Whether the statements regarding the Cauchy Product are true?
a) Exist a divergent series $ sum_{n=0}^{+ infty }x_{n} $ such that series $ (sum_{n=0}^{+ infty }1) odot (sum_{n=0}^{+ infty }x_{n})$ is convergent
b) Exist an absolutely convergent series $ sum_{n=0}^{+ infty }x_{n} $ such that for every conditionally convergent $ sum_{n=0}^{+ infty }a_{n} $ the series $ (sum_{n=0}^{+ infty }a_{n}) odot (sum_{n=0}^{+ infty }x_{n})$ is also conditionally convergent
c) Find a series $ sum_{n=0}^{+ infty }j_{n} $ which is neutral element for a Cauchy Product, so for each series $ sum_{n=0}^{+ infty }a_{n} $ occurs $(sum_{n=0}^{+ infty }a_{n}) odot (sum_{n=0}^{+ infty }j_n)=sum_{n=0}^{+ infty }a_{n} $
a) It is my idea how to do this task, but I am afraid that it is incorrect and I need an assessment of this: $$ (sum_{n=0}^{+ infty }1) odot (sum_{n=0}^{+ infty }x_{n})= sum_{n=0}^{+ infty }( sum_{k=0}^{n} 1 cdot x_{n-k}) =sum_{n=0}^{+ infty }(x_{n}+x_{n-1}+...+x_{0})=sum_{n=0}^{+ infty }x_{n}+sum_{n=0}^{+ infty }x_{n-1}+...+sum_{n=0}^{+ infty }x_{0}$$ If $ sum_{n=0}^{+ infty } x_{n}$ is divergent, then also $sum_{n=0}^{+ infty } x_{n-1}$, $sum_{n=0}^{+ infty } x_{n-2}$,...,$ sum_{n=0}^{+ infty } x_{0}$ are divergent and the sum of divergent series is divergent so
there is no such series which meets the requirements of the task
b),c) Unfortunately I don't have an idea how to deal it and I am really asking for some tips to do this tasks
real-analysis
asked yesterday
MP3129
826
add a comment |
a) Exist a divergent series $ sum_{n=0}^{+ infty }x_{n} $ such that series $ (sum_{n=0}^{+ infty }1) odot (sum_{n=0}^{+ infty }x_{n})$ is convergent
b) Exist an absolutely convergent series $ sum_{n=0}^{+ infty }x_{n} $ such that for every conditionally convergent $ sum_{n=0}^{+ infty }a_{n} $ the series $ (sum_{n=0}^{+ infty }a_{n}) odot (sum_{n=0}^{+ infty }x_{n})$ is also conditionally convergent
c) Find a series $ sum_{n=0}^{+ infty }j_{n} $ which is neutral element for a Cauchy Product, so for each series $ sum_{n=0}^{+ infty }a_{n} $ occurs $(sum_{n=0}^{+ infty }a_{n}) odot (sum_{n=0}^{+ infty }j_n)=sum_{n=0}^{+ infty }a_{n} $
a) It is my idea how to do this task, but I am afraid that it is incorrect and I need an assessment of this: $$ (sum_{n=0}^{+ infty }1) odot (sum_{n=0}^{+ infty }x_{n})= sum_{n=0}^{+ infty }( sum_{k=0}^{n} 1 cdot x_{n-k}) =sum_{n=0}^{+ infty }(x_{n}+x_{n-1}+...+x_{0})=sum_{n=0}^{+ infty }x_{n}+sum_{n=0}^{+ infty }x_{n-1}+...+sum_{n=0}^{+ infty }x_{0}$$ If $ sum_{n=0}^{+ infty } x_{n}$ is divergent, then also $sum_{n=0}^{+ infty } x_{n-1}$, $sum_{n=0}^{+ infty } x_{n-2}$,...,$ sum_{n=0}^{+ infty } x_{0}$ are divergent and the sum of divergent series is divergent so
there is no such series which meets the requirements of the task
b),c) Unfortunately I don't have an idea how to deal it and I am really asking for some tips to do this tasks
real-analysis
asked yesterday
MP3129
826
Can you explain what your symbol with a circle and a dot inside means?
