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Is $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$ absolutely convergent?
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up vote
2
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Consider the series
$$sum_n=1^inftyfrac(-1)^nnlog^2(n+1).$$
Determine whether it converges absolutely or conditionally.
My attempt
S=$sum_n=1^infty( -1)^n$ an
an 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.
calculus real-analysis sequences-and-series
edited Jul 30 at 12:58
user 108128
19k41544
asked Jul 30 at 10:08
blue boy
528211
add a comment |Â
up vote
2
down vote
favorite
Consider the series
$$sum_n=1^inftyfrac(-1)^nnlog^2(n+1).$$
Determine whether it converges absolutely or conditionally.
My attempt
S=$sum_n=1^infty( -1)^n$ an
an 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.
calculus real-analysis sequences-and-series
edited Jul 30 at 12:58
user 108128
19k41544
asked Jul 30 at 10:08
blue boy
528211
1
Is the series $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$?
– José Carlos Santos
Jul 30 at 10:12
To test the absolute convergence, you could try Cauchy's integral test.
– xbh
Jul 30 at 10:15
Heh. When I first read this question, I assumed that $log^2$ was the iterated logarithm ($logcirclog$), not the square. I am not sure if there is any firmly established convention in this area.
– Harald Hanche-Olsen
Jul 30 at 11:18
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the series
$$sum_n=1^inftyfrac(-1)^nnlog^2(n+1).$$
Determine whether it converges absolutely or conditionally.
My attempt
S=$sum_n=1^infty( -1)^n$ an
an 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.
calculus real-analysis sequences-and-series
edited Jul 30 at 12:58
user 108128
19k41544
asked Jul 30 at 10:08
blue boy
528211
Consider the series
$$sum_n=1^inftyfrac(-1)^nnlog^2(n+1).$$
Determine whether it converges absolutely or conditionally.
My attempt
S=$sum_n=1^infty( -1)^n$ an
an 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.
calculus real-analysis sequences-and-series
edited Jul 30 at 12:58
user 108128
19k41544
asked Jul 30 at 10:08
blue boy
528211
edited Jul 30 at 12:58
user 108128
19k41544
edited Jul 30 at 12:58
user 108128
19k41544
edited Jul 30 at 12:58
user 108128
19k41544
19k41544
asked Jul 30 at 10:08
blue boy
528211
asked Jul 30 at 10:08
blue boy
528211
asked Jul 30 at 10:08
blue boy
528211
528211
1
Is the series $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$?
– José Carlos Santos
Jul 30 at 10:12
To test the absolute convergence, you could try Cauchy's integral test.
– xbh
Jul 30 at 10:15
Heh. When I first read this question, I assumed that $log^2$ was the iterated logarithm ($logcirclog$), not the square. I am not sure if there is any firmly established convention in this area.
– Harald Hanche-Olsen
Jul 30 at 11:18
add a comment |Â
1
Is the series $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$?
– José Carlos Santos
Jul 30 at 10:12
To test the absolute convergence, you could try Cauchy's integral test.
– xbh
Jul 30 at 10:15
Heh. When I first read this question, I assumed that $log^2$ was the iterated logarithm ($logcirclog$), not the square. I am not sure if there is any firmly established convention in this area.
– Harald Hanche-Olsen
Jul 30 at 11:18
1
1
Is the series $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$?
– José Carlos Santos
Jul 30 at 10:12
Is the series $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$?
– José Carlos Santos
Jul 30 at 10:12
To test the absolute convergence, you could try Cauchy's integral test.
– xbh
Jul 30 at 10:15
To test the absolute convergence, you could try Cauchy's integral test.
– xbh
Jul 30 at 10:15
Heh. When I first read this question, I assumed that $log^2$ was the iterated logarithm ($logcirclog$), not the square. I am not sure if there is any firmly established convention in this area.
– Harald Hanche-Olsen
Jul 30 at 11:18
Heh. When I first read this question, I assumed that $log^2$ was the iterated logarithm ($logcirclog$), not the square. I am not sure if there is any firmly established convention in this area.
– Harald Hanche-Olsen
Jul 30 at 11:18
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
For the absolute convergence by cauchy condensation test we can consider the convergenge of the condensed series $sum 2^n a_2^n$ that is
$$sum frac2^n2^n(log^2(2^n+1))=sum frac1log^2(2^n+1)$$
which converges by limit comparison test with $sum frac1n^2$ indeed
$$frac1log^2(2^n+1)simfrac1n^2log^2 2$$
answered Jul 30 at 10:13
gimusi
64.5k73482
add a comment |Â
up vote
3
down vote
Yes, you are correct, ratio test fails here. Hint: note that for $ngeq 3$,
$$0leq frac1n(log^2(n+1))leq frac1n(log^2(n))leq int_n-1^nfracdxx(log^2(x))=frac1log(n-1)-frac1log(n).$$
What may we conclude?
answered Jul 30 at 10:21
Robert Z
83.6k954122
add a comment |Â
up vote
2
down vote
Simply we have
$$sum_n=2^inftyfrac1nlog^2(n+1)<sum_n=2^inftyfrac1nlog^2(n)$$
and one may use the integral test for evaluating $displaystyleint_2^inftyfracdxxlog^2x=dfrac12$.
