Convergence of $sum_{n=1}^{infty} n^{alpha}logleft(1+frac{1}{n^2+2n}right)$ where $0<alpha <1$ [closed]
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$$sum_{n=1}^{infty} n^{alpha}logleft(1+frac{1}{n^2+2n}right) text{ where } 0<alpha <1$$
According to Wolfram, the series seems to converge by comparison test. However, i struggle to find a suitable function to compare it against. Any suggestions?
sequences-and-series
edited Dec 4 at 20:34
Lorenzo B.
1,8222519
asked Dec 4 at 19:28
Timothy
1738
closed as off-topic by Nosrati, John B, Rebellos, Vidyanshu Mishra, Shailesh Dec 5 at 10:50
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up vote
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down vote
favorite
$$sum_{n=1}^{infty} n^{alpha}logleft(1+frac{1}{n^2+2n}right) text{ where } 0<alpha <1$$
According to Wolfram, the series seems to converge by comparison test. However, i struggle to find a suitable function to compare it against. Any suggestions?
sequences-and-series
edited Dec 4 at 20:34
Lorenzo B.
1,8222519
asked Dec 4 at 19:28
Timothy
1738
closed as off-topic by Nosrati, John B, Rebellos, Vidyanshu Mishra, Shailesh Dec 5 at 10:50
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, John B, Rebellos, Vidyanshu Mishra, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$sum_{n=1}^{infty} n^{alpha}logleft(1+frac{1}{n^2+2n}right) text{ where } 0<alpha <1$$
According to Wolfram, the series seems to converge by comparison test. However, i struggle to find a suitable function to compare it against. Any suggestions?
sequences-and-series
edited Dec 4 at 20:34
Lorenzo B.
1,8222519
asked Dec 4 at 19:28
Timothy
1738
$$sum_{n=1}^{infty} n^{alpha}logleft(1+frac{1}{n^2+2n}right) text{ where } 0<alpha <1$$
According to Wolfram, the series seems to converge by comparison test. However, i struggle to find a suitable function to compare it against. Any suggestions?
sequences-and-series
sequences-and-series
edited Dec 4 at 20:34
Lorenzo B.
1,8222519
asked Dec 4 at 19:28
Timothy
1738
edited Dec 4 at 20:34
Lorenzo B.
1,8222519
asked Dec 4 at 19:28
Timothy
1738
edited Dec 4 at 20:34
Lorenzo B.
1,8222519
edited Dec 4 at 20:34
Lorenzo B.
1,8222519
edited Dec 4 at 20:34
Lorenzo B.
1,8222519
1,8222519
asked Dec 4 at 19:28
Timothy
1738
asked Dec 4 at 19:28
Timothy
1738
asked Dec 4 at 19:28
Timothy
1738
1738
closed as off-topic by Nosrati, John B, Rebellos, Vidyanshu Mishra, Shailesh Dec 5 at 10:50
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, John B, Rebellos, Vidyanshu Mishra, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, John B, Rebellos, Vidyanshu Mishra, Shailesh Dec 5 at 10:50
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, John B, Rebellos, Vidyanshu Mishra, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Follows directly from
$0 < ln(1+x) < x$
for $x > 0$.
Each term is then bounded by
$dfrac1{n^{2-alpha}}$
which converges for
$alpha < 1$.
answered Dec 4 at 19:33
marty cohen
72k547126
1
Typo here $dfrac1{n^{2-alpha}}$. Maybe the asker was looking only for a suggestion not a full answer.
– gimusi
Dec 4 at 19:52
Thanks. Fixed and upvoted.
