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How could I prove that as $ \epsilon \to 0 $ the regularized series goes as

$$ \sum _{n=1}^{\infty} \frac{\exp(-n\epsilon)}{n}=-\log(1-e^{-\epsilon}) $$

and how could I prove that the finite part

$$ \text{F.P}\sum _{n=1}^{\infty} \frac{\exp(-n\epsilon)}{n}= \gamma $$

Euler-Mascheroni constant.

how about the series $ \text{F.P} \sum _{n=1}^{\infty} n^{s}\exp(-n\epsilon)= \zeta (-s) $

Martin Sleziak
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Jose Garcia
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    This series $\sum_{n=1}^\infty \frac{\exp(-n\epsilon)}{n}$ IS convergent to a finite value for any $\epsilon>0$, what do you mean by its finite part? – Adam Latosiński Jun 30 '19 at 21:36

1 Answers1

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For the first part: \begin{align} \sum_{n=1}^\infty \frac{x^n}{n} &= \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} \\ &= \sum_{n=0}^\infty \int_0^x t^n \mathrm{d}t \\ &= \int_0^x \sum_{n=0}^\infty t^n \mathrm{d}t \\ &= \int_0^x \frac{1}{1-t} \mathrm{d}t \\ &= \int_0^x \mathrm{d}(-\log(1-t)) \\ &= -\log(1-x) \end{align}

Plugging in $x = \mathrm{e}^{-\varepsilon}$ yields \begin{align} \sum_{n=1}^\infty \frac{\mathrm{e}^{-\varepsilon n}}{n} &= -\log(1-\mathrm{e}^{-\varepsilon}) \end{align}

As for regularizing the harmonic series, I described one approach here.

user76284
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