answered Jul 30 at 11:43
user 108128
19k41544
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
For the absolute convergence by cauchy condensation test we can consider the convergenge of the condensed series $sum 2^n a_2^n$ that is
$$sum frac2^n2^n(log^2(2^n+1))=sum frac1log^2(2^n+1)$$
which converges by limit comparison test with $sum frac1n^2$ indeed
$$frac1log^2(2^n+1)simfrac1n^2log^2 2$$
answered Jul 30 at 10:13
gimusi
64.5k73482
add a comment |Â
up vote
4
down vote
accepted
For the absolute convergence by cauchy condensation test we can consider the convergenge of the condensed series $sum 2^n a_2^n$ that is
$$sum frac2^n2^n(log^2(2^n+1))=sum frac1log^2(2^n+1)$$
which converges by limit comparison test with $sum frac1n^2$ indeed
$$frac1log^2(2^n+1)simfrac1n^2log^2 2$$
answered Jul 30 at 10:13
gimusi
64.5k73482
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
For the absolute convergence by cauchy condensation test we can consider the convergenge of the condensed series $sum 2^n a_2^n$ that is
$$sum frac2^n2^n(log^2(2^n+1))=sum frac1log^2(2^n+1)$$
which converges by limit comparison test with $sum frac1n^2$ indeed
$$frac1log^2(2^n+1)simfrac1n^2log^2 2$$
answered Jul 30 at 10:13
gimusi
64.5k73482
For the absolute convergence by cauchy condensation test we can consider the convergenge of the condensed series $sum 2^n a_2^n$ that is
$$sum frac2^n2^n(log^2(2^n+1))=sum frac1log^2(2^n+1)$$
which converges by limit comparison test with $sum frac1n^2$ indeed
$$frac1log^2(2^n+1)simfrac1n^2log^2 2$$
answered Jul 30 at 10:13
gimusi
64.5k73482
edited Jul 30 at 10:21
answered Jul 30 at 10:13
gimusi
64.5k73482
answered Jul 30 at 10:13
gimusi
64.5k73482
answered Jul 30 at 10:13
gimusi
64.5k73482
64.5k73482
add a comment |Â
add a comment |Â
up vote
3
down vote
Yes, you are correct, ratio test fails here. Hint: note that for $ngeq 3$,
$$0leq frac1n(log^2(n+1))leq frac1n(log^2(n))leq int_n-1^nfracdxx(log^2(x))=frac1log(n-1)-frac1log(n).$$
What may we conclude?
answered Jul 30 at 10:21
Robert Z
83.6k954122
add a comment |Â
up vote
3
down vote
Yes, you are correct, ratio test fails here. Hint: note that for $ngeq 3$,
$$0leq frac1n(log^2(n+1))leq frac1n(log^2(n))leq int_n-1^nfracdxx(log^2(x))=frac1log(n-1)-frac1log(n).$$
What may we conclude?
answered Jul 30 at 10:21
Robert Z
83.6k954122
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Yes, you are correct, ratio test fails here. Hint: note that for $ngeq 3$,
$$0leq frac1n(log^2(n+1))leq frac1n(log^2(n))leq int_n-1^nfracdxx(log^2(x))=frac1log(n-1)-frac1log(n).$$
What may we conclude?
answered Jul 30 at 10:21
Robert Z
83.6k954122
Yes, you are correct, ratio test fails here. Hint: note that for $ngeq 3$,
$$0leq frac1n(log^2(n+1))leq frac1n(log^2(n))leq int_n-1^nfracdxx(log^2(x))=frac1log(n-1)-frac1log(n).$$
What may we conclude?
answered Jul 30 at 10:21
Robert Z
83.6k954122
edited Jul 30 at 10:30
answered Jul 30 at 10:21
Robert Z
83.6k954122
answered Jul 30 at 10:21
Robert Z
83.6k954122
answered Jul 30 at 10:21
Robert Z
83.6k954122
83.6k954122
add a comment |Â
add a comment |Â
up vote
2
down vote
Simply we have
$$sum_n=2^inftyfrac1nlog^2(n+1)<sum_n=2^inftyfrac1nlog^2(n)$$
and one may use the integral test for evaluating $displaystyleint_2^inftyfracdxxlog^2x=dfrac12$.
answered Jul 30 at 11:43
user 108128
19k41544
add a comment |Â
up vote
2
down vote
Simply we have
$$sum_n=2^inftyfrac1nlog^2(n+1)<sum_n=2^inftyfrac1nlog^2(n)$$
and one may use the integral test for evaluating $displaystyleint_2^inftyfracdxxlog^2x=dfrac12$.
answered Jul 30 at 11:43
user 108128
19k41544
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Simply we have
$$sum_n=2^inftyfrac1nlog^2(n+1)<sum_n=2^inftyfrac1nlog^2(n)$$
and one may use the integral test for evaluating $displaystyleint_2^inftyfracdxxlog^2x=dfrac12$.
answered Jul 30 at 11:43
user 108128
19k41544
Simply we have
$$sum_n=2^inftyfrac1nlog^2(n+1)<sum_n=2^inftyfrac1nlog^2(n)$$
and one may use the integral test for evaluating $displaystyleint_2^inftyfracdxxlog^2x=dfrac12$.
answered Jul 30 at 11:43
user 108128
19k41544
edited Jul 30 at 12:57
answered Jul 30 at 11:43
user 108128
19k41544
answered Jul 30 at 11:43
user 108128
19k41544
answered Jul 30 at 11:43
user 108128
19k41544
19k41544
add a comment |Â
add a comment |Â
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1
Is the series $sum_n=1^inftyfrac(-1)^nnlog^2(n+1)$?
– José Carlos Santos
Jul 30 at 10:12
To test the absolute convergence, you could try Cauchy's integral test.
– xbh
Jul 30 at 10:15
Heh. When I first read this question, I assumed that $log^2$ was the iterated logarithm ($logcirclog$), not the square. I am not sure if there is any firmly established convention in this area.
– Harald Hanche-Olsen
Jul 30 at 11:18