How to prove series convergence by comparison [closed]
0
$begingroup$
I have tried to do it by comparison to
but I'm not sure if it's strong enough
$$sum_{n=2}^infty frac{sqrt n+1}{n^2-3}$$
convergence
edited Jan 7 at 19:38
John Doe
11.1k11238
asked Jan 7 at 19:33
QasparQaspar
1
New contributor
Qaspar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
closed as off-topic by José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Chris Custer, Cesareo, amWhy Jan 8 at 18:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Chris Custer, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
$begingroup$
I have tried to do it by comparison to
but I'm not sure if it's strong enough
$$sum_{n=2}^infty frac{sqrt n+1}{n^2-3}$$
convergence
edited Jan 7 at 19:38
John Doe
11.1k11238
asked Jan 7 at 19:33
QasparQaspar
1
New contributor
Qaspar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
closed as off-topic by José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Chris Custer, Cesareo, amWhy Jan 8 at 18:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Chris Custer, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Could you edit your question to include what you have tried so we can check it?
$endgroup$
– John Doe
Jan 7 at 19:38
$begingroup$
Usually, it is difficult to determine if this kind of series is convergent by using comparison test. I recommend that you use 'limit comparison test' en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– LeB
Jan 7 at 19:45
$begingroup$
oh, sorry, thought it was there, would sum{i=2}^infty frac{(sqrt(i))*8}/{i^2} work ?
$endgroup$
– Qaspar
Jan 7 at 20:02
add a comment |
0
0
0
$begingroup$
I have tried to do it by comparison to
but I'm not sure if it's strong enough
$$sum_{n=2}^infty frac{sqrt n+1}{n^2-3}$$
convergence
edited Jan 7 at 19:38
John Doe
11.1k11238
asked Jan 7 at 19:33
QasparQaspar
1
New contributor
Qaspar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have tried to do it by comparison to
but I'm not sure if it's strong enough
$$sum_{n=2}^infty frac{sqrt n+1}{n^2-3}$$
convergence
convergence
edited Jan 7 at 19:38
John Doe
11.1k11238
asked Jan 7 at 19:33
QasparQaspar
1
New contributor
Qaspar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 7 at 19:38
John Doe
11.1k11238
asked Jan 7 at 19:33
QasparQaspar
1
New contributor
Qaspar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 7 at 19:38
John Doe
11.1k11238
edited Jan 7 at 19:38
John Doe
11.1k11238
edited Jan 7 at 19:38
John Doe
11.1k11238
11.1k11238
asked Jan 7 at 19:33
QasparQaspar
1
New contributor
Qaspar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 7 at 19:33
QasparQaspar
1
asked Jan 7 at 19:33
QasparQaspar
1
1
New contributor
Qaspar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Qaspar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Qaspar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as off-topic by José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Chris Custer, Cesareo, amWhy Jan 8 at 18:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Chris Custer, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Chris Custer, Cesareo, amWhy Jan 8 at 18:31
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, GNUSupporter 8964民主女神 地下教會, Chris Custer, Cesareo, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Could you edit your question to include what you have tried so we can check it?
$endgroup$
– John Doe
Jan 7 at 19:38
$begingroup$
Usually, it is difficult to determine if this kind of series is convergent by using comparison test. I recommend that you use 'limit comparison test' en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– LeB
Jan 7 at 19:45
$begingroup$
oh, sorry, thought it was there, would sum{i=2}^infty frac{(sqrt(i))*8}/{i^2} work ?
$endgroup$
– Qaspar
Jan 7 at 20:02
add a comment |
2
$begingroup$
Could you edit your question to include what you have tried so we can check it?
$endgroup$
– John Doe
Jan 7 at 19:38
$begingroup$
Usually, it is difficult to determine if this kind of series is convergent by using comparison test. I recommend that you use 'limit comparison test' en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– LeB
Jan 7 at 19:45
$begingroup$
oh, sorry, thought it was there, would sum{i=2}^infty frac{(sqrt(i))*8}/{i^2} work ?
