Study the convergence of the following series:
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{\left(1+\frac{1}{\ln(\ln(n))}\right)}}$$
The absolute convergence is giving me hard times, eventually I was following this path, any tips?
$\left( n^{\left(1+\frac{1}{\ln(\ln(n))}\right)}\right)^{-1} = \left( e^{\ln(n)+\frac{1}{\ln(\ln(n))}\ln(n)}\right)^{-1} = \left( e^{\ln(n)\left(1+\frac{1}{\ln(\ln(n))}\right)}\right)^{-1} =\,\,\,\,$?
Since $\dfrac{x}{\ln(x)} \to \infty $ as $x \to \infty$, it follows that $\dfrac{f(x)}{f(\ln(x))} \to \infty $ as $x \to \infty$ where $f$ is any positive, strictly increasing, continuous, unbounded function.
Take $f(x) = \ln(x)$ to get the desired result.