Convergence series problem. How to show that $$\sum_{n=3}^{\infty }\frac{1}{n(\ln n)(\ln \ln n)^p}$$ converges if and only if $p>1$ ?
By integral test, 
$$\sum_{n=3}^{\infty }\frac{1}{n(\ln n)(\ln \ln n)^p}$$
$$f(x)=\frac{1}{x(\ln x)(\ln \ln x)^P}$$
$$\int_{3}^{\infty }\frac{1}{x(\ln x)(\ln \ln x)^p}$$
I stucked at here.

$$\int_3^{\infty} \frac{dx}{x(\log x)(\log \log x)^p}$$Let $u=\log \log x$.  Then, $du=\frac{dx}{x \log x}$.
$$\int_3^{\infty} \frac{dx}{x(\log x)(\log \log x)^p}=\int_{\log \log 3}^{\infty} \frac{du}{u^p}$$This indefinite integral equals $\frac{u^{1-p}}{1-p}$ for $p \ne 1$ and equals $\log u$ for $p=1$.  If $p>1$, then $$\lim_{u\to \infty} \frac{u^{1-p}}{1-p}=0$$  Thus, the integral converges as an improper Riemann integral.  And by the integral test for series convergence, the series of interest converges also!
