Analysis of "Absolute Convergence" of $\sum_{n=2}^{\infty}\frac{1}{(lnn)^{n}}$

[rocomath]

\sum_{n=2}^{\infty}\frac{1}{(lnn)^{n}}

If I treat it as a geometric series, then when n=3, r is smaller than 1

\sum_{n=2}^{\infty}(\frac{1}{ln3})^3

What matters is that it's ultimately decreasing, would that be the correct approach? Even problem, no answer!

Thanks.

 

[EnumaElish]

What is r?

 

[rocomath]

Well, as for an answer I would like to say that:

When n=3, r equals

|\frac{1}{ln3}|<1

Which is true, and so, what matters is that it is ultimately decreasing.

 

[EnumaElish]

In the ratio test r = Lim_n-->infinity|Ln(n+1)^-(n+1)/Ln(n)^-n|.

If r < 1 then the series converges.

But ultimately the ratio test is about whether you have a decreasing sequence; so I think you can use the ratio test here.

Also, for n > 1, absolute convergence is the same as convergence because Ln(n) > 0.