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I would like to perform an asymptotic expansion of the function $$f(x) = \sum_{n=1}^{\infty}\frac{1}{(nx)^2}K_2(n x),$$ where $K_2(x)$ is the modified Bessel function of the second kind, around $x=0$. One way of doing it is to expand $K_2(nx)$ for small $x$ and then sum over $n$. However, doing this, we are implicity assuming $nx \ll 1$, while, on the other hand, $n$ goes all the way to infinity in the sum.

Is there a smart way of expanding this function?

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    $\begingroup$ Try this, for x close to zero $ f(x) \sim \frac{\pi^4}{45 x^2} $. $\endgroup$ – Mhenni Benghorbal Mar 7 '13 at 15:17
  • $\begingroup$ I managed to calculate the leading term. I'm interested in the terms up to (and including) x^5. $\endgroup$ – user54031 Mar 7 '13 at 17:19

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