Prove that $sum_{n=1}^∞frac{left(ln nright)^3}{n^3}$ is a convergent series by using comparison test...
I proved by using the integral test that the series is convergent but can't find a way to prove by using the comparison test, which was required.
real-analysis calculus sequences-and-series convergence
asked Dec 10 '18 at 5:24
Pratik Apshinge
395
closed as off-topic by Isaac Browne, RRL, user10354138, user21820, Jyrki Lahtonen Dec 12 '18 at 5:18
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." – Isaac Browne, RRL, user10354138, user21820, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I proved by using the integral test that the series is convergent but can't find a way to prove by using the comparison test, which was required.
real-analysis calculus sequences-and-series convergence
asked Dec 10 '18 at 5:24
Pratik Apshinge
395
closed as off-topic by Isaac Browne, RRL, user10354138, user21820, Jyrki Lahtonen Dec 12 '18 at 5:18
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." – Isaac Browne, RRL, user10354138, user21820, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I proved by using the integral test that the series is convergent but can't find a way to prove by using the comparison test, which was required.
real-analysis calculus sequences-and-series convergence
asked Dec 10 '18 at 5:24
Pratik Apshinge
395
I proved by using the integral test that the series is convergent but can't find a way to prove by using the comparison test, which was required.
real-analysis calculus sequences-and-series convergence
real-analysis calculus sequences-and-series convergence
asked Dec 10 '18 at 5:24
Pratik Apshinge
395
asked Dec 10 '18 at 5:24
Pratik Apshinge
395
edited Dec 10 '18 at 6:58
asked Dec 10 '18 at 5:24
Pratik Apshinge
395
asked Dec 10 '18 at 5:24
Pratik Apshinge
395
asked Dec 10 '18 at 5:24
Pratik Apshinge
395
395
closed as off-topic by Isaac Browne, RRL, user10354138, user21820, Jyrki Lahtonen Dec 12 '18 at 5:18
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." – Isaac Browne, RRL, user10354138, user21820, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Isaac Browne, RRL, user10354138, user21820, Jyrki Lahtonen Dec 12 '18 at 5:18
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." – Isaac Browne, RRL, user10354138, user21820, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
Use L'Hopital's Rule show that $frac {log, x } {x^{1/3}} to 0$as $ x to infty$. This gives $frac {log, n } {n^{1/3}} <1$ for $n$ sufficiently large. Hence $(log, n)^{3} < n$ for such $n$. Now compare the given series with $sum frac 1 {n^{2}}$.
edited Dec 11 '18 at 7:14
Martin Sleziak
44.7k7115270
answered Dec 10 '18 at 5:33
Kavi Rama Murthy
50.4k31854
Sorry Sir, I need an advice about my answer. Do you consider that equivalent to yours one. In particular is direct comparison test to be considered equivalent to limit comparison test? Thanks
– gimusi
Dec 10 '18 at 17:59
add a comment |
HINT:
Note that for all interger $nge1$
$$log(n)le sqrt n$$
answered Dec 10 '18 at 5:34
Mark Viola
130k1274170
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Dec 10 '18 at 5:53
Tht's the best way in my opinion.
– gimusi
Dec 10 '18 at 7:32
add a comment |
Remember that $log$ grows very slowly, so $frac{log(x)^3}{x^3}$ "looks a lot like $frac{1}{x^3}$": it will eventually be beaten by $frac{1}{x^2}$.
answered Dec 10 '18 at 5:29
Patrick Stevens
28.5k52874
Thank you! makes sense
– Pratik Apshinge
Dec 10 '18 at 5:34
That's far to be a correct proof, I think it would be marked as wrong.
– gimusi
Dec 10 '18 at 7:23
1
@PratikApshinge If you intention is give a hint or a guide, as I suppose, for an intuition is fine but you should explain that. Otherwise the aker could misunderstand your indication.
– gimusi
Dec 10 '18 at 7:29
2
@gimusi Obviously it's an incomplete proof. I don't think anyone would believe that "$f$ will eventually be beaten by $g$" is a rigorous exam-worthy answer.
– Patrick Stevens
Dec 10 '18 at 8:37
1
@PatrickStevens That's ok, but you should explain that in my opinion in your answer as a disclaimer, otherwise it could be not clear to some (less skilled) reader.
