$\sum_{n=2}^\infty a^{-\ln(\ln(n))}$ for every $a>1$ I want to determine if this sum is convergent or divergent: $\sum_{n=2}^\infty a^{-\ln(\ln(n))}$ for every $a>1$.
Define $f(x)=a^{-\ln(\ln(x))}$ for every $x \ge 2$.
$f$ is decreasing (since $a^{\ln(\ln(x))}$ is increasing), so we can use the Integral test,
But I find it hard to calculate $\int_2^\infty a^{-\ln(\ln(n))}$.
Any hints to solve this integral (hints only please)?
Thanks a lot!
 A: By repeated use of L'Hôpital's rule, for any $\ r>0,$
$$\displaystyle\lim_{u\to\infty} \frac{u^r}{e^u} = \displaystyle\lim_{u\to\infty} \frac{ru^{r-1}}{e^u}=\ldots= \displaystyle\lim_{u\to\infty} \frac{r(r-1)\ldots (r-\lfloor r\rfloor)u^{r-\lceil r \rceil}}{e^u} = \frac{0}{+\infty} = 0.\qquad (1)$$
Let $\ x=e^u,\ $ so that $\ \large{\frac{u^r}{e^u}\  = \frac{(\ln x)^r}{x}}.\quad$ We have: $\ u\to\infty\iff x\to\infty.\ $ Therefore:
$$ \displaystyle\lim_{x\to\infty} \frac{(\ln x)^r}{x} = 0. $$
So regardless of the choice of $\ r>0,\ x\ $ is eventually greater than $\ (\ln x)^r.\qquad (2)$
$\large{a^{-\ln(\ln(x))} \equiv \frac{1}{\ln(x)^{ln(a)}} }$.
Since $\ a>1,\ \ln(a)>0.$
From $\ (2),\ \exists\ c>2\ $ s.t.
$$(\ln x)^r < x,\quad \therefore\ \frac{1}{(\ln x)^r} > \frac{1}{x}>0\qquad \forall x\geq c\qquad (3)$$
Therefore,
$$ \displaystyle\int_2^\infty \frac{1}{\ln(x)^{ln(a)}} dx = \underset{\Large{=\ b\ \in\ \mathbb{R}}}{\underbrace{\displaystyle\int_2^c \frac{1}{\ln(x)^{ln(a)}} dx}} + \displaystyle\int_c^\infty \frac{1}{\ln(x)^{ln(a)}} dx = b + \displaystyle\int_c^\infty\frac{1}{\ln(x)^{ln(a)}} dx \geq b + \displaystyle\int_c^\infty\frac{1}{x} dx = \infty.$$
