An infinite series involving the mod-parity of Euler's totient function

Question

Can you prove or disprove the following claim:

First, define the function $\xi(n)$ as follows: $$\xi(n)=\begin{cases}-1, & \text{if }\varphi(n) \equiv 0 \pmod{4} \\ 1, & \text{if }\varphi(n) \equiv 2 \pmod{4} \\ 0, & \text{if otherwise } \end{cases}$$ where $\varphi(n)$ denotes Euler's totient function. Then, $$\frac{\pi^2}{72}=\displaystyle\sum_{n=1}^{\infty}\frac{\xi(n)}{n^2}$$

The SageMath cell that demonstrates this claim can be found here.

Answer 1

The only odd values of $\phi(n)$ are $\phi(1)=\phi(2)=1$.

$\phi(n)$ is even but not divisible by $4$ when:

1. $n=4$

2. $n=2^{\left\{0,1\right\}}p^m$, where $p=4k+3$ is prime, $m=1,2,3,...$

We have $$ \frac{\pi^2}{6}=1+\frac14+\sum_{\substack{n=1\\\phi(n)\equiv 0}}^\infty\frac{1}{n^2}+\sum_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}. $$ (congruences are modulo $4$)

The claim in the question reads $$ \frac{\pi^2}{72}=-\sum_{\substack{n=1\\\phi(n)\equiv 0}}^\infty\frac{1}{n^2}+\sum_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}. $$

Combining these two we get a hypothetical identity $$ \sum_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}=\frac{13\pi^2}{144}-\frac{5}{8}. $$ However $$ \sum_{\substack{n=1\\\phi(n)\equiv 2}}^\infty\frac{1}{n^2}=\frac{1}{16}+\left(1+\frac14\right)\sum_{m=1}^\infty\sum_{p\equiv 3}\frac{1}{p^{2m}}=\frac{1}{16}+\frac54\sum_{p\equiv 3}\frac{1}{p^2-1}. $$ Thus the conjecture is equivalent to $$ \sum_{p\equiv 3}\frac{1}{p^2-1}=\frac{1}{20} \left(\frac{13 \pi ^2}{9}-11\right).\tag{*} $$ The article on sums over primes on mathworld https://mathworld.wolfram.com/PrimeSums.html does not list any sums of this kind. Numerical checks show that the claim is false. For example summing over the first $N=500000$ primes gives for the ratio of the LHS of (*) to the RHS: $$ 1.000153116 $$ with the truncation error term of the order $$ \int_{N}^\infty \frac{dx}{x^2\ln^2(x)}\sim\frac{1}{N\ln^2(N)}\sim 10^{-10}. $$

Answer 2

Reinforcing Nemo's negative answer.

$$ \sum_{n=1}^{10^4} \frac{\xi(n)}{n^2} - \sum_{n=10^4+1}^\infty \frac{1}{n^2} \leq \sum_{n=1}^{\infty} \frac{\xi(n)}{n^2} \leq \sum_{n=1}^{10^4} \frac{\xi(n)}{n^2} + \sum_{n=10^4+1}^\infty \frac{1}{n^2} $$

The first and third of these are directly computable:

$$ 0.13712{\dots} \leq \sum_{n=1}^{\infty} \frac{\xi(n)}{n^2} \leq 0.13732\dots \text{.} $$ However, $$ \frac{\pi^2}{72} = 0.137077{\dots} \text{,} $$ which does not fit between those bounds.

(Aside: There may be smaller cutoffs than $10^4$ that exhibit this negative result. $10^3$ does not.)