– gimusi
Dec 10 '18 at 8:40
add a comment |
HINT
Recall that $forall a,b>0$
$$frac{log^ an}{n^b} to 0$$
then use limit comparison test with $sum frac 1{n^2}$.
Indeed
$$frac{frac{left(ln nright)^3}{n^3}}{frac 1{n^2}}=frac{left(ln nright)^3}{n}to 0$$
and the given series converges.
For the difference between limit comparison test and comparison test refer also to
- The Comparison test or The limit comparison test
answered Dec 11 '18 at 20:54
gimusi
1
@T.Bongers Just to clarify the objective mathematical truth, I've also added a reference on the distincion between direct comparison test and limit comparison test.
– gimusi
Dec 11 '18 at 20:57
@user21820 Is it now fine?
– gimusi
Dec 11 '18 at 21:09
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Use L'Hopital's Rule show that $frac {log, x } {x^{1/3}} to 0$as $ x to infty$. This gives $frac {log, n } {n^{1/3}} <1$ for $n$ sufficiently large. Hence $(log, n)^{3} < n$ for such $n$. Now compare the given series with $sum frac 1 {n^{2}}$.
edited Dec 11 '18 at 7:14
Martin Sleziak
44.7k7115270
answered Dec 10 '18 at 5:33
Kavi Rama Murthy
50.4k31854
Sorry Sir, I need an advice about my answer. Do you consider that equivalent to yours one. In particular is direct comparison test to be considered equivalent to limit comparison test? Thanks
– gimusi
Dec 10 '18 at 17:59
add a comment |
Use L'Hopital's Rule show that $frac {log, x } {x^{1/3}} to 0$as $ x to infty$. This gives $frac {log, n } {n^{1/3}} <1$ for $n$ sufficiently large. Hence $(log, n)^{3} < n$ for such $n$. Now compare the given series with $sum frac 1 {n^{2}}$.
edited Dec 11 '18 at 7:14
Martin Sleziak
44.7k7115270
answered Dec 10 '18 at 5:33
Kavi Rama Murthy
50.4k31854
Sorry Sir, I need an advice about my answer. Do you consider that equivalent to yours one. In particular is direct comparison test to be considered equivalent to limit comparison test? Thanks
– gimusi
Dec 10 '18 at 17:59
add a comment |
Use L'Hopital's Rule show that $frac {log, x } {x^{1/3}} to 0$as $ x to infty$. This gives $frac {log, n } {n^{1/3}} <1$ for $n$ sufficiently large. Hence $(log, n)^{3} < n$ for such $n$. Now compare the given series with $sum frac 1 {n^{2}}$.
edited Dec 11 '18 at 7:14
Martin Sleziak
44.7k7115270
answered Dec 10 '18 at 5:33
Kavi Rama Murthy
50.4k31854
Use L'Hopital's Rule show that $frac {log, x } {x^{1/3}} to 0$as $ x to infty$. This gives $frac {log, n } {n^{1/3}} <1$ for $n$ sufficiently large. Hence $(log, n)^{3} < n$ for such $n$. Now compare the given series with $sum frac 1 {n^{2}}$.
edited Dec 11 '18 at 7:14
Martin Sleziak
44.7k7115270
answered Dec 10 '18 at 5:33
Kavi Rama Murthy
50.4k31854
edited Dec 11 '18 at 7:14
Martin Sleziak
44.7k7115270
edited Dec 11 '18 at 7:14
Martin Sleziak
44.7k7115270
edited Dec 11 '18 at 7:14
Martin Sleziak
44.7k7115270
44.7k7115270
answered Dec 10 '18 at 5:33
Kavi Rama Murthy
50.4k31854
answered Dec 10 '18 at 5:33
Kavi Rama Murthy
50.4k31854
answered Dec 10 '18 at 5:33
Kavi Rama Murthy
50.4k31854
50.4k31854
Sorry Sir, I need an advice about my answer. Do you consider that equivalent to yours one. In particular is direct comparison test to be considered equivalent to limit comparison test? Thanks
– gimusi
Dec 10 '18 at 17:59
add a comment |
Sorry Sir, I need an advice about my answer. Do you consider that equivalent to yours one. In particular is direct comparison test to be considered equivalent to limit comparison test? Thanks
– gimusi
Dec 10 '18 at 17:59
Sorry Sir, I need an advice about my answer. Do you consider that equivalent to yours one. In particular is direct comparison test to be considered equivalent to limit comparison test? Thanks
– gimusi
Dec 10 '18 at 17:59
Sorry Sir, I need an advice about my answer. Do you consider that equivalent to yours one. In particular is direct comparison test to be considered equivalent to limit comparison test? Thanks
– gimusi
Dec 10 '18 at 17:59
add a comment |
HINT:
Note that for all interger $nge1$
$$log(n)le sqrt n$$
answered Dec 10 '18 at 5:34
Mark Viola
130k1274170
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Dec 10 '18 at 5:53
Tht's the best way in my opinion.
