15 '18 at 16:32
TheoryQuest1TheoryQuest1
1498
asked Dec 15 '18 at 16:32
TheoryQuest1TheoryQuest1
1498
1498
3
$begingroup$
Well, the product (in the sense you specify) doesn't even need to converge. Say $a_n=b_n=frac {(-1)^n}{sqrt n}$.
$endgroup$
– lulu
Dec 15 '18 at 16:35
1
$begingroup$
It could also be zero if $a_n$ is zero on odd integers and $b_n$ on even integers. I don't think you can say anything about it.
$endgroup$
– Yanko
Dec 15 '18 at 16:36
$begingroup$
Thank you. As a follow up: in case, if both $k_0$ and $K_1$ are 0, still its unanswerable?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:39
add a comment |
3
$begingroup$
Well, the product (in the sense you specify) doesn't even need to converge. Say $a_n=b_n=frac {(-1)^n}{sqrt n}$.
$endgroup$
– lulu
Dec 15 '18 at 16:35
1
$begingroup$
It could also be zero if $a_n$ is zero on odd integers and $b_n$ on even integers. I don't think you can say anything about it.
$endgroup$
– Yanko
Dec 15 '18 at 16:36
$begingroup$
Thank you. As a follow up: in case, if both $k_0$ and $K_1$ are 0, still its unanswerable?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:39
3
3
$begingroup$
Well, the product (in the sense you specify) doesn't even need to converge. Say $a_n=b_n=frac {(-1)^n}{sqrt n}$.
$endgroup$
– lulu
Dec 15 '18 at 16:35
$begingroup$
Well, the product (in the sense you specify) doesn't even need to converge. Say $a_n=b_n=frac {(-1)^n}{sqrt n}$.
$endgroup$
– lulu
Dec 15 '18 at 16:35
1
1
$begingroup$
It could also be zero if $a_n$ is zero on odd integers and $b_n$ on even integers. I don't think you can say anything about it.
$endgroup$
– Yanko
Dec 15 '18 at 16:36
$begingroup$
It could also be zero if $a_n$ is zero on odd integers and $b_n$ on even integers. I don't think you can say anything about it.
$endgroup$
– Yanko
Dec 15 '18 at 16:36
$begingroup$
Thank you. As a follow up: in case, if both $k_0$ and $K_1$ are 0, still its unanswerable?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:39
$begingroup$
Thank you. As a follow up: in case, if both $k_0$ and $K_1$ are 0, still its unanswerable?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I am fairly sure that nothing can be said about your product series; consider the sequences below with $alpha_1,alpha_2,beta_1,beta_2,C_0,C_1inmathbb{C}$:
$$
a_n = left{
begin{array}{ll}
alpha_1 & quad n=-3 \
0 & quad n=-2 \
alpha_2 & quad n=-1 \
0 & quad n=2kgeq0 \
frac{C_0}{k^2} & quad n=2k+1geq1
end{array}
right.
$$
$$
b_n = left{
begin{array}{ll}
beta_1 & quad n=-3 \
beta_2 & quad n=-2 \
0 & quad n=-1\
frac{C_1}{k^2} & quad n=2kgeq0 \
0 & quad n=2k+1geq1
end{array}
right.
$$
Then each series ($sum a_n$, $sum b_n$, and $sum a_ncdot b_n$) can be made to converge absolutely to any three complex numbers you wish, even in the case that you wish to force the first two series to be $0$.
One of the comments on the OP also demonstrates that the product series need not converge at all.
answered Dec 15 '18 at 16:43
RandomMathDudeRandomMathDude
1763
$endgroup$
$begingroup$
Thanks a lot. I understand the general point, but i am unsure about the case (0, 0). If at all possible, can you please help with a an example/argument?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:55
1
$begingroup$
@TheoryQuest1 Sorry, I may have made a typo, so I have just gone and made the answer even more carefully constructed. Now, you use $alpha_2$ to cancel the contribution of the $C_0/k^2$ part, subtract $epsilon$ from $alpha_2$ and add it to $alpha_1$. Do the same thing to $beta_1$, $C_1/k^2$, and $beta_2$. Now, your two original series sum absolutely to $0$, but the product series sums to $epsilon^2$. Every complex number has a "square root", and then we are done.
