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Determine whether $sum_n=1^inftyfrac(-1)^nn log^2(n+1)$ converges absolutely or conditionally. [duplicate]
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Is $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$ absolutely convergent?
3 answers
Problem
Let $S = sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$.
Determine the series converges absolutely or conditionally.
Attempt
$S=sum_n=1^infty( -1)^n a_n$
$a_n$ is monotonically decreasing and it approaches zero when $n$ approaches infinity. So series is convergent .
Doubt
How to check for absolute convergence?
Ratio test fails here. Root test is of no use.
I have attempted comparison tests using the fact that $n>log(n)$, but no success there also.
real-analysis sequences-and-series convergence
edited Jul 30 at 11:07
Blue
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asked Jul 30 at 10:11
blue boy
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marked as duplicate by Fabian, Robert ZÂ real-analysis
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Jul 30 at 10:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
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down vote
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This question already has an answer here:
Is $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$ absolutely convergent?
3 answers
Problem
Let $S = sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$.
Determine the series converges absolutely or conditionally.
Attempt
$S=sum_n=1^infty( -1)^n a_n$
$a_n$ is monotonically decreasing and it approaches zero when $n$ approaches infinity. So series is convergent .
Doubt
How to check for absolute convergence?
Ratio test fails here. Root test is of no use.
I have attempted comparison tests using the fact that $n>log(n)$, but no success there also.
real-analysis sequences-and-series convergence
edited Jul 30 at 11:07
Blue
43.6k868141
asked Jul 30 at 10:11
blue boy
528211
marked as duplicate by Fabian, Robert ZÂ real-analysis
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Jul 30 at 10:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Do you know the integral test?
– Fabian
Jul 30 at 10:14
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
This question already has an answer here:
Is $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$ absolutely convergent?
3 answers
Problem
Let $S = sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$.
Determine the series converges absolutely or conditionally.
Attempt
$S=sum_n=1^infty( -1)^n a_n$
$a_n$ is monotonically decreasing and it approaches zero when $n$ approaches infinity. So series is convergent .
Doubt
How to check for absolute convergence?
Ratio test fails here. Root test is of no use.
I have attempted comparison tests using the fact that $n>log(n)$, but no success there also.
real-analysis sequences-and-series convergence
edited Jul 30 at 11:07
Blue
43.6k868141
asked Jul 30 at 10:11
blue boy
528211
This question already has an answer here:
Is $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$ absolutely convergent?
3 answers
Problem
Let $S = sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$.
Determine the series converges absolutely or conditionally.
Attempt
$S=sum_n=1^infty( -1)^n a_n$
$a_n$ is monotonically decreasing and it approaches zero when $n$ approaches infinity. So series is convergent .
Doubt
How to check for absolute convergence?
Ratio test fails here. Root test is of no use.
I have attempted comparison tests using the fact that $n>log(n)$, but no success there also.
This question already has an answer here:
Is $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$ absolutely convergent?
3 answers
real-analysis sequences-and-series convergence
edited Jul 30 at 11:07
Blue
43.6k868141
asked Jul 30 at 10:11
blue boy
528211
edited Jul 30 at 11:07
Blue
43.6k868141
edited Jul 30 at 11:07
Blue
43.6k868141
edited Jul 30 at 11:07
Blue
43.6k868141
43.6k868141
asked Jul 30 at 10:11
blue boy
528211
asked Jul 30 at 10:11
blue boy
528211
asked Jul 30 at 10:11
blue boy
528211
528211
marked as duplicate by Fabian, Robert ZÂ real-analysis
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Jul 30 at 10:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Fabian, Robert ZÂ real-analysis
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Jul 30 at 10:21
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Do you know the integral test?
– Fabian
Jul 30 at 10:14
add a comment |Â
Do you know the integral test?
– Fabian
Jul 30 at 10:14
Do you know the integral test?
– Fabian
Jul 30 at 10:14
Do you know the integral test?
– Fabian
Jul 30 at 10:14
add a comment |Â
1 Answer
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One option is the condensation test, which says $sum_ngeq 1frac1n(log(n+1))^2$ converges if and only if $sum_ngeq 0frac2^n2^n(log (2^n+1))^2$ does.
answered Jul 30 at 10:19
Especially Lime
19k22252
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
One option is the condensation test, which says $sum_ngeq 1frac1n(log(n+1))^2$ converges if and only if $sum_ngeq 0frac2^n2^n(log (2^n+1))^2$ does.
answered Jul 30 at 10:19
Especially Lime
19k22252
add a comment |Â
up vote
0
down vote
One option is the condensation test, which says $sum_ngeq 1frac1n(log(n+1))^2$ converges if and only if $sum_ngeq 0frac2^n2^n(log (2^n+1))^2$ does.
answered Jul 30 at 10:19
Especially Lime
19k22252
add a comment |Â
up vote
0
down vote
up vote
0
down vote
One option is the condensation test, which says $sum_ngeq 1frac1n(log(n+1))^2$ converges if and only if $sum_ngeq 0frac2^n2^n(log (2^n+1))^2$ does.
answered Jul 30 at 10:19
Especially Lime
19k22252
One option is the condensation test, which says $sum_ngeq 1frac1n(log(n+1))^2$ converges if and only if $sum_ngeq 0frac2^n2^n(log (2^n+1))^2$ does.
answered Jul 30 at 10:19
Especially Lime
19k22252
answered Jul 30 at 10:19
Especially Lime
19k22252
answered Jul 30 at 10:19
Especially Lime
19k22252
answered Jul 30 at 10:19
Especially Lime
19k22252
19k22252
add a comment |Â
add a comment |Â
Do you know the integral test?
– Fabian
Jul 30 at 10:14