Check if series converges $sum_{n=1}^infty frac{n-4}{sqrt{n^3+n^2+8}}$ [on hold]
Help me check if that series converges:
$sum_{n=1}^infty frac{n-4}{sqrt{n^3+n^2+8}}$
A quick hint will help me out!
sequences-and-series convergence
asked 2 days ago
Theodossis PapadopoulosTheodossis Papadopoulos
186
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Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd 2 days ago
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." – Nosrati, Saad, max_zorn, metamorphy, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
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Help me check if that series converges:
$sum_{n=1}^infty frac{n-4}{sqrt{n^3+n^2+8}}$
A quick hint will help me out!
sequences-and-series convergence
asked 2 days ago
Theodossis PapadopoulosTheodossis Papadopoulos
186
New contributor
Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd 2 days ago
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." – Nosrati, Saad, max_zorn, metamorphy, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
2
Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
– Olivier Oloa
2 days ago
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
– Theodossis Papadopoulos
2 days ago
Do you know when $sum frac1{n^a}$ converges?
– Olivier Oloa
2 days ago
1
Thus your initial series is divergent, by the limit comparison test.
– Olivier Oloa
2 days ago
1
@OlivierOloa thanks for clearing things out!
– Theodossis Papadopoulos
2 days ago
|
show 3 more comments
Help me check if that series converges:
$sum_{n=1}^infty frac{n-4}{sqrt{n^3+n^2+8}}$
A quick hint will help me out!
sequences-and-series convergence
asked 2 days ago
Theodossis PapadopoulosTheodossis Papadopoulos
186
New contributor
Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Help me check if that series converges:
$sum_{n=1}^infty frac{n-4}{sqrt{n^3+n^2+8}}$
A quick hint will help me out!
sequences-and-series convergence
sequences-and-series convergence
asked 2 days ago
Theodossis PapadopoulosTheodossis Papadopoulos
186
New contributor
Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Theodossis PapadopoulosTheodossis Papadopoulos
186
New contributor
Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Theodossis PapadopoulosTheodossis Papadopoulos
186
New contributor
Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Theodossis PapadopoulosTheodossis Papadopoulos
186
asked 2 days ago
Theodossis PapadopoulosTheodossis Papadopoulos
186
186
New contributor
Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Theodossis Papadopoulos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd 2 days ago
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." – Nosrati, Saad, max_zorn, metamorphy, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd 2 days ago
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." – Nosrati, Saad, max_zorn, metamorphy, Abcd
If this question can be reworded to fit the rules in the help center, please edit the question.
2
Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
– Olivier Oloa
2 days ago
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
– Theodossis Papadopoulos
2 days ago
Do you know when $sum frac1{n^a}$ converges?
– Olivier Oloa
2 days ago
1
Thus your initial series is divergent, by the limit comparison test.
– Olivier Oloa
2 days ago
1
@OlivierOloa thanks for clearing things out!
– Theodossis Papadopoulos
2 days ago
|
show 3 more comments
2
Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
– Olivier Oloa
2 days ago
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
– Theodossis Papadopoulos
2 days ago
Do you know when $sum frac1{n^a}$ converges?
– Olivier Oloa
2 days ago
1
Thus your initial series is divergent, by the limit comparison test.
– Olivier Oloa
2 days ago
1
@OlivierOloa thanks for clearing things out!
– Theodossis Papadopoulos
2 days ago
2
2
Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
– Olivier Oloa
2 days ago
Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
– Olivier Oloa
2 days ago
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
– Theodossis Papadopoulos
2 days ago
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
– Theodossis Papadopoulos
2 days ago
Do you know when $sum frac1{n^a}$ converges?
– Olivier Oloa
2 days ago
Do you know when $sum frac1{n^a}$ converges?
– Olivier Oloa
2 days ago
1
1
Thus your initial series is divergent, by the limit comparison test.
– Olivier Oloa
2 days ago
Thus your initial series is divergent, by the limit comparison test.
– Olivier Oloa
2 days ago
1
1
@OlivierOloa thanks for clearing things out!
– Theodossis Papadopoulos
2 days ago
@OlivierOloa thanks for clearing things out!
– Theodossis Papadopoulos
2 days ago
|
show 3 more comments
3 Answers
3
active
oldest
votes
As $n to infty$, one has
$$
frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}
$$ then use the limit comparison test.
answered 2 days ago
Olivier OloaOlivier Oloa
108k17176293
add a comment |
Hint
For large enough $n$ we have $${n-4over sqrt {n^3+n^2+8}}{>{n-4over sqrt {n^3+3n^2+3n+1}}\>{n-4over {(n+1)}sqrt{n+1}}\={1over sqrt {n+1}}-{5over {(n+1)}sqrt{n+1}}}$$
answered 2 days ago
Mostafa AyazMostafa Ayaz
14.1k3937
add a comment |
Without fiddling with factorisation, &c., it' simpler to use equivalents:
A polynomial is asymptotically equivalent to its leading term, so
$$frac{n-4}{sqrt{n^3+n^2+8}}sim_inftyfrac{n}{sqrt{n^3}}=frac1{sqrt n},$$
and the latter is the general term of a divergent $p$-series.
answered 2 days ago
BernardBernard
118k639112
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
As $n to infty$, one has
$$
frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}
$$ then use the limit comparison test.
