Whether the statements regarding the Cauchy Product are true?












2














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







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  • Can you explain what your symbol with a circle and a dot inside means?
    – mathcounterexamples.net
    Jan 4 at 22:02










  • $odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
    – MP3129
    Jan 4 at 22:28
















2














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







share|cite|improve this question






















  • Can you explain what your symbol with a circle and a dot inside means?
    – mathcounterexamples.net
    Jan 4 at 22:02










  • $odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
    – MP3129
    Jan 4 at 22:28














2












2








2







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







share|cite|improve this question













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






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share|cite|improve this question











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asked Jan 4 at 21:58









MP3129MP3129

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956












  • Can you explain what your symbol with a circle and a dot inside means?
    – mathcounterexamples.net
    Jan 4 at 22:02










  • $odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
    – MP3129
    Jan 4 at 22:28


















  • Can you explain what your symbol with a circle and a dot inside means?
    – mathcounterexamples.net
    Jan 4 at 22:02










  • $odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
    – MP3129
    Jan 4 at 22:28
















Can you explain what your symbol with a circle and a dot inside means?
– mathcounterexamples.net
Jan 4 at 22:02




Can you explain what your symbol with a circle and a dot inside means?
– mathcounterexamples.net
Jan 4 at 22:02












$odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
– MP3129
Jan 4 at 22:28




$odot$ is a Cauchy Product for me and dot inside means ordinary multiplication how for example $2cdot2=4$
– MP3129
Jan 4 at 22:28










1 Answer
1






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oldest

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2














(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.






share|cite|improve this answer





















  • 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
    Jan 4 at 23:43








  • 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
    Jan 5 at 0:17











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1 Answer
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1 Answer
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2














(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.






share|cite|improve this answer





















  • 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
    Jan 4 at 23:43








  • 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
    Jan 5 at 0:17
















2














(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.






share|cite|improve this answer





















  • 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
    Jan 4 at 23:43








  • 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
    Jan 5 at 0:17














2












2








2






(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.






share|cite|improve this answer












(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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 22:34









Mike EarnestMike Earnest

20.8k11950




20.8k11950












  • 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
    Jan 4 at 23:43








  • 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
    Jan 5 at 0:17


















  • 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
    Jan 4 at 23:43








  • 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
    Jan 5 at 0:17
















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
Jan 4 at 23:43






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
Jan 4 at 23:43






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
Jan 5 at 0:17




@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
Jan 5 at 0:17


















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