# Evaluating $\sum_{n=1}^{\infty} 1/\phi(n)^2$

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)

• At least it can be written as an Euler product, $$\prod_p \bigg( 1+\frac{p^2}{(p-1)^3 (p+1)} \bigg) \approx 3.390642.$$ – Greg Martin Jul 22 at 3:03
• It's given to 105 decimal places at oeis.org/A109695 – no closed form is given there, which is good evidence that no closed form is known. – Gerry Myerson Jul 22 at 3:46
• @GregMartin, amusingly, if you forget (as I did) that $\phi(p^k)=p^k(1-1/p)$ does not hold for $k=0$, you get the very nice, but utterly nonsensical, Euler product $$\prod_p{1\over(1-p^{-1})^2(1-p^{-2})}$$ i.e., $\zeta(1)^2\zeta(2)$. It took me an embarrassingly long time to figure out where my mistake was. – Barry Cipra Jul 22 at 13:09
• using $\phi(p^k)=p^k(1-\frac1p)$ is fine for $k=0$, you just then have to use it for $k\le-1$ as well ;) – Greg Martin Jul 22 at 16:49

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.