Closed form for the sum $\sum_{\boldsymbol{k}\in \mathbb{Z}^d} \frac{1}{\|\boldsymbol{k}\|^{\alpha}_2}$? I am wondering if there is a closed form for  
\begin{align*}\sum_{\boldsymbol{k}\in \mathbb{Z}^d \setminus \{0\}}  \frac{1}{\|\boldsymbol{k}\|^{\alpha}_2},
\end{align*}
where $d$ is a finite integer, $\alpha>2$ and the norm is l2 norm. I know this can be bounded by using $\zeta(\alpha/2)$ since
\begin{align*}
\frac{1}{k_1^2+...+k_d^2}\leq\frac{1}{d}(\frac{1}{k_1^2}+...+\frac{1}{k_d^2}).
\end{align*} 
Further I am interested in the convergence speed of the following finite sum to the given one above: 
\begin{align*}S_M=\sum_{\boldsymbol{k}\in \mathbb{Z}^d, \|\boldsymbol{k}\|_2\leq M}  \frac{1}{\|\boldsymbol{k}\|^{\alpha}_2},
\end{align*}　
and I am guessing the convergence rate is something like
\begin{align*}|\sum_{\boldsymbol{k}\in \mathbb{Z}^d}  \frac{1}{\|\boldsymbol{k}\|^{\alpha}_2}-S_M|\to O(M^{-\alpha}).\end{align*}
Can anyone help with this?
I would not be surprised if this has already been studied but I cannot find any reference or appropriate words to search. 
Thanks!
 A: Yes there is a closed-form for $\sum_{k \in \mathbb{Z}^d \setminus \{0\}} \|k\|^{-2m}$.
The derivation is quite long, do you know the theory of the Riemann zeta function ? 
Let $$\Theta(x) = \sum_{n=-\infty}^\infty e^{-\pi n^2 x} \qquad \Theta(x)^d = \sum_{k \in \mathbb{Z}^d} e^{-\pi \|k\|^2 x}, \qquad \mathfrak{Z}_d(s)=\sum_{k \in \mathbb{Z}^d \setminus \{0\}} \|k\|^{-s}$$
Then $\Lambda_d(s) = \int_0^\infty x^{s/2-1} (\Theta(x)^d-1) dx = \pi^{-s/2}\Gamma(s/2)\mathfrak{Z}_d(s)$.
The poisson summation formula yields $\Theta(x) = x^{-1/2} \Theta(1/x)$ from which we obtain $$\Lambda_d(s) = \frac{1}{s-d}-\frac{1}{-s}+\int_1^\infty (x^{s/2-1}+x^{(d-s)/2})(\Theta(x)^d-1)dx= \Lambda_d(d-s) $$ and $\Theta(x) = \sum_{m=-1}^{M-1} c_m x^{m/2}+ \mathcal{O}(x^{M/2})$, 
which means $\Lambda(s)-\int_0^1 x^{s/2-1}\sum_{m=-1}^{M-1} c_m x^{m/2}  dx=\Lambda(s)-\sum_{m=-1}^{M-1} c_m \frac{2}{s+m}$ is analytic for $\Re(s) > -M$.
And the same holds with $\Theta(x)^d = (\sum_{m=-1}^{M-1} c_m x^{m/2}+ \mathcal{O}(x^{M/2}))^d=\sum_{m=-d}^{M-d} c_m(d) x^{m/2}+\mathcal{O}(x^{(M-d)/2}$. 
Whence $\mathfrak{Z}_d(-2m) = \lim_{s\to -2m} \frac{\pi^{m}}{\Gamma(s/2)}\Lambda_d(s) = \lim_{s\to -2m} \frac{\pi^{m}}{\Gamma(s/2)}\sum_{m=-1}^{M-1} c_m(d) \frac{2}{s+m} = \pi^m \frac{m!}{(-1)^m}c_{2m}(d)$
Finally we use the same method to evaluate from $\Gamma(s)\zeta(s) = \int_0^\infty x^{s-2}\frac{x}{e^x-1}dx$ that $\zeta(-m) = (-1)^m\frac{B_{m+1}}{m+1}$ where $\frac{x}{e^x-1}= \sum_{m=0}^\infty \frac{B_m}{m!} x^m$ 
And from $\mathfrak{Z}_1(s) = 2 \zeta(s)$ we evaluate the $c_{m}$ and hence the $c_{m}(d)$ and $\mathfrak{Z}_d(-2m)$ and $\mathfrak{Z}_d(d+2m)$.