– gimusi
Dec 10 '18 at 7:32
add a comment |
HINT:
Note that for all interger $nge1$
$$log(n)le sqrt n$$
answered Dec 10 '18 at 5:34
Mark Viola
130k1274170
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Dec 10 '18 at 5:53
Tht's the best way in my opinion.
– gimusi
Dec 10 '18 at 7:32
add a comment |
HINT:
Note that for all interger $nge1$
$$log(n)le sqrt n$$
answered Dec 10 '18 at 5:34
Mark Viola
130k1274170
HINT:
Note that for all interger $nge1$
$$log(n)le sqrt n$$
answered Dec 10 '18 at 5:34
Mark Viola
130k1274170
answered Dec 10 '18 at 5:34
Mark Viola
130k1274170
answered Dec 10 '18 at 5:34
Mark Viola
130k1274170
answered Dec 10 '18 at 5:34
Mark Viola
130k1274170
130k1274170
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Dec 10 '18 at 5:53
Tht's the best way in my opinion.
– gimusi
Dec 10 '18 at 7:32
add a comment |
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Dec 10 '18 at 5:53
Tht's the best way in my opinion.
– gimusi
Dec 10 '18 at 7:32
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Dec 10 '18 at 5:53
Please let me know how I can improve my answer. I really want to give you the best answer I can.
– Mark Viola
Dec 10 '18 at 5:53
Tht's the best way in my opinion.
– gimusi
Dec 10 '18 at 7:32
Tht's the best way in my opinion.
– gimusi
Dec 10 '18 at 7:32
add a comment |
Remember that $log$ grows very slowly, so $frac{log(x)^3}{x^3}$ "looks a lot like $frac{1}{x^3}$": it will eventually be beaten by $frac{1}{x^2}$.
answered Dec 10 '18 at 5:29
Patrick Stevens
28.5k52874
Thank you! makes sense
– Pratik Apshinge
Dec 10 '18 at 5:34
That's far to be a correct proof, I think it would be marked as wrong.
– gimusi
Dec 10 '18 at 7:23
1
@PratikApshinge If you intention is give a hint or a guide, as I suppose, for an intuition is fine but you should explain that. Otherwise the aker could misunderstand your indication.
– gimusi
Dec 10 '18 at 7:29
2
@gimusi Obviously it's an incomplete proof. I don't think anyone would believe that "$f$ will eventually be beaten by $g$" is a rigorous exam-worthy answer.
– Patrick Stevens
Dec 10 '18 at 8:37
1
@PatrickStevens That's ok, but you should explain that in my opinion in your answer as a disclaimer, otherwise it could be not clear to some (less skilled) reader.
– gimusi
Dec 10 '18 at 8:40
add a comment |
Remember that $log$ grows very slowly, so $frac{log(x)^3}{x^3}$ "looks a lot like $frac{1}{x^3}$": it will eventually be beaten by $frac{1}{x^2}$.
answered Dec 10 '18 at 5:29
Patrick Stevens
28.5k52874
Thank you! makes sense
– Pratik Apshinge
Dec 10 '18 at 5:34
That's far to be a correct proof, I think it would be marked as wrong.
– gimusi
Dec 10 '18 at 7:23
1
@PratikApshinge If you intention is give a hint or a guide, as I suppose, for an intuition is fine but you should explain that. Otherwise the aker could misunderstand your indication.
