Check if series converges $sum_{n=1}^infty frac{n-4}{sqrt{n^3+n^2+8}}$ [closed]
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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 Jan 5 at 13:34
Theodossis PapadopoulosTheodossis Papadopoulos
186
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closed as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd Jan 5 at 18:45
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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$begingroup$
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 Jan 5 at 13:34
Theodossis PapadopoulosTheodossis Papadopoulos
186
$endgroup$
closed as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd Jan 5 at 18:45
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
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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 $n to infty$.
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– Olivier Oloa
Jan 5 at 13:40
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@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
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– Theodossis Papadopoulos
Jan 5 at 13:42
$begingroup$
Do you know when $sum frac1{n^a}$ converges?
$endgroup$
– Olivier Oloa
Jan 5 at 13:42
1
$begingroup$
Thus your initial series is divergent, by the limit comparison test.
$endgroup$
– Olivier Oloa
Jan 5 at 13:47
1
$begingroup$
@OlivierOloa thanks for clearing things out!
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:48
|
show 3 more comments
$begingroup$
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 Jan 5 at 13:34
Theodossis PapadopoulosTheodossis Papadopoulos
186
$endgroup$
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 Jan 5 at 13:34
Theodossis PapadopoulosTheodossis Papadopoulos
186
asked Jan 5 at 13:34
Theodossis PapadopoulosTheodossis Papadopoulos
186
asked Jan 5 at 13:34
Theodossis PapadopoulosTheodossis Papadopoulos
186
asked Jan 5 at 13:34
Theodossis PapadopoulosTheodossis Papadopoulos
186
asked Jan 5 at 13:34
Theodossis PapadopoulosTheodossis Papadopoulos
186
186
closed as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd Jan 5 at 18:45
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.
closed as off-topic by Nosrati, Saad, max_zorn, metamorphy, Abcd Jan 5 at 18:45
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
$begingroup$
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 $n to infty$.
$endgroup$
– Olivier Oloa
Jan 5 at 13:40
$begingroup$
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:42
$begingroup$
Do you know when $sum frac1{n^a}$ converges?
$endgroup$
– Olivier Oloa
Jan 5 at 13:42
1
$begingroup$
Thus your initial series is divergent, by the limit comparison test.
$endgroup$
– Olivier Oloa
Jan 5 at 13:47
1
$begingroup$
@OlivierOloa thanks for clearing things out!
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:48
|
show 3 more comments
2
$begingroup$
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 $n to infty$.
$endgroup$
– Olivier Oloa
Jan 5 at 13:40
$begingroup$
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:42
$begingroup$
Do you know when $sum frac1{n^a}$ converges?
$endgroup$
– Olivier Oloa
Jan 5 at 13:42
1
$begingroup$
Thus your initial series is divergent, by the limit comparison test.
$endgroup$
– Olivier Oloa
Jan 5 at 13:47
1
$begingroup$
@OlivierOloa thanks for clearing things out!
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:48
2
2
$begingroup$
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 $n to infty$.
$endgroup$
– Olivier Oloa
Jan 5 at 13:40
$begingroup$
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 $n to infty$.
$endgroup$
– Olivier Oloa
Jan 5 at 13:40
$begingroup$
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:42
$begingroup$
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:42
$begingroup$
Do you know when $sum frac1{n^a}$ converges?
$endgroup$
– Olivier Oloa
Jan 5 at 13:42
$begingroup$
Do you know when $sum frac1{n^a}$ converges?
$endgroup$
– Olivier Oloa
Jan 5 at 13:42
1
1
$begingroup$
Thus your initial series is divergent, by the limit comparison test.
$endgroup$
– Olivier Oloa
Jan 5 at 13:47
$begingroup$
Thus your initial series is divergent, by the limit comparison test.
$endgroup$
– Olivier Oloa
Jan 5 at 13:47
1
1
$begingroup$
@OlivierOloa thanks for clearing things out!
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:48
$begingroup$
@OlivierOloa thanks for clearing things out!
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:48
|
show 3 more comments
3 Answers
3
active
oldest
votes
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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 Jan 5 at 13:49
Olivier OloaOlivier Oloa
108k17176293
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add a comment |
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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 Jan 5 at 14:43
Mostafa AyazMostafa Ayaz
14.8k3938
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 14:03
BernardBernard
119k639112
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Jan 5 at 13:49
Olivier OloaOlivier Oloa
108k17176293
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 13:49
Olivier OloaOlivier Oloa
108k17176293
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 13:49
Olivier OloaOlivier Oloa
108k17176293
$endgroup$
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 Jan 5 at 13:49
Olivier OloaOlivier Oloa
108k17176293
answered Jan 5 at 13:49
Olivier OloaOlivier Oloa
108k17176293
answered Jan 5 at 13:49
Olivier OloaOlivier Oloa
108k17176293
answered Jan 5 at 13:49
Olivier OloaOlivier Oloa
108k17176293
108k17176293
add a comment |
add a comment |
$begingroup$
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 Jan 5 at 14:43
Mostafa AyazMostafa Ayaz
14.8k3938
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 14:43
Mostafa AyazMostafa Ayaz
14.8k3938
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 14:43
Mostafa AyazMostafa Ayaz
14.8k3938
$endgroup$
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 Jan 5 at 14:43
Mostafa AyazMostafa Ayaz
14.8k3938
answered Jan 5 at 14:43
Mostafa AyazMostafa Ayaz
14.8k3938
answered Jan 5 at 14:43
Mostafa AyazMostafa Ayaz
14.8k3938
answered Jan 5 at 14:43
Mostafa AyazMostafa Ayaz
14.8k3938
14.8k3938
add a comment |
add a comment |
$begingroup$
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 Jan 5 at 14:03
BernardBernard
119k639112
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 14:03
BernardBernard
119k639112
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 14:03
BernardBernard
119k639112
$endgroup$
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 Jan 5 at 14:03
BernardBernard
119k639112
answered Jan 5 at 14:03
BernardBernard
119k639112
answered Jan 5 at 14:03
BernardBernard
119k639112
answered Jan 5 at 14:03
BernardBernard
119k639112
119k639112
add a comment |
add a comment |
2
$begingroup$
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 $n to infty$.
$endgroup$
– Olivier Oloa
Jan 5 at 13:40
$begingroup$
@OlivierOloa I have already tried this, but I don't see anything. Oh you edited.
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:42
$begingroup$
Do you know when $sum frac1{n^a}$ converges?
$endgroup$
– Olivier Oloa
Jan 5 at 13:42
1
$begingroup$
Thus your initial series is divergent, by the limit comparison test.
$endgroup$
– Olivier Oloa
Jan 5 at 13:47
1
$begingroup$
@OlivierOloa thanks for clearing things out!
$endgroup$
– Theodossis Papadopoulos
Jan 5 at 13:48