Is it possible to prove that for the regularized Harmonic series
$$ \tag 1\sum_{n=1}^{\infty} \frac{e^{-n\epsilon}}{n}=\gamma + 1/\epsilon $$
if epsilon is very small $ \epsilon \to 0 $ i can use $ e^{-n\epsilon}=1-n\epsilon $ , to the Harmonic series appear
however if i use the Polylogarithm i get
$$ \sum_{n=1}^{\infty} \frac{e^{-n\epsilon}}{n}=log(1-e^{-\epsilon})$$
so no constant appear, how can I prove $(1)$?
On
In general, you can consider the power series $$f(x) = \sum_{n=1}^\infty \frac{x^n}{n}$$ which is defined (and $\mathcal{C}^\infty$) on the open ball $(-1,1)$. Using this, and standard theorems about power series, you have $$f^\prime(x) = \sum_{n=1}^\infty \frac{d}{dx}\frac{x^n}{n} = \sum_{n=1}^\infty x^{n-1} = \sum_{n=0}^\infty x^{n} = \frac{1}{1-x}$$ and since $f(0)= 0$ (using the very first expression as an infinite series), you get $$f(x) = \int_{0}^x \frac{dt}{1-t} = -\ln(1-x)$$ Now, plugging $x=e^{-\epsilon}\in(0,1)$ (for $\epsilon > 0$), you get $$\sum_{n=1}^\infty \frac{e^{-n\epsilon}}{n} = f(e^{-\epsilon}) = -\ln(1-e^{-\epsilon}) \operatorname*{=}_{\epsilon\to 0^+} \log \frac{1}{\epsilon}+\frac{\epsilon}{2}+o(\epsilon)$$
Therefore, if I have understood correctly your question, the answer is no: you cannot prove (1) — as it is not true.
You cannot prove (1) because it is indeed not true. In fact $$\sum_{n=1}^\infty \frac{e^{-n\varepsilon}}{n}\sim -\log(\varepsilon)$$ as $\varepsilon\to 0^+$, so certainly this cannot be equivalent to $1/\varepsilon$, as would follow from (1).
The above equivalence is proved as follows: as you mentioned, $\sum_{1}^\infty \frac{e^{-n\varepsilon}}{n}=-\log(1-e^{-\varepsilon})$; but $1-e^{-\varepsilon}=\varepsilon+O(\varepsilon^2)$ because $e^{-\varepsilon}=1-\varepsilon +O(\varepsilon^2)$, so $$\log(1-e^{-\varepsilon} )=\log(\varepsilon+O(\varepsilon^2))=\log(\varepsilon)+\log(1+O(\varepsilon))\sim\log(\varepsilon)\, ,$$ which gives the result.