answered 2 days ago
Olivier OloaOlivier Oloa
108k17176293
add a comment |
As $n to infty$, one has
$$
frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}
$$ then use the limit comparison test.
answered 2 days ago
Olivier OloaOlivier Oloa
108k17176293
add a comment |
As $n to infty$, one has
$$
frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}
$$ then use the limit comparison test.
answered 2 days ago
Olivier OloaOlivier Oloa
108k17176293
As $n to infty$, one has
$$
frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}
$$ then use the limit comparison test.
answered 2 days ago
Olivier OloaOlivier Oloa
108k17176293
answered 2 days ago
Olivier OloaOlivier Oloa
108k17176293
answered 2 days ago
Olivier OloaOlivier Oloa
108k17176293
answered 2 days ago
Olivier OloaOlivier Oloa
108k17176293
108k17176293
add a comment |
add a comment |
Hint
For large enough $n$ we have $${n-4over sqrt {n^3+n^2+8}}{>{n-4over sqrt {n^3+3n^2+3n+1}}\>{n-4over {(n+1)}sqrt{n+1}}\={1over sqrt {n+1}}-{5over {(n+1)}sqrt{n+1}}}$$
answered 2 days ago
Mostafa AyazMostafa Ayaz
14.1k3937
add a comment |
Hint
For large enough $n$ we have $${n-4over sqrt {n^3+n^2+8}}{>{n-4over sqrt {n^3+3n^2+3n+1}}\>{n-4over {(n+1)}sqrt{n+1}}\={1over sqrt {n+1}}-{5over {(n+1)}sqrt{n+1}}}$$
answered 2 days ago
Mostafa AyazMostafa Ayaz
14.1k3937
add a comment |
Hint
For large enough $n$ we have $${n-4over sqrt {n^3+n^2+8}}{>{n-4over sqrt {n^3+3n^2+3n+1}}\>{n-4over {(n+1)}sqrt{n+1}}\={1over sqrt {n+1}}-{5over {(n+1)}sqrt{n+1}}}$$
answered 2 days ago
Mostafa AyazMostafa Ayaz
14.1k3937
Hint
For large enough $n$ we have $${n-4over sqrt {n^3+n^2+8}}{>{n-4over sqrt {n^3+3n^2+3n+1}}\>{n-4over {(n+1)}sqrt{n+1}}\={1over sqrt {n+1}}-{5over {(n+1)}sqrt{n+1}}}$$
answered 2 days ago
Mostafa AyazMostafa Ayaz
14.1k3937
answered 2 days ago
Mostafa AyazMostafa Ayaz
14.1k3937
answered 2 days ago
Mostafa AyazMostafa Ayaz
14.1k3937
answered 2 days ago
Mostafa AyazMostafa Ayaz
14.1k3937
14.1k3937
add a comment |
add a comment |
Without fiddling with factorisation, &c., it' simpler to use equivalents:
A polynomial is asymptotically equivalent to its leading term, so
$$frac{n-4}{sqrt{n^3+n^2+8}}sim_inftyfrac{n}{sqrt{n^3}}=frac1{sqrt n},$$
and the latter is the general term of a divergent $p$-series.
answered 2 days ago
BernardBernard
118k639112
add a comment |
Without fiddling with factorisation, &c., it' simpler to use equivalents:
A polynomial is asymptotically equivalent to its leading term, so
$$frac{n-4}{sqrt{n^3+n^2+8}}sim_inftyfrac{n}{sqrt{n^3}}=frac1{sqrt n},$$
and the latter is the general term of a divergent $p$-series.
answered 2 days ago
BernardBernard
118k639112
add a comment |
Without fiddling with factorisation, &c., it' simpler to use equivalents:
A polynomial is asymptotically equivalent to its leading term, so
$$frac{n-4}{sqrt{n^3+n^2+8}}sim_inftyfrac{n}{sqrt{n^3}}=frac1{sqrt n},$$
and the latter is the general term of a divergent $p$-series.
answered 2 days ago
BernardBernard
118k639112
Without fiddling with factorisation, &c., it' simpler to use equivalents:
A polynomial is asymptotically equivalent to its leading term, so
$$frac{n-4}{sqrt{n^3+n^2+8}}sim_inftyfrac{n}{sqrt{n^3}}=frac1{sqrt n},$$
and the latter is the general term of a divergent $p$-series.
answered 2 days ago
BernardBernard
118k639112
answered 2 days ago
BernardBernard
118k639112
answered 2 days ago
BernardBernard
118k639112
answered 2 days ago
BernardBernard
118k639112
118k639112
add a comment |
add a comment |
2
Observe that $frac{n-4}{sqrt{n^3+n^2+8}}=frac{1-frac4n}{sqrt{n}sqrt{1+frac1n+frac8{n^3}}}sim frac1{sqrt{n}}$ as $n to infty$.
– Olivier Oloa
2 days ago
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
– Theodossis Papadopoulos
2 days ago
Do you know when $sum frac1{n^a}$ converges?
– Olivier Oloa
2 days ago
1
Thus your initial series is divergent, by the limit comparison test.
– Olivier Oloa
2 days ago
1
@OlivierOloa thanks for clearing things out!
– Theodossis Papadopoulos
2 days ago