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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: enter image description here.

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)

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    $\begingroup$ At least it can be written as an Euler product, $$\prod_p \bigg( 1+\frac{p^2}{(p-1)^3 (p+1)} \bigg) \approx 3.390642.$$ $\endgroup$ – Greg Martin Jul 22 at 3:03
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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jul 22 at 3:46
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    $\begingroup$ @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. $\endgroup$ – Barry Cipra Jul 22 at 13:09
  • $\begingroup$ 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 ;) $\endgroup$ – Greg Martin Jul 22 at 16:49
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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.

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