$endgroup$
– Qaspar
Jan 7 at 20:02
2
2
$begingroup$
Could you edit your question to include what you have tried so we can check it?
$endgroup$
– John Doe
Jan 7 at 19:38
$begingroup$
Could you edit your question to include what you have tried so we can check it?
$endgroup$
– John Doe
Jan 7 at 19:38
$begingroup$
Usually, it is difficult to determine if this kind of series is convergent by using comparison test. I recommend that you use 'limit comparison test' en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– LeB
Jan 7 at 19:45
$begingroup$
Usually, it is difficult to determine if this kind of series is convergent by using comparison test. I recommend that you use 'limit comparison test' en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– LeB
Jan 7 at 19:45
$begingroup$
oh, sorry, thought it was there, would sum{i=2}^infty frac{(sqrt(i))*8}/{i^2} work ?
$endgroup$
– Qaspar
Jan 7 at 20:02
$begingroup$
oh, sorry, thought it was there, would sum{i=2}^infty frac{(sqrt(i))*8}/{i^2} work ?
$endgroup$
– Qaspar
Jan 7 at 20:02
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
Hint:
$$sum_{n=1}^infty n^{-1.1}leint_1^infty x^{-1.1}dx=lim_{xtoinfty}left(-frac{x^{-0.1}}{0.1}+frac{1^{-0.1}}{0.1}right)=10$$
answered Jan 7 at 19:48
ajotatxeajotatxe
53.7k23890
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Hint:
$$sum_{n=1}^infty n^{-1.1}leint_1^infty x^{-1.1}dx=lim_{xtoinfty}left(-frac{x^{-0.1}}{0.1}+frac{1^{-0.1}}{0.1}right)=10$$
answered Jan 7 at 19:48
ajotatxeajotatxe
53.7k23890
$endgroup$
add a comment |
0
$begingroup$
Hint:
$$sum_{n=1}^infty n^{-1.1}leint_1^infty x^{-1.1}dx=lim_{xtoinfty}left(-frac{x^{-0.1}}{0.1}+frac{1^{-0.1}}{0.1}right)=10$$
answered Jan 7 at 19:48
ajotatxeajotatxe
53.7k23890
$endgroup$
add a comment |
0
0
0
$begingroup$
Hint:
$$sum_{n=1}^infty n^{-1.1}leint_1^infty x^{-1.1}dx=lim_{xtoinfty}left(-frac{x^{-0.1}}{0.1}+frac{1^{-0.1}}{0.1}right)=10$$
answered Jan 7 at 19:48
ajotatxeajotatxe
53.7k23890
$endgroup$
Hint:
$$sum_{n=1}^infty n^{-1.1}leint_1^infty x^{-1.1}dx=lim_{xtoinfty}left(-frac{x^{-0.1}}{0.1}+frac{1^{-0.1}}{0.1}right)=10$$
answered Jan 7 at 19:48
ajotatxeajotatxe
53.7k23890
answered Jan 7 at 19:48
ajotatxeajotatxe
53.7k23890
answered Jan 7 at 19:48
ajotatxeajotatxe
53.7k23890
answered Jan 7 at 19:48
ajotatxeajotatxe
53.7k23890
53.7k23890
add a comment |
add a comment |
2
$begingroup$
Could you edit your question to include what you have tried so we can check it?
$endgroup$
– John Doe
Jan 7 at 19:38
$begingroup$
Usually, it is difficult to determine if this kind of series is convergent by using comparison test. I recommend that you use 'limit comparison test' en.wikipedia.org/wiki/Limit_comparison_test
$endgroup$
– LeB
Jan 7 at 19:45
$begingroup$
oh, sorry, thought it was there, would sum{i=2}^infty frac{(sqrt(i))*8}/{i^2} work ?
$endgroup$
– Qaspar
Jan 7 at 20:02