– gimusi
Dec 10 '18 at 7:29
2
@gimusi Obviously it's an incomplete proof. I don't think anyone would believe that "$f$ will eventually be beaten by $g$" is a rigorous exam-worthy answer.
– Patrick Stevens
Dec 10 '18 at 8:37
1
@PatrickStevens That's ok, but you should explain that in my opinion in your answer as a disclaimer, otherwise it could be not clear to some (less skilled) reader.
– gimusi
Dec 10 '18 at 8:40
add a comment |
Remember that $log$ grows very slowly, so $frac{log(x)^3}{x^3}$ "looks a lot like $frac{1}{x^3}$": it will eventually be beaten by $frac{1}{x^2}$.
answered Dec 10 '18 at 5:29
Patrick Stevens
28.5k52874
Remember that $log$ grows very slowly, so $frac{log(x)^3}{x^3}$ "looks a lot like $frac{1}{x^3}$": it will eventually be beaten by $frac{1}{x^2}$.
answered Dec 10 '18 at 5:29
Patrick Stevens
28.5k52874
answered Dec 10 '18 at 5:29
Patrick Stevens
28.5k52874
answered Dec 10 '18 at 5:29
Patrick Stevens
28.5k52874
answered Dec 10 '18 at 5:29
Patrick Stevens
28.5k52874
28.5k52874
Thank you! makes sense
– Pratik Apshinge
Dec 10 '18 at 5:34
That's far to be a correct proof, I think it would be marked as wrong.
– gimusi
Dec 10 '18 at 7:23
1
@PratikApshinge If you intention is give a hint or a guide, as I suppose, for an intuition is fine but you should explain that. Otherwise the aker could misunderstand your indication.
– gimusi
Dec 10 '18 at 7:29
2
@gimusi Obviously it's an incomplete proof. I don't think anyone would believe that "$f$ will eventually be beaten by $g$" is a rigorous exam-worthy answer.
– Patrick Stevens
Dec 10 '18 at 8:37
1
@PatrickStevens That's ok, but you should explain that in my opinion in your answer as a disclaimer, otherwise it could be not clear to some (less skilled) reader.
– gimusi
Dec 10 '18 at 8:40
add a comment |
Thank you! makes sense
– Pratik Apshinge
Dec 10 '18 at 5:34
That's far to be a correct proof, I think it would be marked as wrong.
– gimusi
Dec 10 '18 at 7:23
1
@PratikApshinge If you intention is give a hint or a guide, as I suppose, for an intuition is fine but you should explain that. Otherwise the aker could misunderstand your indication.
– gimusi
Dec 10 '18 at 7:29
2
@gimusi Obviously it's an incomplete proof. I don't think anyone would believe that "$f$ will eventually be beaten by $g$" is a rigorous exam-worthy answer.
– Patrick Stevens
Dec 10 '18 at 8:37
1
@PatrickStevens That's ok, but you should explain that in my opinion in your answer as a disclaimer, otherwise it could be not clear to some (less skilled) reader.
– gimusi
Dec 10 '18 at 8:40
Thank you! makes sense
– Pratik Apshinge
Dec 10 '18 at 5:34
Thank you! makes sense
– Pratik Apshinge
Dec 10 '18 at 5:34
That's far to be a correct proof, I think it would be marked as wrong.
– gimusi
Dec 10 '18 at 7:23
That's far to be a correct proof, I think it would be marked as wrong.
– gimusi
Dec 10 '18 at 7:23
1
1
@PratikApshinge If you intention is give a hint or a guide, as I suppose, for an intuition is fine but you should explain that. Otherwise the aker could misunderstand your indication.
– gimusi
Dec 10 '18 at 7:29
@PratikApshinge If you intention is give a hint or a guide, as I suppose, for an intuition is fine but you should explain that. Otherwise the aker could misunderstand your indication.
– gimusi
Dec 10 '18 at 7:29
2
2
@gimusi Obviously it's an incomplete proof. I don't think anyone would believe that "$f$ will eventually be beaten by $g$" is a rigorous exam-worthy answer.
– Patrick Stevens
Dec 10 '18 at 8:37
@gimusi Obviously it's an incomplete proof. I don't think anyone would believe that "$f$ will eventually be beaten by $g$" is a rigorous exam-worthy answer.
