# Alternating 1D lattice sum

Are there any equivalent representations of the following (real valued) sum, in particular that are suitable for evaluation as $$z\rightarrow0$$ ? $$S=\sum_{k=-\infty}^\infty \frac{i^k(z-2ik)}{(\rho^2+(z-2ik)^2)^{3/2}}$$

I am aware that $$S$$ resembles a coulomb force sum which can be rearranged using Lekner summation into a seires of Bessel functions, but the factor $$i^k$$ seems to prohibit this transformation.

• I don't understand the statement about the terms decreasing as $k^{-1/2}$. Isn't $k/(k^2)^{3/2} = k^{-2}$? – Michael Engelhardt Feb 11 at 4:28
• Thanks for pointing this out, I don't know how I reached that conclusion. I have reworded the question appropriately. – Matt Majic Feb 11 at 20:32

For $$\rho$$ equal to an even integer the sum $$S$$ diverges as $$1/z^{3/2}$$ when $$z\rightarrow 0$$. For $$\rho$$ unequal to an even integer and $$z>0$$, one has $$S_0=\lim_{z\rightarrow 0}\sum_{k=-\infty}^\infty \frac{i^k(z-2ik)}{(\rho^2+(z-2ik)^2)^{3/2}}=-4\,\Re\sum_{k=1}^\infty\frac{ i^k k}{\left(4 k^2-\rho^2\right)^{3/2}}.$$
• Yes this techincally is the limit as $z\rightarrow0$, but I would like to find an expression that isn't conditionally convergent - I have updated the question to specify this. – Matt Majic Feb 11 at 4:02
• @MattMajic -- please clarify for me: the sum I wrote down has terms that decrease as $1/k^2$ --- why is this only "conditionally convergent" ? – Carlo Beenakker Feb 11 at 6:43
• you're right, I mistakenly took them to go as $k^{-1/2}$, which would be conditional. – Matt Majic Feb 11 at 20:18
$$S$$ may be expressed as a series of Bessel functions:
$$S=-\pi^2\sum_{n=1}^\infty \left(n-1/4\right) \exp[-(n-1/4)\pi z] J_0[(n-1/4)\pi\rho],$$
however this form diverges for $$z=0$$.