– mathcounterexamples.net
yesterday
$odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
– MP3129
yesterday
add a comment |
a) Exist a divergent series $ sum_{n=0}^{+ infty }x_{n} $ such that series $ (sum_{n=0}^{+ infty }1) odot (sum_{n=0}^{+ infty }x_{n})$ is convergent
b) Exist an absolutely convergent series $ sum_{n=0}^{+ infty }x_{n} $ such that for every conditionally convergent $ sum_{n=0}^{+ infty }a_{n} $ the series $ (sum_{n=0}^{+ infty }a_{n}) odot (sum_{n=0}^{+ infty }x_{n})$ is also conditionally convergent
c) Find a series $ sum_{n=0}^{+ infty }j_{n} $ which is neutral element for a Cauchy Product, so for each series $ sum_{n=0}^{+ infty }a_{n} $ occurs $(sum_{n=0}^{+ infty }a_{n}) odot (sum_{n=0}^{+ infty }j_n)=sum_{n=0}^{+ infty }a_{n} $
a) It is my idea how to do this task, but I am afraid that it is incorrect and I need an assessment of this: $$ (sum_{n=0}^{+ infty }1) odot (sum_{n=0}^{+ infty }x_{n})= sum_{n=0}^{+ infty }( sum_{k=0}^{n} 1 cdot x_{n-k}) =sum_{n=0}^{+ infty }(x_{n}+x_{n-1}+...+x_{0})=sum_{n=0}^{+ infty }x_{n}+sum_{n=0}^{+ infty }x_{n-1}+...+sum_{n=0}^{+ infty }x_{0}$$ If $ sum_{n=0}^{+ infty } x_{n}$ is divergent, then also $sum_{n=0}^{+ infty } x_{n-1}$, $sum_{n=0}^{+ infty } x_{n-2}$,...,$ sum_{n=0}^{+ infty } x_{0}$ are divergent and the sum of divergent series is divergent so
there is no such series which meets the requirements of the task
b),c) Unfortunately I don't have an idea how to deal it and I am really asking for some tips to do this tasks
real-analysis
asked yesterday
MP3129
826
a) Exist a divergent series $ sum_{n=0}^{+ infty }x_{n} $ such that series $ (sum_{n=0}^{+ infty }1) odot (sum_{n=0}^{+ infty }x_{n})$ is convergent
b) Exist an absolutely convergent series $ sum_{n=0}^{+ infty }x_{n} $ such that for every conditionally convergent $ sum_{n=0}^{+ infty }a_{n} $ the series $ (sum_{n=0}^{+ infty }a_{n}) odot (sum_{n=0}^{+ infty }x_{n})$ is also conditionally convergent
c) Find a series $ sum_{n=0}^{+ infty }j_{n} $ which is neutral element for a Cauchy Product, so for each series $ sum_{n=0}^{+ infty }a_{n} $ occurs $(sum_{n=0}^{+ infty }a_{n}) odot (sum_{n=0}^{+ infty }j_n)=sum_{n=0}^{+ infty }a_{n} $
a) It is my idea how to do this task, but I am afraid that it is incorrect and I need an assessment of this: $$ (sum_{n=0}^{+ infty }1) odot (sum_{n=0}^{+ infty }x_{n})= sum_{n=0}^{+ infty }( sum_{k=0}^{n} 1 cdot x_{n-k}) =sum_{n=0}^{+ infty }(x_{n}+x_{n-1}+...+x_{0})=sum_{n=0}^{+ infty }x_{n}+sum_{n=0}^{+ infty }x_{n-1}+...+sum_{n=0}^{+ infty }x_{0}$$ If $ sum_{n=0}^{+ infty } x_{n}$ is divergent, then also $sum_{n=0}^{+ infty } x_{n-1}$, $sum_{n=0}^{+ infty } x_{n-2}$,...,$ sum_{n=0}^{+ infty } x_{0}$ are divergent and the sum of divergent series is divergent so
there is no such series which meets the requirements of the task
b),c) Unfortunately I don't have an idea how to deal it and I am really asking for some tips to do this tasks
real-analysis
real-analysis
asked yesterday
MP3129
826
asked yesterday
MP3129
826
asked yesterday
MP3129
826
asked yesterday
MP3129
826
asked yesterday
MP3129
826
826
Can you explain what your symbol with a circle and a dot inside means?
– mathcounterexamples.net
yesterday
$odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
– MP3129
yesterday
add a comment |
Can you explain what your symbol with a circle and a dot inside means?
– mathcounterexamples.net
yesterday
$odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
– MP3129
yesterday
Can you explain what your symbol with a circle and a dot inside means?
