Question: Test the convergence of the following integral$$\int\_{e^2}^\infty {dx\over x\log\log x}$$

I understand that the problem is only at $\infty$ how to proceed ?

Answer: 
**Hint:** Your function $\frac{1}{x\log\log x}$ goes to $0$ more slowly than $\frac{1}{x\log x}$. You can comfortably find $\int\_{e^2}^M \frac{dx}{x\log x}$ explicitly, and show that it (sedately) blows up as $M\to \infty$.
The integral converges iff the series

$$\sum\_{n>e^2}\frac1{n\log\log n}\;\;\text{converges}$$

Applying the condensation test to this series, we get it converges iff

$$\sum\_{n>e^2}\frac{2^n}{2^n\log\log2^n}=\sum\_{n>e^2}\frac1{\log n+\log\log2}$$

and this last series diverges...