– marty cohen
Dec 4 at 19:53
add a comment |
up vote
1
down vote
Equivalents is the simplest way, as this is a series with positive terms:
Near $0$, $log(1+u)sim u$. Near $infty$, $n^2+2nsim n^2$, hence
$$logbiggl(1+frac1{n^2+2n}biggr)sim_{infty}frac1{n^2},$$
so that
$$n^alphalogbiggl(1+frac1{n^2+2n}biggr)sim_{infty}frac1{n^{2-alpha}}, $$
which converges if and only if $2-alpha >1iffalpha<1$.
answered Dec 4 at 19:36
Bernard
117k637109
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0
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We have that
$$n^{alpha}logleft(1+dfrac{1}{n^2+2n}right)=frac{n^{alpha}}{n^2+2n}cdotlogleft(1+dfrac{1}{n^2+2n}right)^{n^2+2n}simfrac{n^{alpha}}{n^2+2n}$$
then refer to limit comparison test.
answered Dec 4 at 19:36
gimusi
92.5k94495
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Follows directly from
$0 < ln(1+x) < x$
for $x > 0$.
Each term is then bounded by
$dfrac1{n^{2-alpha}}$
which converges for
$alpha < 1$.
answered Dec 4 at 19:33
marty cohen
72k547126
1
Typo here $dfrac1{n^{2-alpha}}$. Maybe the asker was looking only for a suggestion not a full answer.
– gimusi
Dec 4 at 19:52
Thanks. Fixed and upvoted.
– marty cohen
Dec 4 at 19:53
add a comment |
up vote
1
down vote
accepted
Follows directly from
$0 < ln(1+x) < x$
for $x > 0$.
Each term is then bounded by
$dfrac1{n^{2-alpha}}$
which converges for
$alpha < 1$.
answered Dec 4 at 19:33
marty cohen
72k547126
1
Typo here $dfrac1{n^{2-alpha}}$. Maybe the asker was looking only for a suggestion not a full answer.
– gimusi
Dec 4 at 19:52
Thanks. Fixed and upvoted.
– marty cohen
Dec 4 at 19:53
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Follows directly from
$0 < ln(1+x) < x$
for $x > 0$.
Each term is then bounded by
$dfrac1{n^{2-alpha}}$
which converges for
$alpha < 1$.
answered Dec 4 at 19:33
marty cohen
72k547126
Follows directly from
$0 < ln(1+x) < x$
for $x > 0$.
Each term is then bounded by
$dfrac1{n^{2-alpha}}$
which converges for
$alpha < 1$.
answered Dec 4 at 19:33
marty cohen
72k547126
edited Dec 4 at 19:54
answered Dec 4 at 19:33
marty cohen
72k547126
answered Dec 4 at 19:33
marty cohen
72k547126
answered Dec 4 at 19:33
marty cohen
72k547126
72k547126
1
Typo here $dfrac1{n^{2-alpha}}$. Maybe the asker was looking only for a suggestion not a full answer.
– gimusi
Dec 4 at 19:52
Thanks. Fixed and upvoted.
– marty cohen
Dec 4 at 19:53
add a comment |
1
Typo here $dfrac1{n^{2-alpha}}$. Maybe the asker was looking only for a suggestion not a full answer.
– gimusi
Dec 4 at 19:52
Thanks. Fixed and upvoted.
– marty cohen
Dec 4 at 19:53
1
1
Typo here $dfrac1{n^{2-alpha}}$. Maybe the asker was looking only for a suggestion not a full answer.
– gimusi
Dec 4 at 19:52
Typo here $dfrac1{n^{2-alpha}}$. Maybe the asker was looking only for a suggestion not a full answer.
– gimusi
Dec 4 at 19:52
Thanks. Fixed and upvoted.
– marty cohen
Dec 4 at 19:53
Thanks. Fixed and upvoted.