– mathcounterexamples.net
yesterday
Can you explain what your symbol with a circle and a dot inside means?
– mathcounterexamples.net
yesterday
$odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
– MP3129
yesterday
$odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
– MP3129
yesterday
add a comment |
1 Answer
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(c) The series $1+0+0+0+dots$ is neutral.
(b) Are you asking if there exists a series $sum_n x_n$ for which this is true? If so, just let $sum_n x_n$ be the neutral series from part (c).
If you are instead asking if this is true for every series $sum_n x_n$, the answer is also yes. See Merten's Theorem.
(a) The argument you wrote does not really work out.
It is true that no such series $sum_n x_n$ exists. Suppose that the Cauchy product of $sum_n x_n$ and $sum_n 1$ is $sum_n y_n$. You can show that this implies that $x_0=y_0$, and for all $nge 1$, $x_n=y_n-y_{n-1}$. This means that $sum_n x_n$ is telescoping, so that $sum_n x_n=lim_n y_n$, provided this limit exists. As long as $sum y_n$ converges, you will have $lim_n y_n=0$, so $sum_n x_n$ converges; therefore, it is impossible to have $sum_n x_n$ diverging and $sum_n y_n$ converging.
answered yesterday
Mike Earnest
20.5k11950
I am sorry, but I don't understand how you know that in (a) $x_0=y_0$ and for all $nge1, x_{n}=y_{n}-y_{n-1}$. Could you explain?
– MP3129
yesterday
1
@MP3129 $y_n = x_0+x_1+dots+x_n$, and $y_{n-1}=x_0+x_1+dots+x_{n-1}$. Therefore, $y_n-y_{n-1}=dots$.
– Mike Earnest
yesterday
add a comment |
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(c) The series $1+0+0+0+dots$ is neutral.
(b) Are you asking if there exists a series $sum_n x_n$ for which this is true? If so, just let $sum_n x_n$ be the neutral series from part (c).
If you are instead asking if this is true for every series $sum_n x_n$, the answer is also yes. See Merten's Theorem.
(a) The argument you wrote does not really work out.
It is true that no such series $sum_n x_n$ exists. Suppose that the Cauchy product of $sum_n x_n$ and $sum_n 1$ is $sum_n y_n$. You can show that this implies that $x_0=y_0$, and for all $nge 1$, $x_n=y_n-y_{n-1}$. This means that $sum_n x_n$ is telescoping, so that $sum_n x_n=lim_n y_n$, provided this limit exists. As long as $sum y_n$ converges, you will have $lim_n y_n=0$, so $sum_n x_n$ converges; therefore, it is impossible to have $sum_n x_n$ diverging and $sum_n y_n$ converging.
answered yesterday
Mike Earnest
20.5k11950
I am sorry, but I don't understand how you know that in (a) $x_0=y_0$ and for all $nge1, x_{n}=y_{n}-y_{n-1}$. Could you explain?
– MP3129
yesterday
1
@MP3129 $y_n = x_0+x_1+dots+x_n$, and $y_{n-1}=x_0+x_1+dots+x_{n-1}$. Therefore, $y_n-y_{n-1}=dots$.
– Mike Earnest
yesterday
add a comment |
(c) The series $1+0+0+0+dots$ is neutral.
(b) Are you asking if there exists a series $sum_n x_n$ for which this is true? If so, just let $sum_n x_n$ be the neutral series from part (c).
If you are instead asking if this is true for every series $sum_n x_n$, the answer is also yes. See Merten's Theorem.
(a) The argument you wrote does not really work out.
It is true that no such series $sum_n x_n$ exists. Suppose that the Cauchy product of $sum_n x_n$ and $sum_n 1$ is $sum_n y_n$. You can show that this implies that $x_0=y_0$, and for all $nge 1$, $x_n=y_n-y_{n-1}$. This means that $sum_n x_n$ is telescoping, so that $sum_n x_n=lim_n y_n$, provided this limit exists. As long as $sum y_n$ converges, you will have $lim_n y_n=0$, so $sum_n x_n$ converges; therefore, it is impossible to have $sum_n x_n$ diverging and $sum_n y_n$ converging.
answered yesterday
Mike Earnest
20.5k11950
I am sorry, but I don't understand how you know that in (a) $x_0=y_0$ and for all $nge1, x_{n}=y_{n}-y_{n-1}$. Could you explain?