$endgroup$
– RandomMathDude
Dec 15 '18 at 17:01
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I am fairly sure that nothing can be said about your product series; consider the sequences below with $alpha_1,alpha_2,beta_1,beta_2,C_0,C_1inmathbb{C}$:
$$
a_n = left{
begin{array}{ll}
alpha_1 & quad n=-3 \
0 & quad n=-2 \
alpha_2 & quad n=-1 \
0 & quad n=2kgeq0 \
frac{C_0}{k^2} & quad n=2k+1geq1
end{array}
right.
$$
$$
b_n = left{
begin{array}{ll}
beta_1 & quad n=-3 \
beta_2 & quad n=-2 \
0 & quad n=-1\
frac{C_1}{k^2} & quad n=2kgeq0 \
0 & quad n=2k+1geq1
end{array}
right.
$$
Then each series ($sum a_n$, $sum b_n$, and $sum a_ncdot b_n$) can be made to converge absolutely to any three complex numbers you wish, even in the case that you wish to force the first two series to be $0$.
One of the comments on the OP also demonstrates that the product series need not converge at all.
answered Dec 15 '18 at 16:43
RandomMathDudeRandomMathDude
1763
$endgroup$
$begingroup$
Thanks a lot. I understand the general point, but i am unsure about the case (0, 0). If at all possible, can you please help with a an example/argument?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:55
1
$begingroup$
@TheoryQuest1 Sorry, I may have made a typo, so I have just gone and made the answer even more carefully constructed. Now, you use $alpha_2$ to cancel the contribution of the $C_0/k^2$ part, subtract $epsilon$ from $alpha_2$ and add it to $alpha_1$. Do the same thing to $beta_1$, $C_1/k^2$, and $beta_2$. Now, your two original series sum absolutely to $0$, but the product series sums to $epsilon^2$. Every complex number has a "square root", and then we are done.
$endgroup$
– RandomMathDude
Dec 15 '18 at 17:01
add a comment |
$begingroup$
I am fairly sure that nothing can be said about your product series; consider the sequences below with $alpha_1,alpha_2,beta_1,beta_2,C_0,C_1inmathbb{C}$:
$$
a_n = left{
begin{array}{ll}
alpha_1 & quad n=-3 \
0 & quad n=-2 \
alpha_2 & quad n=-1 \
0 & quad n=2kgeq0 \
frac{C_0}{k^2} & quad n=2k+1geq1
end{array}
right.
$$
$$
b_n = left{
begin{array}{ll}
beta_1 & quad n=-3 \
beta_2 & quad n=-2 \
0 & quad n=-1\
frac{C_1}{k^2} & quad n=2kgeq0 \
0 & quad n=2k+1geq1
end{array}
right.
$$
Then each series ($sum a_n$, $sum b_n$, and $sum a_ncdot b_n$) can be made to converge absolutely to any three complex numbers you wish, even in the case that you wish to force the first two series to be $0$.
One of the comments on the OP also demonstrates that the product series need not converge at all.
answered Dec 15 '18 at 16:43
RandomMathDudeRandomMathDude
1763
$endgroup$
$begingroup$
Thanks a lot. I understand the general point, but i am unsure about the case (0, 0). If at all possible, can you please help with a an example/argument?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:55
1
$begingroup$
@TheoryQuest1 Sorry, I may have made a typo, so I have just gone and made the answer even more carefully constructed. Now, you use $alpha_2$ to cancel the contribution of the $C_0/k^2$ part, subtract $epsilon$ from $alpha_2$ and add it to $alpha_1$. Do the same thing to $beta_1$, $C_1/k^2$, and $beta_2$. Now, your two original series sum absolutely to $0$, but the product series sums to $epsilon^2$. Every complex number has a "square root", and then we are done.
$endgroup$
– RandomMathDude
Dec 15 '18 at 17:01
add a comment |
$begingroup$
I am fairly sure that nothing can be said about your product series; consider the sequences below with $alpha_1,alpha_2,beta_1,beta_2,C_0,C_1inmathbb{C}$:
$$
a_n = left{
begin{array}{ll}
alpha_1 & quad n=-3 \
0 & quad n=-2 \
alpha_2 & quad n=-1 \
0 & quad n=2kgeq0 \
frac{C_0}{k^2} & quad n=2k+1geq1
end{array}
right.
$$
$$
b_n = left{
begin{array}{ll}
beta_1 & quad n=-3 \
beta_2 & quad n=-2 \
0 & quad n=-1\
frac{C_1}{k^2} & quad n=2kgeq0 \
0 & quad n=2k+1geq1
end{array}
right.
$$
Then each series ($sum a_n$, $sum b_n$, and $sum a_ncdot b_n$) can be made to converge absolutely to any three complex numbers you wish, even in the case that you wish to force the first two series to be $0$.
