Convergence/Divergence of an Infinite Series with Natural Logarithms
Clash Royale CLAN TAG#URR8PPP
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I've spent a good week and half manipulating and trying different tests to find the convergence or divergence of this series:
$$sum_n=0^infty frac1(ln n)^ln n$$
I've tried all the Convergence Tests I know, and have tried to exhaust the Ratio Test for many different manipulations of the above, including
$ln n^-ln n$, $ln n^ln(1/n)$, and, using the relationship $a^x = e^xln a$, $e^ln(1/n)*ln(ln(n))implies(1/n)^ln(ln n)$.
Basically, I'm just burnt out trying to solve this, and came for any help someone on here can offer me.
Cheers.
EDIT: the only hint we're given is to note that $n > e^2$.
calculus sequences-and-series convergence logarithms divergent-series
edited Aug 16 at 22:33
Eclipse Sun
6,6221336
asked Aug 16 at 22:23
anyone
133
add a comment |Â
up vote
2
down vote
favorite
1
I've spent a good week and half manipulating and trying different tests to find the convergence or divergence of this series:
$$sum_n=0^infty frac1(ln n)^ln n$$
I've tried all the Convergence Tests I know, and have tried to exhaust the Ratio Test for many different manipulations of the above, including
$ln n^-ln n$, $ln n^ln(1/n)$, and, using the relationship $a^x = e^xln a$, $e^ln(1/n)*ln(ln(n))implies(1/n)^ln(ln n)$.
Basically, I'm just burnt out trying to solve this, and came for any help someone on here can offer me.
Cheers.
EDIT: the only hint we're given is to note that $n > e^2$.
calculus sequences-and-series convergence logarithms divergent-series
edited Aug 16 at 22:33
Eclipse Sun
6,6221336
asked Aug 16 at 22:23
anyone
133
2
Perhaps that you meant $displaystylesum_n=2^inftyfrac1(log n)^log n$.
– José Carlos Santos
Aug 16 at 22:25
1
No, the question uses natural logarithms.
– anyone
Aug 16 at 22:26
1
@anyone The base of the logarithm is (probably) immaterial. Compare the comment with the sum in the question: they are entirely different.
– Umberto P.
Aug 16 at 22:27
3
Hint: try Cauchy-condensation test. It is very good for series with logarithms.
– Mark
Aug 16 at 22:33
4
I believe @JoséCarlosSantos is pointing out that at $n=0$ and $n=1$, the summand is undefined. (Also $log$ is a common notation for natural log)
– Simply Beautiful Art
Aug 17 at 0:10
add a comment |Â
up vote
2
down vote
favorite
1
up vote
2
down vote
favorite
1
1
I've spent a good week and half manipulating and trying different tests to find the convergence or divergence of this series:
$$sum_n=0^infty frac1(ln n)^ln n$$
I've tried all the Convergence Tests I know, and have tried to exhaust the Ratio Test for many different manipulations of the above, including
$ln n^-ln n$, $ln n^ln(1/n)$, and, using the relationship $a^x = e^xln a$, $e^ln(1/n)*ln(ln(n))implies(1/n)^ln(ln n)$.
Basically, I'm just burnt out trying to solve this, and came for any help someone on here can offer me.
Cheers.
EDIT: the only hint we're given is to note that $n > e^2$.
calculus sequences-and-series convergence logarithms divergent-series
edited Aug 16 at 22:33
Eclipse Sun
6,6221336
asked Aug 16 at 22:23
anyone
133
I've spent a good week and half manipulating and trying different tests to find the convergence or divergence of this series:
$$sum_n=0^infty frac1(ln n)^ln n$$
I've tried all the Convergence Tests I know, and have tried to exhaust the Ratio Test for many different manipulations of the above, including
$ln n^-ln n$, $ln n^ln(1/n)$, and, using the relationship $a^x = e^xln a$, $e^ln(1/n)*ln(ln(n))implies(1/n)^ln(ln n)$.
Basically, I'm just burnt out trying to solve this, and came for any help someone on here can offer me.
Cheers.
