# Convergence of Series

## 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
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!

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## 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
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.

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

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.