TITLE: Evaluating $\sum_{n=1}^{\infty} 1/\phi(n)^2$
QUESTION [8 upvotes]: Wolfram Alpha gives $$\sum_{n=1}^{10000} 1/\phi(n)^2\approx 3.3901989747265619591157$$ and a graph of partial sums indicates fairly clearly this converges: .
It's well-known that the sum of the inverse squares $\sum_{n=1}^{\infty} 1/n^2=\pi^2/6$. Is there a closed form for $\sum_{n=1}^{\infty} 1/\phi(n)^2$?
(Note: $\phi$ refers to Euler's totient function)

REPLY [9 votes]: Proving convergence at least is easy. Note that due to Ramanujan, we have some definite lower bounds for Euler's totient. In particular we have
$$\varphi(n)\geq 2\left(\frac{n}{6}\right)^{2/3}$$
So
$$\sum_{n\in\mathbb N}\frac{1}{\varphi(n)^2}\leq\frac{6^{2/3}}{2}\zeta(4/3)$$
Where it is very well known that the sum representation of $\zeta(s)$ converges for $\operatorname{Re}s>1$.
Finding a closed form I suspect is completely impossible. Even the much simpler sum
$$\sum_{n\in\mathbb N}\frac{1}{{p_n}^2}$$
Where $p_n$ denotes the $n$th prime does not have a known closed form in terms of well known mathematical functions and constants.