– marty cohen
Dec 4 at 19:53
add a comment |
up vote
1
down vote
Equivalents is the simplest way, as this is a series with positive terms:
Near $0$, $log(1+u)sim u$. Near $infty$, $n^2+2nsim n^2$, hence
$$logbiggl(1+frac1{n^2+2n}biggr)sim_{infty}frac1{n^2},$$
so that
$$n^alphalogbiggl(1+frac1{n^2+2n}biggr)sim_{infty}frac1{n^{2-alpha}}, $$
which converges if and only if $2-alpha >1iffalpha<1$.
answered Dec 4 at 19:36
Bernard
117k637109
add a comment |
up vote
1
down vote
Equivalents is the simplest way, as this is a series with positive terms:
Near $0$, $log(1+u)sim u$. Near $infty$, $n^2+2nsim n^2$, hence
$$logbiggl(1+frac1{n^2+2n}biggr)sim_{infty}frac1{n^2},$$
so that
$$n^alphalogbiggl(1+frac1{n^2+2n}biggr)sim_{infty}frac1{n^{2-alpha}}, $$
which converges if and only if $2-alpha >1iffalpha<1$.
answered Dec 4 at 19:36
Bernard
117k637109
add a comment |
up vote
1
down vote
up vote
1
down vote
Equivalents is the simplest way, as this is a series with positive terms:
Near $0$, $log(1+u)sim u$. Near $infty$, $n^2+2nsim n^2$, hence
$$logbiggl(1+frac1{n^2+2n}biggr)sim_{infty}frac1{n^2},$$
so that
$$n^alphalogbiggl(1+frac1{n^2+2n}biggr)sim_{infty}frac1{n^{2-alpha}}, $$
which converges if and only if $2-alpha >1iffalpha<1$.
answered Dec 4 at 19:36
Bernard
117k637109
Equivalents is the simplest way, as this is a series with positive terms:
Near $0$, $log(1+u)sim u$. Near $infty$, $n^2+2nsim n^2$, hence
$$logbiggl(1+frac1{n^2+2n}biggr)sim_{infty}frac1{n^2},$$
so that
$$n^alphalogbiggl(1+frac1{n^2+2n}biggr)sim_{infty}frac1{n^{2-alpha}}, $$
which converges if and only if $2-alpha >1iffalpha<1$.
answered Dec 4 at 19:36
Bernard
117k637109
answered Dec 4 at 19:36
Bernard
117k637109
answered Dec 4 at 19:36
Bernard
117k637109
answered Dec 4 at 19:36
Bernard
117k637109
117k637109
add a comment |
add a comment |
up vote
0
down vote
We have that
$$n^{alpha}logleft(1+dfrac{1}{n^2+2n}right)=frac{n^{alpha}}{n^2+2n}cdotlogleft(1+dfrac{1}{n^2+2n}right)^{n^2+2n}simfrac{n^{alpha}}{n^2+2n}$$
then refer to limit comparison test.
answered Dec 4 at 19:36
gimusi
92.5k94495
add a comment |
up vote
0
down vote
We have that
$$n^{alpha}logleft(1+dfrac{1}{n^2+2n}right)=frac{n^{alpha}}{n^2+2n}cdotlogleft(1+dfrac{1}{n^2+2n}right)^{n^2+2n}simfrac{n^{alpha}}{n^2+2n}$$
then refer to limit comparison test.
answered Dec 4 at 19:36
gimusi
92.5k94495
add a comment |
up vote
0
down vote
up vote
0
down vote
We have that
$$n^{alpha}logleft(1+dfrac{1}{n^2+2n}right)=frac{n^{alpha}}{n^2+2n}cdotlogleft(1+dfrac{1}{n^2+2n}right)^{n^2+2n}simfrac{n^{alpha}}{n^2+2n}$$
then refer to limit comparison test.
answered Dec 4 at 19:36
gimusi
92.5k94495
We have that
$$n^{alpha}logleft(1+dfrac{1}{n^2+2n}right)=frac{n^{alpha}}{n^2+2n}cdotlogleft(1+dfrac{1}{n^2+2n}right)^{n^2+2n}simfrac{n^{alpha}}{n^2+2n}$$
then refer to limit comparison test.
answered Dec 4 at 19:36
gimusi
92.5k94495
edited Dec 4 at 19:44
answered Dec 4 at 19:36
gimusi
92.5k94495
answered Dec 4 at 19:36
gimusi
92.5k94495
answered Dec 4 at 19:36
gimusi
92.5k94495
92.5k94495
add a comment |
add a comment |
