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Convergence of Series

by limddavid
Tags: convergence, series
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limddavid
#1
Nov4-11, 12:42 AM
P: 6
1. The problem statement, all variables and given/known data

"Determine whether the following series converge:

[itex]\sum_{n \geq 2} \frac{n^{ln (n)}}{ln(n)^{n}}[/itex]

and

[itex]\sum_{n \geq 2} \frac{1}{(ln(n))^{ln(n)}}[/itex]

2. Relevant equations

The convergence/divergence tests (EXCEPT INTEGRAL TEST):

Ratio
Dyadic
Comparison
P-test
Cauchy Criterion
Root Criterion
Alternating Series Test/Leibniz Criterion
Abel's Criterion

3. The attempt at a solution

My TA said it was helpful to use the Dyadic Criterion to solve series involving logs... I believe this is an exception. It made the equation really convoluted:

[itex]\sum_{n \geq 2} \frac{2^{2k}*k*ln(2)}{(k*ln(2))^{2^{k}}}[/itex]

I'm sure I have to use some combination of the tests, but I kind of need to be pointed in the right direction... I have no idea how to work with that series..

Thank you!
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shaon0
#2
Nov4-11, 01:49 AM
P: 48
Quote Quote by limddavid View Post
1. The problem statement, all variables and given/known data

"Determine whether the following series converge:

[itex]\sum_{n \geq 2} \frac{n^{ln (n)}}{ln(n)^{n}}[/itex]

and

[itex]\sum_{n \geq 2} \frac{1}{(ln(n))^{ln(n)}}[/itex]

2. Relevant equations

The convergence/divergence tests (EXCEPT INTEGRAL TEST):

Ratio
Dyadic
Comparison
P-test
Cauchy Criterion
Root Criterion
Alternating Series Test/Leibniz Criterion
Abel's Criterion

3. The attempt at a solution

My TA said it was helpful to use the Dyadic Criterion to solve series involving logs... I believe this is an exception. It made the equation really convoluted:

[itex]\sum_{n \geq 2} \frac{2^{2k}*k*ln(2)}{(k*ln(2))^{2^{k}}}[/itex]

I'm sure I have to use some combination of the tests, but I kind of need to be pointed in the right direction... I have no idea how to work with that series..

Thank you!
Try the root test; C=lim{n->inf} sup n^(ln(n)/n)/ln(n). Then Let u=ln(n) and substitute this into the root test. Answer should converge to C=0. So the series converges absolutely.
limddavid
#3
Nov4-11, 03:15 AM
P: 6
Quote Quote by shaon0 View Post
Try the root test; C=lim{n->inf} sup n^(ln(n)/n)/ln(n). Then Let u=ln(n) and substitute this into the root test. Answer should converge to C=0. So the series converges absolutely.
Ok.. I tried to root test, but I'm not sure how I can take the limsup of what I get:

n^(u/n)/u

shaon0
#4
Nov4-11, 04:24 AM
P: 48
Convergence of Series

Quote Quote by limddavid View Post
Ok.. I tried to root test, but I'm not sure how I can take the limsup of what I get:

n^(u/n)/u
u=ln(n) → eu2e-u/u and so you get convergence to 0.


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