I am wondering if my reasoning is correct.
I want to determine if the following series converges or not: \begin{equation} \sum_{n=1}^\infty\frac{1}{(\ln p_n)^2} \end{equation} where $p_n$ is the $n$-th prime number. I am aware that the series of the reciprocals of the prime numbers diverges, i.e. \begin{equation} \sum_{n=1}^\infty\frac{1}{p_n}=\infty \end{equation} Since we also have that for any $\alpha>0$ \begin{equation} \lim_{x\rightarrow\infty}\frac{\ln x}{x^\alpha}=0 \end{equation} then, for all $\alpha>0$, \begin{equation} \sum_{k=1}^\infty\frac{1}{k^\alpha}<\sum_{k=1}^\infty\frac{1}{(\ln k)^2} \end{equation} Now, the tricky part is the one I am unsure about. Can I just conclude that the original series diverges by using the last result with the $k$'s replaced by the primes and setting $\alpha$ to $1$? I guess it's a stupid question but I am missing something here. Thank you!
It suffices to prove that $\ln^2x < x$ for all $x>1$, since this gives you $$\sum_{n=1}^N \frac1{\ln^2 p_n} > \sum_{n=1}^N \frac1{p_n} \stackrel{N\to\infty}\longrightarrow \infty$$ To show this see that $\ln^2 1 = 0 < 1$ and $$\frac{\mathrm d}{\mathrm dx} \ln^2 x = 2\frac{\ln x}x < 1$$