Mixing
$sum_limitsn=1^inftyfracnx(1+x)(1+2x)…(1+nx)$ convergence domain.
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Problem: Study the absolute and conditional convergence domain of the series: $$sum_limitsn=1^inftyfracnx(1+x)(1+2x)...(1+nx)$$
1) for $0leqslant xleqslant epsilon$,$epsilon>0$ and 2) $epsilonleqslant x<+infty$.
I chose to apply the Alambert critera:$lim_ntoinftyfracfrac(n+1)x(1+x)(1+2x)...(1+(n+1)x)fracnx(1+x)(1+2x)...(1+nx)=limntoinftyfracnx+xnx+n^2x^2+nx^2=lim_ntoinfty=fracx+fracxnx+nx^2+x^2=0$
Question:
Is my attempt right? Because if it is. Why would the author ask the question for different convergence domain("1) for $0leqslant xleqslant epsilon$,$epsilon>0$ and 2) $epsilonleqslant x<+infty$")?
Thanks in advance!
calculus real-analysis sequences-and-series
asked Jul 15 at 12:43
Pedro Gomes
1,3192618
add a comment |Â
up vote
2
down vote
favorite
Problem: Study the absolute and conditional convergence domain of the series: $$sum_limitsn=1^inftyfracnx(1+x)(1+2x)...(1+nx)$$
1) for $0leqslant xleqslant epsilon$,$epsilon>0$ and 2) $epsilonleqslant x<+infty$.
I chose to apply the Alambert critera:$lim_ntoinftyfracfrac(n+1)x(1+x)(1+2x)...(1+(n+1)x)fracnx(1+x)(1+2x)...(1+nx)=limntoinftyfracnx+xnx+n^2x^2+nx^2=lim_ntoinfty=fracx+fracxnx+nx^2+x^2=0$
Question:
Is my attempt right? Because if it is. Why would the author ask the question for different convergence domain("1) for $0leqslant xleqslant epsilon$,$epsilon>0$ and 2) $epsilonleqslant x<+infty$")?
Thanks in advance!
calculus real-analysis sequences-and-series
asked Jul 15 at 12:43
Pedro Gomes
1,3192618
If that's the actual problem, your proof is fine. Something tells me that's not the problem however. I sense that the real question is about uniform convergence.
– zhw.
Jul 15 at 20:12
@zhw. Thanks for the observation. I guess you are right. However I have never dealt with uniform convergence on series. What are you thinking? I would be grateful if you formalized your statement.
– Pedro Gomes
Jul 15 at 21:01
You can find the definition online. It's exactly what you think it should be.
– zhw.
Jul 16 at 23:18
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Problem: Study the absolute and conditional convergence domain of the series: $$sum_limitsn=1^inftyfracnx(1+x)(1+2x)...(1+nx)$$
1) for $0leqslant xleqslant epsilon$,$epsilon>0$ and 2) $epsilonleqslant x<+infty$.
I chose to apply the Alambert critera:$lim_ntoinftyfracfrac(n+1)x(1+x)(1+2x)...(1+(n+1)x)fracnx(1+x)(1+2x)...(1+nx)=limntoinftyfracnx+xnx+n^2x^2+nx^2=lim_ntoinfty=fracx+fracxnx+nx^2+x^2=0$
Question:
Is my attempt right? Because if it is. Why would the author ask the question for different convergence domain("1) for $0leqslant xleqslant epsilon$,$epsilon>0$ and 2) $epsilonleqslant x<+infty$")?
Thanks in advance!
calculus real-analysis sequences-and-series
asked Jul 15 at 12:43
Pedro Gomes
1,3192618
Problem: Study the absolute and conditional convergence domain of the series: $$sum_limitsn=1^inftyfracnx(1+x)(1+2x)...(1+nx)$$
1) for $0leqslant xleqslant epsilon$,$epsilon>0$ and 2) $epsilonleqslant x<+infty$.
I chose to apply the Alambert critera:$lim_ntoinftyfracfrac(n+1)x(1+x)(1+2x)...(1+(n+1)x)fracnx(1+x)(1+2x)...(1+nx)=limntoinftyfracnx+xnx+n^2x^2+nx^2=lim_ntoinfty=fracx+fracxnx+nx^2+x^2=0$
Question:
Is my attempt right? Because if it is. Why would the author ask the question for different convergence domain("1) for $0leqslant xleqslant epsilon$,$epsilon>0$ and 2) $epsilonleqslant x<+infty$")?
Thanks in advance!
calculus real-analysis sequences-and-series
asked Jul 15 at 12:43
Pedro Gomes
1,3192618
edited Jul 15 at 18:08
asked Jul 15 at 12:43
Pedro Gomes
1,3192618
asked Jul 15 at 12:43
Pedro Gomes
1,3192618
asked Jul 15 at 12:43
Pedro Gomes
1,3192618
1,3192618
If that's the actual problem, your proof is fine. Something tells me that's not the problem however. I sense that the real question is about uniform convergence.
– zhw.
Jul 15 at 20:12
@zhw. Thanks for the observation. I guess you are right. However I have never dealt with uniform convergence on series. What are you thinking? I would be grateful if you formalized your statement.
– Pedro Gomes
Jul 15 at 21:01
You can find the definition online. It's exactly what you think it should be.
– zhw.
Jul 16 at 23:18
add a comment |Â
If that's the actual problem, your proof is fine. Something tells me that's not the problem however. I sense that the real question is about uniform convergence.
– zhw.
Jul 15 at 20:12
@zhw. Thanks for the observation. I guess you are right. However I have never dealt with uniform convergence on series. What are you thinking? I would be grateful if you formalized your statement.
– Pedro Gomes
Jul 15 at 21:01
You can find the definition online. It's exactly what you think it should be.
– zhw.
Jul 16 at 23:18
If that's the actual problem, your proof is fine. Something tells me that's not the problem however. I sense that the real question is about uniform convergence.
– zhw.
Jul 15 at 20:12
If that's the actual problem, your proof is fine. Something tells me that's not the problem however. I sense that the real question is about uniform convergence.
– zhw.
Jul 15 at 20:12
@zhw. Thanks for the observation. I guess you are right. However I have never dealt with uniform convergence on series. What are you thinking? I would be grateful if you formalized your statement.
– Pedro Gomes
Jul 15 at 21:01
@zhw. Thanks for the observation. I guess you are right. However I have never dealt with uniform convergence on series. What are you thinking? I would be grateful if you formalized your statement.
– Pedro Gomes
Jul 15 at 21:01
You can find the definition online. It's exactly what you think it should be.
– zhw.
Jul 16 at 23:18
You can find the definition online. It's exactly what you think it should be.
– zhw.
Jul 16 at 23:18
add a comment |Â
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If that's the actual problem, your proof is fine. Something tells me that's not the problem however. I sense that the real question is about uniform convergence.
– zhw.
Jul 15 at 20:12
@zhw. Thanks for the observation. I guess you are right. However I have never dealt with uniform convergence on series. What are you thinking? I would be grateful if you formalized your statement.
– Pedro Gomes
Jul 15 at 21:01
You can find the definition online. It's exactly what you think it should be.
– zhw.
Jul 16 at 23:18