calculus - Can I show the series convergence by this relabeling method?

The series is

$$\sum_2^{\infty}\frac {1}{(\log(n))^{\log(n)}}$$

Can I simply relabel the $\log(n)$ with the variable $y$ and use the ratio test on $\frac{1}{y^y}$? I'd get 0 in the ratio test limit and I conclude that the series converges.

Is this method valid?

Or should I really stick with the integral test instead? I tried the integral test too, but the integral is hard to evaluate.

Also, the Cauchy Condensation Test is usually helpful for series of logarithms but it doesn't seem useful in this example.

Any hints or solutions are welcome.

Thanks,

Answer

Note that $$\sum_{n\geq2}\frac{1}{\left(\log\left(n\right)\right)^{\log\left(n\right)}}=\sum_{n\geq2}\frac{1}{\exp\left(\log\left(\left(\log\left(n\right)\right)^{\log\left(n\right)}\right)\right)}=\sum_{n\geq2}\frac{1}{n^{\log\left(\log\left(n\right)\right)}}

$$ and so, for example, for a sufficient large $N
$ we have $\forall n\geq N
$ that $n^{\log\left(\log\left(n\right)\right)}>n^{2}
$ hence $$\sum_{n\geq N}\frac{1}{n^{\log\left(\log\left(n\right)\right)}}\leq\sum_{n\geq N}\frac{1}{n^{2}}<\infty.
$$