– Patrick Stevens
Dec 10 '18 at 8:37
1
1
@PatrickStevens That's ok, but you should explain that in my opinion in your answer as a disclaimer, otherwise it could be not clear to some (less skilled) reader.
– gimusi
Dec 10 '18 at 8:40
@PatrickStevens That's ok, but you should explain that in my opinion in your answer as a disclaimer, otherwise it could be not clear to some (less skilled) reader.
– gimusi
Dec 10 '18 at 8:40
add a comment |
HINT
Recall that $forall a,b>0$
$$frac{log^ an}{n^b} to 0$$
then use limit comparison test with $sum frac 1{n^2}$.
Indeed
$$frac{frac{left(ln nright)^3}{n^3}}{frac 1{n^2}}=frac{left(ln nright)^3}{n}to 0$$
and the given series converges.
For the difference between limit comparison test and comparison test refer also to
- The Comparison test or The limit comparison test
answered Dec 11 '18 at 20:54
gimusi
1
@T.Bongers Just to clarify the objective mathematical truth, I've also added a reference on the distincion between direct comparison test and limit comparison test.
– gimusi
Dec 11 '18 at 20:57
@user21820 Is it now fine?
– gimusi
Dec 11 '18 at 21:09
add a comment |
HINT
Recall that $forall a,b>0$
$$frac{log^ an}{n^b} to 0$$
then use limit comparison test with $sum frac 1{n^2}$.
Indeed
$$frac{frac{left(ln nright)^3}{n^3}}{frac 1{n^2}}=frac{left(ln nright)^3}{n}to 0$$
and the given series converges.
For the difference between limit comparison test and comparison test refer also to
- The Comparison test or The limit comparison test
answered Dec 11 '18 at 20:54
gimusi
1
@T.Bongers Just to clarify the objective mathematical truth, I've also added a reference on the distincion between direct comparison test and limit comparison test.
– gimusi
Dec 11 '18 at 20:57
@user21820 Is it now fine?
– gimusi
Dec 11 '18 at 21:09
add a comment |
HINT
Recall that $forall a,b>0$
$$frac{log^ an}{n^b} to 0$$
then use limit comparison test with $sum frac 1{n^2}$.
Indeed
$$frac{frac{left(ln nright)^3}{n^3}}{frac 1{n^2}}=frac{left(ln nright)^3}{n}to 0$$
and the given series converges.
For the difference between limit comparison test and comparison test refer also to
- The Comparison test or The limit comparison test
answered Dec 11 '18 at 20:54
gimusi
1
HINT
Recall that $forall a,b>0$
$$frac{log^ an}{n^b} to 0$$
then use limit comparison test with $sum frac 1{n^2}$.
Indeed
$$frac{frac{left(ln nright)^3}{n^3}}{frac 1{n^2}}=frac{left(ln nright)^3}{n}to 0$$
and the given series converges.
For the difference between limit comparison test and comparison test refer also to
- The Comparison test or The limit comparison test
answered Dec 11 '18 at 20:54
gimusi
1
answered Dec 11 '18 at 20:54
gimusi
1
answered Dec 11 '18 at 20:54
gimusi
1
answered Dec 11 '18 at 20:54
gimusi
1
1
@T.Bongers Just to clarify the objective mathematical truth, I've also added a reference on the distincion between direct comparison test and limit comparison test.
– gimusi
Dec 11 '18 at 20:57
@user21820 Is it now fine?
– gimusi
Dec 11 '18 at 21:09
add a comment |
@T.Bongers Just to clarify the objective mathematical truth, I've also added a reference on the distincion between direct comparison test and limit comparison test.
– gimusi
Dec 11 '18 at 20:57
@user21820 Is it now fine?
– gimusi
Dec 11 '18 at 21:09
@T.Bongers Just to clarify the objective mathematical truth, I've also added a reference on the distincion between direct comparison test and limit comparison test.
– gimusi
Dec 11 '18 at 20:57
@T.Bongers Just to clarify the objective mathematical truth, I've also added a reference on the distincion between direct comparison test and limit comparison test.
– gimusi
Dec 11 '18 at 20:57
@user21820 Is it now fine?
– gimusi
Dec 11 '18 at 21:09
@user21820 Is it now fine?
– gimusi
Dec 11 '18 at 21:09
add a comment |
