Convergence of Series: Dyadic Criterion for Series Involving Logs

[limddavid]

Homework Statement

"Determine whether the following series converge:

$\sum_{n \geq 2} \frac{n^{ln (n)}}{ln(n)^{n}}$

and

$\sum_{n \geq 2} \frac{1}{(ln(n))^{ln(n)}}$

Homework 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

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:

$\sum_{n \geq 2} \frac{2^{2k}*k*ln(2)}{(k*ln(2))^{2^{k}}}$

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!

 

[shaon0]

Homework Statement

"Determine whether the following series converge:

$\sum_{n \geq 2} \frac{n^{ln (n)}}{ln(n)^{n}}$

and

$\sum_{n \geq 2} \frac{1}{(ln(n))^{ln(n)}}$

Homework 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

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:

$\sum_{n \geq 2} \frac{2^{2k}*k*ln(2)}{(k*ln(2))^{2^{k}}}$

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]

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]

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) → e^u2e-u/u and so you get convergence to 0.