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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.

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    $\begingroup$ I don't understand the statement about the terms decreasing as $k^{-1/2} $. Isn't $k/(k^2)^{3/2} = k^{-2} $? $\endgroup$ – Michael Engelhardt Feb 11 at 4:28
  • $\begingroup$ Thanks for pointing this out, I don't know how I reached that conclusion. I have reworded the question appropriately. $\endgroup$ – Matt Majic Feb 11 at 20:32
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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 $$S_0=\lim_{z\rightarrow 0}\sum_{k=-\infty}^\infty \frac{i^k(z-2ik)}{(\rho^2+(z-2ik)^2)^{3/2}}=-4\,\Re\sum_{k=1}^\infty\frac{ i^k k}{\left(4 k^2-\rho^2\right)^{3/2}}.$$

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  • $\begingroup$ 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. $\endgroup$ – Matt Majic Feb 11 at 4:02
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    $\begingroup$ @MattMajic -- please clarify for me: the sum I wrote down has terms that decrease as $1/k^2$ --- why is this only "conditionally convergent" ? $\endgroup$ – Carlo Beenakker Feb 11 at 6:43
  • $\begingroup$ you're right, I mistakenly took them to go as $k^{-1/2}$, which would be conditional. $\endgroup$ – Matt Majic Feb 11 at 20:18
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$S$ 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 $z=0$.

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