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let be the sum

$$ f(\epsilon) = \sum_{n=1}^{\infty}e^{-n\epsilon} $$

  • $ f(0)= \infty $ diverges

  • for any positive epsilon $ \epsilon >0 $ the sum converges

assume we know the value of $ f(\epsilon) $ as $ \epsilon \to 0 $

then the asymptotics of the partial sum of the coefficients has the asymptotics

$$ \sum_{n=1}^{x}a(n) \sim L^{-1}[f(\epsilon)/\epsilon](x) $$

where $ L^{-1} [f(s)](x) $ is the inverse laplace transform

Jose Garcia
  • 8,506
  • Indeed we know the value of $f(\epsilon) = e^{-\epsilon}/(1-e^{-\epsilon})$ (it is the sum of a geometric series). – Rigel May 19 '17 at 17:41
  • Continuing Rigel's comment: "... and its limit's value when $;\epsilon\to0;$ isn't finite" . – DonAntonio May 19 '17 at 17:44
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    More generally these kinds of results fall under the umbrella of the Hardy Littlewood Tauberian theorems. – Alex R. May 19 '17 at 17:44

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