EDIT: the only hint we're given is to note that $n > e^2$.
calculus sequences-and-series convergence logarithms divergent-series
calculus sequences-and-series convergence logarithms divergent-series
edited Aug 16 at 22:33
Eclipse Sun
6,6221336
asked Aug 16 at 22:23
anyone
133
edited Aug 16 at 22:33
Eclipse Sun
6,6221336
asked Aug 16 at 22:23
anyone
133
edited Aug 16 at 22:33
Eclipse Sun
6,6221336
edited Aug 16 at 22:33
Eclipse Sun
6,6221336
edited Aug 16 at 22:33
Eclipse Sun
6,6221336
6,6221336
asked Aug 16 at 22:23
anyone
133
asked Aug 16 at 22:23
anyone
133
asked Aug 16 at 22:23
anyone
133
133
2
Perhaps that you meant $displaystylesum_n=2^inftyfrac1(log n)^log n$.
– José Carlos Santos
Aug 16 at 22:25
1
No, the question uses natural logarithms.
– anyone
Aug 16 at 22:26
1
@anyone The base of the logarithm is (probably) immaterial. Compare the comment with the sum in the question: they are entirely different.
– Umberto P.
Aug 16 at 22:27
3
Hint: try Cauchy-condensation test. It is very good for series with logarithms.
– Mark
Aug 16 at 22:33
4
I believe @JoséCarlosSantos is pointing out that at $n=0$ and $n=1$, the summand is undefined. (Also $log$ is a common notation for natural log)
– Simply Beautiful Art
Aug 17 at 0:10
add a comment |Â
2
Perhaps that you meant $displaystylesum_n=2^inftyfrac1(log n)^log n$.
– José Carlos Santos
Aug 16 at 22:25
1
No, the question uses natural logarithms.
– anyone
Aug 16 at 22:26
1
@anyone The base of the logarithm is (probably) immaterial. Compare the comment with the sum in the question: they are entirely different.
– Umberto P.
Aug 16 at 22:27
3
Hint: try Cauchy-condensation test. It is very good for series with logarithms.
– Mark
Aug 16 at 22:33
4
I believe @JoséCarlosSantos is pointing out that at $n=0$ and $n=1$, the summand is undefined. (Also $log$ is a common notation for natural log)
– Simply Beautiful Art
Aug 17 at 0:10
2
2
Perhaps that you meant $displaystylesum_n=2^inftyfrac1(log n)^log n$.
– José Carlos Santos
Aug 16 at 22:25
Perhaps that you meant $displaystylesum_n=2^inftyfrac1(log n)^log n$.
– José Carlos Santos
Aug 16 at 22:25
1
1
No, the question uses natural logarithms.
– anyone
Aug 16 at 22:26
No, the question uses natural logarithms.
– anyone
Aug 16 at 22:26
1
1
@anyone The base of the logarithm is (probably) immaterial. Compare the comment with the sum in the question: they are entirely different.
– Umberto P.
Aug 16 at 22:27
@anyone The base of the logarithm is (probably) immaterial. Compare the comment with the sum in the question: they are entirely different.
– Umberto P.
Aug 16 at 22:27
3
3
Hint: try Cauchy-condensation test. It is very good for series with logarithms.
– Mark
Aug 16 at 22:33
Hint: try Cauchy-condensation test. It is very good for series with logarithms.
– Mark
Aug 16 at 22:33
4
4
I believe @JoséCarlosSantos is pointing out that at $n=0$ and $n=1$, the summand is undefined. (Also $log$ is a common notation for natural log)
– Simply Beautiful Art
Aug 17 at 0:10
I believe @JoséCarlosSantos is pointing out that at $n=0$ and $n=1$, the summand is undefined. (Also $log$ is a common notation for natural log)
– Simply Beautiful Art
Aug 17 at 0:10
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
9
down vote
accepted
HINT:
$$(ln n)^ln n = n ^ln ln n > n^2$$
for all $n > n_0$. So
$$frac1n ^ln ln n < frac1n^2$$
Then use comparison test
answered Aug 16 at 22:33
Ahmad Bazzi
6,1991624
I assume you meant "for every $n$ greater than some $n_0$." (Not "for some $n>n_0$," which does not imply the conclusion...)
– Clement C.
