1. ## Series' Convergence?

Hello

Determine whether the following series is convergent or divergent:

$\displaystyle \displaystyle \sum_{n=2}^{\infty} \dfrac{1}{ln^e(n)}$

This one makes me crazy!
How to deal with it ?!
The power e is on the whole ln.

2. I don't know, ratio seems to be of the chart here, root likewise. Have you tried the integral test?

3. Ratio will give 1.
Can not use integral test since $\displaystyle \displaystyle \int \dfrac{1}{ln^e(x)} \, dx$ is an unelemetary integral.

4. Is this true, $\displaystyle \dfrac{1}{{\ln ^e (n)}} > \dfrac{1}{{\ln ^3 (n)}}~?$

Use a comparison test.

5. Originally Posted by Liverpool
Hello

Determine whether the following series is convergent or divergent:

$\displaystyle \displaystyle \sum_{n=2}^{\infty} \dfrac{1}{ln^e(n)}$

This one makes me crazy!
How to deal with it ?!
The power e is on the whole ln.
It's also fair game to use Cauchy's Condensation test. Namely $\displaystyle \displaystyle \frac{2^n}{\log^e\left(2^n\right)}=\frac{2^n}{n^e\ log^e(2)}\to\infty$

6. Originally Posted by Liverpool
Hello

Determine whether the following series is convergent or divergent:

$\displaystyle \displaystyle \sum_{n=2}^{\infty} \dfrac{1}{ln^e(n)}$

This one makes me crazy!
How to deal with it ?!
The power e is on the whole ln.
I may be being a bit dense here, but what does the notation $\displaystyle \ln^e(n)$ denote? Is this a fractionally iterated logarithm, a fractional derivative (not likely I would have thought), abused notation for $\displaystyle (\ln(n))^e$, ... ?

By the way as $\displaystyle \ln(n)$ grows more slowly than any positive power of $\displaystyle $$n, so does \displaystyle (\ln(n))^e and in particular for \displaystyle$$n$ large enough there exists a $\displaystyle$$k$ such that $\displaystyle (\ln(n))^e<k\times n$, and so the series diverges.

CB

7. Thanks.