One of the comments on the OP also demonstrates that the product series need not converge at all.
answered Dec 15 '18 at 16:43
RandomMathDudeRandomMathDude
1763
$endgroup$
I am fairly sure that nothing can be said about your product series; consider the sequences below with $alpha_1,alpha_2,beta_1,beta_2,C_0,C_1inmathbb{C}$:
$$
a_n = left{
begin{array}{ll}
alpha_1 & quad n=-3 \
0 & quad n=-2 \
alpha_2 & quad n=-1 \
0 & quad n=2kgeq0 \
frac{C_0}{k^2} & quad n=2k+1geq1
end{array}
right.
$$
$$
b_n = left{
begin{array}{ll}
beta_1 & quad n=-3 \
beta_2 & quad n=-2 \
0 & quad n=-1\
frac{C_1}{k^2} & quad n=2kgeq0 \
0 & quad n=2k+1geq1
end{array}
right.
$$
Then each series ($sum a_n$, $sum b_n$, and $sum a_ncdot b_n$) can be made to converge absolutely to any three complex numbers you wish, even in the case that you wish to force the first two series to be $0$.
One of the comments on the OP also demonstrates that the product series need not converge at all.
answered Dec 15 '18 at 16:43
RandomMathDudeRandomMathDude
1763
edited Dec 15 '18 at 16:58
answered Dec 15 '18 at 16:43
RandomMathDudeRandomMathDude
1763
answered Dec 15 '18 at 16:43
RandomMathDudeRandomMathDude
1763
answered Dec 15 '18 at 16:43
RandomMathDudeRandomMathDude
1763
1763
$begingroup$
Thanks a lot. I understand the general point, but i am unsure about the case (0, 0). If at all possible, can you please help with a an example/argument?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:55
1
$begingroup$
@TheoryQuest1 Sorry, I may have made a typo, so I have just gone and made the answer even more carefully constructed. Now, you use $alpha_2$ to cancel the contribution of the $C_0/k^2$ part, subtract $epsilon$ from $alpha_2$ and add it to $alpha_1$. Do the same thing to $beta_1$, $C_1/k^2$, and $beta_2$. Now, your two original series sum absolutely to $0$, but the product series sums to $epsilon^2$. Every complex number has a "square root", and then we are done.
$endgroup$
– RandomMathDude
Dec 15 '18 at 17:01
add a comment |
$begingroup$
Thanks a lot. I understand the general point, but i am unsure about the case (0, 0). If at all possible, can you please help with a an example/argument?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:55
1
$begingroup$
@TheoryQuest1 Sorry, I may have made a typo, so I have just gone and made the answer even more carefully constructed. Now, you use $alpha_2$ to cancel the contribution of the $C_0/k^2$ part, subtract $epsilon$ from $alpha_2$ and add it to $alpha_1$. Do the same thing to $beta_1$, $C_1/k^2$, and $beta_2$. Now, your two original series sum absolutely to $0$, but the product series sums to $epsilon^2$. Every complex number has a "square root", and then we are done.
$endgroup$
– RandomMathDude
Dec 15 '18 at 17:01
$begingroup$
Thanks a lot. I understand the general point, but i am unsure about the case (0, 0). If at all possible, can you please help with a an example/argument?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:55
$begingroup$
Thanks a lot. I understand the general point, but i am unsure about the case (0, 0). If at all possible, can you please help with a an example/argument?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:55
1
1
$begingroup$
@TheoryQuest1 Sorry, I may have made a typo, so I have just gone and made the answer even more carefully constructed. Now, you use $alpha_2$ to cancel the contribution of the $C_0/k^2$ part, subtract $epsilon$ from $alpha_2$ and add it to $alpha_1$. Do the same thing to $beta_1$, $C_1/k^2$, and $beta_2$. Now, your two original series sum absolutely to $0$, but the product series sums to $epsilon^2$. Every complex number has a "square root", and then we are done.
$endgroup$
– RandomMathDude
Dec 15 '18 at 17:01
$begingroup$
@TheoryQuest1 Sorry, I may have made a typo, so I have just gone and made the answer even more carefully constructed. Now, you use $alpha_2$ to cancel the contribution of the $C_0/k^2$ part, subtract $epsilon$ from $alpha_2$ and add it to $alpha_1$. Do the same thing to $beta_1$, $C_1/k^2$, and $beta_2$. Now, your two original series sum absolutely to $0$, but the product series sums to $epsilon^2$. Every complex number has a "square root", and then we are done.
$endgroup$
– RandomMathDude
Dec 15 '18 at 17:01
add a comment |
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$begingroup$
Well, the product (in the sense you specify) doesn't even need to converge. Say $a_n=b_n=frac {(-1)^n}{sqrt n}$.
$endgroup$
– lulu
Dec 15 '18 at 16:35
1
$begingroup$
It could also be zero if $a_n$ is zero on odd integers and $b_n$ on even integers. I don't think you can say anything about it.
$endgroup$
– Yanko
Dec 15 '18 at 16:36
$begingroup$
Thank you. As a follow up: in case, if both $k_0$ and $K_1$ are 0, still its unanswerable?
$endgroup$
– TheoryQuest1
Dec 15 '18 at 16:39