Aug 17 at 0:36
Linguistically speaking "for all" implies "for some". Vice versa is not true though. In the context of series, i agree that it has to be "for all" yes.
– Ahmad Bazzi
Aug 17 at 0:52
I know what "for all" and "for some" mean; I'm saying you used the wrong one. You need to say "for all sufficiently large $n$"; you said "for some sufficiently large $n$." (The way the sentence is written, the "for some" applies to $n$, not to $n_0$; what you actually want to convey is "for some $n_0$, and all $n > n_0$".)
– Clement C.
Aug 17 at 1:03
I agree @ClementC. What you say is correct.
– Ahmad Bazzi
Aug 17 at 3:21
So... maybe edit your answer accordingly?
– Clement C.
Aug 17 at 4:29
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
HINT:
$$(ln n)^ln n = n ^ln ln n > n^2$$
for all $n > n_0$. So
$$frac1n ^ln ln n < frac1n^2$$
Then use comparison test
answered Aug 16 at 22:33
Ahmad Bazzi
6,1991624
I assume you meant "for every $n$ greater than some $n_0$." (Not "for some $n>n_0$," which does not imply the conclusion...)
– Clement C.
Aug 17 at 0:36
Linguistically speaking "for all" implies "for some". Vice versa is not true though. In the context of series, i agree that it has to be "for all" yes.
– Ahmad Bazzi
Aug 17 at 0:52
I know what "for all" and "for some" mean; I'm saying you used the wrong one. You need to say "for all sufficiently large $n$"; you said "for some sufficiently large $n$." (The way the sentence is written, the "for some" applies to $n$, not to $n_0$; what you actually want to convey is "for some $n_0$, and all $n > n_0$".)
– Clement C.
Aug 17 at 1:03
I agree @ClementC. What you say is correct.
– Ahmad Bazzi
Aug 17 at 3:21
So... maybe edit your answer accordingly?
– Clement C.
Aug 17 at 4:29
add a comment |Â
up vote
9
down vote
accepted
HINT:
$$(ln n)^ln n = n ^ln ln n > n^2$$
for all $n > n_0$. So
$$frac1n ^ln ln n < frac1n^2$$
Then use comparison test
answered Aug 16 at 22:33
Ahmad Bazzi
6,1991624
I assume you meant "for every $n$ greater than some $n_0$." (Not "for some $n>n_0$," which does not imply the conclusion...)
– Clement C.
Aug 17 at 0:36
Linguistically speaking "for all" implies "for some". Vice versa is not true though. In the context of series, i agree that it has to be "for all" yes.
– Ahmad Bazzi
Aug 17 at 0:52
I know what "for all" and "for some" mean; I'm saying you used the wrong one. You need to say "for all sufficiently large $n$"; you said "for some sufficiently large $n$." (The way the sentence is written, the "for some" applies to $n$, not to $n_0$; what you actually want to convey is "for some $n_0$, and all $n > n_0$".)
– Clement C.
Aug 17 at 1:03
I agree @ClementC. What you say is correct.
– Ahmad Bazzi
Aug 17 at 3:21
So... maybe edit your answer accordingly?
– Clement C.
Aug 17 at 4:29
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
HINT:
$$(ln n)^ln n = n ^ln ln n > n^2$$
for all $n > n_0$. So
$$frac1n ^ln ln n < frac1n^2$$
Then use comparison test
answered Aug 16 at 22:33
Ahmad Bazzi
6,1991624
HINT:
$$(ln n)^ln n = n ^ln ln n > n^2$$
for all $n > n_0$. So
$$frac1n ^ln ln n < frac1n^2$$
Then use comparison test
answered Aug 16 at 22:33
Ahmad Bazzi
6,1991624
edited Aug 17 at 11:11
answered Aug 16 at 22:33
Ahmad Bazzi
6,1991624
answered Aug 16 at 22:33
Ahmad Bazzi
6,1991624
answered Aug 16 at 22:33
Ahmad Bazzi
6,1991624
6,1991624
I assume you meant "for every $n$ greater than some $n_0$." (Not "for some $n>n_0$," which does not imply the conclusion...)
– Clement C.
Aug 17 at 0:36
Linguistically speaking "for all" implies "for some". Vice versa is not true though. In the context of series, i agree that it has to be "for all" yes.
