TITLE: An infinite series involving the mod-parity of Euler's totient function
QUESTION [6 upvotes]: 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.

REPLY [9 votes]: 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.)