Convergence/Divergence of an Infinite Series with Natural Logarithms

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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 $n > e^2$.










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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














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 $n > e^2$.










share|cite|improve this question



















  • 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












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 $n > e^2$.










share|cite|improve this question















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






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edited Aug 16 at 22:33









Eclipse Sun

6,6221336




6,6221336










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












  • 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










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






share|cite|improve this answer






















  • 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










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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






share|cite|improve this answer






















  • 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














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






share|cite|improve this answer






















  • 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












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






share|cite|improve this answer














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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 17 at 11:11

























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
















  • 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

















 

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