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.
Another way is to use the Cauchy condensation test: For a non-negative, non-decreasing sequence $a_n$ of reals, we have
$$\sum_{n=1}^\infty a_n<\infty\;\;\Leftrightarrow\;\;\sum_{n=0}^\infty 2^n a_{2^n}<\infty$$
Applied to your situation we get that your series converges if and only if
$$\sum_{n=2}^\infty \frac{1}{n (\ln n)^p}$$
converges, which by applying the Cauchy condensation test again converges if and only if
$$\sum_{n=1}^\infty \frac{1}{n^p}$$
converges. Now for this one you should now that it converges iff $p>1$.
Note that, for the sake of clarity, I ignored constants coming from $\ln 2$ that pop up.
One way to do it is to use the integral test.
$$\int_3^{\infty} \frac{dx}{x(\log x)(\log \log x)^p}$$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!