– MP3129
yesterday
1
@MP3129 $y_n = x_0+x_1+dots+x_n$, and $y_{n-1}=x_0+x_1+dots+x_{n-1}$. Therefore, $y_n-y_{n-1}=dots$.
– Mike Earnest
yesterday
add a comment |
(c) The series $1+0+0+0+dots$ is neutral.
(b) Are you asking if there exists a series $sum_n x_n$ for which this is true? If so, just let $sum_n x_n$ be the neutral series from part (c).
If you are instead asking if this is true for every series $sum_n x_n$, the answer is also yes. See Merten's Theorem.
(a) The argument you wrote does not really work out.
It is true that no such series $sum_n x_n$ exists. Suppose that the Cauchy product of $sum_n x_n$ and $sum_n 1$ is $sum_n y_n$. You can show that this implies that $x_0=y_0$, and for all $nge 1$, $x_n=y_n-y_{n-1}$. This means that $sum_n x_n$ is telescoping, so that $sum_n x_n=lim_n y_n$, provided this limit exists. As long as $sum y_n$ converges, you will have $lim_n y_n=0$, so $sum_n x_n$ converges; therefore, it is impossible to have $sum_n x_n$ diverging and $sum_n y_n$ converging.
answered yesterday
Mike Earnest
20.5k11950
(c) The series $1+0+0+0+dots$ is neutral.
(b) Are you asking if there exists a series $sum_n x_n$ for which this is true? If so, just let $sum_n x_n$ be the neutral series from part (c).
If you are instead asking if this is true for every series $sum_n x_n$, the answer is also yes. See Merten's Theorem.
(a) The argument you wrote does not really work out.
It is true that no such series $sum_n x_n$ exists. Suppose that the Cauchy product of $sum_n x_n$ and $sum_n 1$ is $sum_n y_n$. You can show that this implies that $x_0=y_0$, and for all $nge 1$, $x_n=y_n-y_{n-1}$. This means that $sum_n x_n$ is telescoping, so that $sum_n x_n=lim_n y_n$, provided this limit exists. As long as $sum y_n$ converges, you will have $lim_n y_n=0$, so $sum_n x_n$ converges; therefore, it is impossible to have $sum_n x_n$ diverging and $sum_n y_n$ converging.
answered yesterday
Mike Earnest
20.5k11950
answered yesterday
Mike Earnest
20.5k11950
answered yesterday
Mike Earnest
20.5k11950
answered yesterday
Mike Earnest
20.5k11950
20.5k11950
I am sorry, but I don't understand how you know that in (a) $x_0=y_0$ and for all $nge1, x_{n}=y_{n}-y_{n-1}$. Could you explain?
– MP3129
yesterday
1
@MP3129 $y_n = x_0+x_1+dots+x_n$, and $y_{n-1}=x_0+x_1+dots+x_{n-1}$. Therefore, $y_n-y_{n-1}=dots$.
– Mike Earnest
yesterday
add a comment |
I am sorry, but I don't understand how you know that in (a) $x_0=y_0$ and for all $nge1, x_{n}=y_{n}-y_{n-1}$. Could you explain?
– MP3129
yesterday
1
@MP3129 $y_n = x_0+x_1+dots+x_n$, and $y_{n-1}=x_0+x_1+dots+x_{n-1}$. Therefore, $y_n-y_{n-1}=dots$.
– Mike Earnest
yesterday
I am sorry, but I don't understand how you know that in (a) $x_0=y_0$ and for all $nge1, x_{n}=y_{n}-y_{n-1}$. Could you explain?
– MP3129
yesterday
I am sorry, but I don't understand how you know that in (a) $x_0=y_0$ and for all $nge1, x_{n}=y_{n}-y_{n-1}$. Could you explain?
– MP3129
yesterday
1
1
@MP3129 $y_n = x_0+x_1+dots+x_n$, and $y_{n-1}=x_0+x_1+dots+x_{n-1}$. Therefore, $y_n-y_{n-1}=dots$.
– Mike Earnest
yesterday
@MP3129 $y_n = x_0+x_1+dots+x_n$, and $y_{n-1}=x_0+x_1+dots+x_{n-1}$. Therefore, $y_n-y_{n-1}=dots$.
– Mike Earnest
yesterday
add a comment |
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Can you explain what your symbol with a circle and a dot inside means?
– mathcounterexamples.net
yesterday
$odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
– MP3129
yesterday