– Ahmad Bazzi
Aug 17 at 0:52
I know what "for all" and "for some" mean; I'm saying you used the wrong one. You need to say "for all sufficiently large $n$"; you said "for some sufficiently large $n$." (The way the sentence is written, the "for some" applies to $n$, not to $n_0$; what you actually want to convey is "for some $n_0$, and all $n > n_0$".)
– Clement C.
Aug 17 at 1:03
I agree @ClementC. What you say is correct.
– Ahmad Bazzi
Aug 17 at 3:21
So... maybe edit your answer accordingly?
– Clement C.
Aug 17 at 4:29
add a comment |Â
I assume you meant "for every $n$ greater than some $n_0$." (Not "for some $n>n_0$," which does not imply the conclusion...)
– Clement C.
Aug 17 at 0:36
Linguistically speaking "for all" implies "for some". Vice versa is not true though. In the context of series, i agree that it has to be "for all" yes.
– Ahmad Bazzi
Aug 17 at 0:52
I know what "for all" and "for some" mean; I'm saying you used the wrong one. You need to say "for all sufficiently large $n$"; you said "for some sufficiently large $n$." (The way the sentence is written, the "for some" applies to $n$, not to $n_0$; what you actually want to convey is "for some $n_0$, and all $n > n_0$".)
– Clement C.
Aug 17 at 1:03
I agree @ClementC. What you say is correct.
– Ahmad Bazzi
Aug 17 at 3:21
So... maybe edit your answer accordingly?
– Clement C.
Aug 17 at 4:29
I assume you meant "for every $n$ greater than some $n_0$." (Not "for some $n>n_0$," which does not imply the conclusion...)
– Clement C.
Aug 17 at 0:36
I assume you meant "for every $n$ greater than some $n_0$." (Not "for some $n>n_0$," which does not imply the conclusion...)
– Clement C.
Aug 17 at 0:36
Linguistically speaking "for all" implies "for some". Vice versa is not true though. In the context of series, i agree that it has to be "for all" yes.
– Ahmad Bazzi
Aug 17 at 0:52
Linguistically speaking "for all" implies "for some". Vice versa is not true though. In the context of series, i agree that it has to be "for all" yes.
– Ahmad Bazzi
Aug 17 at 0:52
I know what "for all" and "for some" mean; I'm saying you used the wrong one. You need to say "for all sufficiently large $n$"; you said "for some sufficiently large $n$." (The way the sentence is written, the "for some" applies to $n$, not to $n_0$; what you actually want to convey is "for some $n_0$, and all $n > n_0$".)
– Clement C.
Aug 17 at 1:03
I know what "for all" and "for some" mean; I'm saying you used the wrong one. You need to say "for all sufficiently large $n$"; you said "for some sufficiently large $n$." (The way the sentence is written, the "for some" applies to $n$, not to $n_0$; what you actually want to convey is "for some $n_0$, and all $n > n_0$".)
– Clement C.
Aug 17 at 1:03
I agree @ClementC. What you say is correct.
– Ahmad Bazzi
Aug 17 at 3:21
I agree @ClementC. What you say is correct.
– Ahmad Bazzi
Aug 17 at 3:21
So... maybe edit your answer accordingly?
– Clement C.
Aug 17 at 4:29
So... maybe edit your answer accordingly?
– Clement C.
Aug 17 at 4:29
add a comment |Â
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2
Perhaps that you meant $displaystylesum_n=2^inftyfrac1(log n)^log n$.
– José Carlos Santos
Aug 16 at 22:25
1
No, the question uses natural logarithms.
– anyone
Aug 16 at 22:26
1
@anyone The base of the logarithm is (probably) immaterial. Compare the comment with the sum in the question: they are entirely different.
– Umberto P.
Aug 16 at 22:27
3
Hint: try Cauchy-condensation test. It is very good for series with logarithms.
– Mark
Aug 16 at 22:33
4
I believe @JoséCarlosSantos is pointing out that at $n=0$ and $n=1$, the summand is undefined. (Also $log$ is a common notation for natural log)
– Simply Beautiful Art
Aug 17 at 0:10