Let consider the sequence of numbers $a_n = 0,1,-1,0,1,-1,0,1,-1, ...$ extended periodically ( so it has period $9$, $a_{n+10}=a_n$. In fact, this is a Dirichlet character $a_n = \chi_9(n)$ modulo 9.
EDIT Following Bruno Joyal's suggestion, I tried
$$ \pi \sqrt{3} \cdot \sum_{n=1}^{150,000} \frac{\chi(n)}{n} \approx 0.19244478483 \approx [0, 3, 12092, 3, 1, 2, 1, 10, 1, 39, 6, 1, 1, 1, 1]$$
suggesting $L(\chi,1) = \pi \cdot \frac{1}{\sqrt{3}}\cdot \frac{1}{3}$ and similarly $L(\chi,3) = \pi^3 \cdot \frac{1}{\sqrt{3}}\cdot \frac{4}{81}$.
Yes, I am trying to find generalized Bernoulli numbers for quadratic characters, $\chi: \mathbb{Z}/m\mathbb{Z} \to \{-1,1\}$.
ORIGINAL I calculated the even Dirichlet series numerically to be
$$ \pi^2 \cdot \sum_{n=1}^{150,000} \frac{\chi(n)}{n^2} \approx 0.0159743129254 \approx [0, 62, 1, 1, 1, 1, 79, 6, 7, 1]$$
I can guess the that $L(\chi,2) = \pi^2 \cdot \frac{1}{62}$. The fraction might be $\frac{1}{63}, \frac{2}{125}$.
Which is correct? And how do we calculate the exact value. There doesn't seem to be a closed value for these Dirichlet series. At least, not in $\pi^{2k}\mathbb{Q}$.
Edit: This answered the question before it was edited.
I don't believe these formulas (namely $L(\chi,2) = \frac{\pi^2}{62}$ and $L(\chi,4) = \frac{\pi^2 }{1906}$) are true.
First minor issue is that your Dirichlet character (which has a period of $9$ rather than $10$) is not primitive, and is induced from the unique primitive Dirichlet character of conductor $3$, namely the Legendre symbol
$$\chi = (\cdot/3) :(\mathbf Z/3\mathbf Z)^\times \to \{\pm 1\}.$$
Your $L$-function is $L(\chi, s)$.
Since $(-1/3)=-1$, $\chi$ is an odd quadratic Dirichlet character of conductor $3$, hence the completed $L$-function is
$$\Lambda(\chi, s) = (\pi/3)^{-(s+1)/2}\Gamma((s+1)/2) L(\chi, s)$$
and by Gauss's theorem on the sign of the functional equation, the functional equation is
$$\Lambda(\chi, s) = \Lambda(\chi, 1-s).$$
Because of the presence of the Gamma factor in the definition of $\Lambda(\chi, s)$, it follows that $L(\chi, s)$ has "trivial zeroes" at the odd negative integers. Therefore, the functional equation relates the value of $L(\chi, 2n)$ not with $L(\chi, 1-2n)=0$ but with the derivative $L'(\chi, 1-2n)$. This is expected to be a transcendental number algebraically independent from $\pi$. The issue is that the even positive integers are not "critical" for $L(\chi, s)$, au sens de Deligne (see Deligne's Values of $L$-Functions and Periods of Integrals).
One can say something at positive odd integers. At negative integers, one has
$$L(\chi, 1-n) = -B_{n, \chi}/n \in \mathbf Q$$
where $B_{n, \chi}$ is the generalized Bernoulli number
$$B_{n, \chi} = {3^{n-1}}(B_n(1/3) - B_n(2/3)).$$Using the functional equation we find a formula of the form
$$L(\chi, 2n+1) = \pi^{2n+1} c_n$$
where $c_n \in \mathbf Q$ is an explicit function of $n$ which I'll leave for you to work out (at this point you have all of the ingredients to do so).