double sum $\sum _{j=0}^{\infty } \sum _{k=0}^{\infty } \sqrt{j^2+k^2} z^{j+k}$ I am interesting in the real zeros of the function:
$$\sum _{j=0}^{\infty } \sum _{k=0}^{\infty } \sqrt{j^2+k^2} z^{j+k}$$
looking where change the value but it is hard to compute , using Mathematica
$$\text{NSum}\left[(-1)^{j+k} \sqrt{j^2+k^2},\{j,1,\infty \},\{k,1,\infty \},\text{WorkingPrecision}\to 20\right]$$ gives ComplexInfinity, but i think it is not true cause the values is 0.14540 aproximately calculate with maple use Levin Transform this method it is capable to calculate to -3 to -1, using a double transform I have been calculating the following values

and for the z =1 need a  condensation alternative series and gives the values 0.0137496275139485217544804966782611 it is must be to reals zeros between -4 and -3 and 1 and 2
it is possible to get the same result with other methods??

using $$\sum _{j=1}^{\infty } \sum _{k=1}^{\infty } \frac{(-1)^{j+k}}{\left(j^2+k^2\right)^s}=2^{-2 s} \left(2^{2 s}-2\right) \zeta (2 s)-2^{-3 s} \left(2^s-2\right) \zeta (s) \left(\zeta \left(s,\frac{1}{4}\right)-\zeta \left(s,\frac{3}{4}\right)\right)$$ formula from Wolfram world takins s=-1/2 is equal to 0.145403 ==$$1/4 - 2 Sqrt[
  2] (-2 + 1/Sqrt[2]) Zeta[-(1/2)] (Zeta[-(1/2), 1/4] - 
    Zeta[-(1/2), 3/4])$$ agree with the above result , the rest it is impossible to calculate at the moment
