Finding an asymptotic equivalent of a series $\forall n \in \mathbb N^*$, let $f_n(x) = n^x\exp(-nx)$.
Find an asymptotic equivalent of the series when $x \to 0^+$ :
$$\sum_{n=1}^{+\infty}f_n(x)$$
Let $f(x) = \sum_{n=1}^{+\infty}f_n(x) $, I found that $f$ is define on $\mathbb R^+_*$.
But after that, it is difficult to find something. I have though one hint given : Use $n \ln(n)\sim n$.
I want to know if there is any documentation about ways to find an asymptotic equivalent of a series of functions.
Thanks for your help !
Edit 1 : I don't know if it is helpful but I noticed that $f$ is convex on $\mathbb R^+_*$
Edit 2 : I looked over the integral $\int_{1}^{+\infty}t^x\exp(-tx)dt$. I found that it is equal to $\dfrac{1}{x^{x+1}}(x^x \exp(-x)+x\Gamma(x)-x\int_{0}^x u^{x-1}\exp(-u)du)$.
I don't know how $x\int_{0}^x u^{x-1}\exp(-u)du$ goes for $x\to0^+$. And, I want to prove it (if possible) without the $\Gamma$ function.
 A: [EDIT: see below for an elementary argument (not using the Gamma function) for the leading order $f(x)\sim 1/x$ as $x\to 0_+$. The argument in the original answer below can also be turned into a fully rigorous answer, at least to leading order, by the Euler-Maclaurin formula. Finally, as Gary mentions in the comment, the full asymptotic expansion also follows by general results about polylogarithms, https://en.wikipedia.org/wiki/Polylogarithm#Series_representations.]
Maybe an easy answer does not exist. This is not a full answer, but probably one can make analytic sense of the following (maybe with the Euler-Maclaurin formula).
Playing with divergent series we have
$$
\sum_{n\geq 0}n^xe^{-nx}=\sum_{n,m\geq 0}n^{x+m}\frac{(-x)^m}{m!}\mbox{"="}\sum_{m\geq 0}\zeta(-m-x)\frac{(-x)^m}{m!}.
$$
On the other hand, it seems natural to regularize such sum by removing the total integral
$$
\int_0^{+\infty}\nu^x \exp(-\nu x)\mathrm d\nu=x^{-x}\Gamma(x).
$$
The asymptotic expansion
\begin{align*}
\sum_{n\geq 0}n^xe^{-nx}&=x^{-x}\Gamma(x)+\sum_{m\geq 0}\zeta(-m-x)\frac{(-x)^m}{m!}
\\
&=\frac{1}{x}+\left(-\log (x)-\frac{1}{2}-\gamma \right)
\\
&\qquad+\frac{1}{12} x \left(6 \log (x) (\log (x)+2 \gamma )+6 \gamma ^2+\pi ^2+1+6 \log (2 \pi )\right)
\\
&\qquad+O(x^2(\log x)^3)
\end{align*}
seems numerically to work very well: in yellow $\sum_{n\geq 0}n^xe^{-nx}$, in blue the asymptotics truncated to order $k$ for $1\leq k\leq 5$.





EDIT: A summation by parts approach to the leading order asymptotics $f(x)\sim 1/x$ as $x\to 0_+$.
The summation by parts formula (https://en.wikipedia.org/wiki/Summation_by_parts) implies that
$$
\sum_{n=k}^{+\infty} p_n(g_{n+1}-g_n) = -p_kg_k-\sum_{n=k+1}^{+\infty}g_n(p_n-p_{n-1})
$$
whenever $\lim_{n\to\infty} p_ng_{n+1}=0$. Let us set $g_n:=-\frac{e^{-nx}}{1-e^{-x}}$. One can check easily that $g_{n+1}-g_n=e^{-nx}$ such that
\begin{align*}
f(x)&=\sum_{n=0}^{+\infty} n^xe^{-nx} \\
&= \frac{1}{1-e^{-x}}\sum_{n=1}^{+\infty}(n^x-(n-1)^x)e^{-nx}.
\end{align*}
Next, $n^x-(n-1)^x\leq x(n-1)^{x-1}$ by the mean value theorem, for all $0<x<1$ and all integers $n\geq 2$.
With a bit of calculus, the function $x\in(0,1)\mapsto x(n-1)^{x-1}e^{-nx}$ attains its maximum $m(n)=\frac{e^{\frac{n}{\log (n-1)-n}} (n-1)^{\frac{1}{n-\log (n-1)}-1}}{n-\log (n-1)}$ at $x=\frac 1{n-\log (n-1)}$.
Since
$$
m(n)\sim \frac 1{en^2},\quad n\to +\infty,
$$
it follows that we can apply Lebesgue dominated convergence theorem to $\sum_{n=1}^{+\infty}(n^x-(n-1)^x)e^{-nx}$ in the limit $x\to 0_+$. In this limit, only the term $n=1$ survives, yielding
$$
\sum_{n=1}^{+\infty}(n^x-(n-1)^x)e^{-nx}\to 1.
$$
Therefore $f(x)(1-e^{-x})\to 1$ as $x\to 0_+$, giving $f(x)\sim 1/x$ as $x\to 0_+$.
A: If $$f_n(x) = n^x\,\exp(-nx)\quad \implies \quad 
f(x)=\sum_{n=1}^{\infty}f_n(x)=\text{Li}_{-x}\left(e^{-x}\right)$$ where appears the polylogarithm function
$$g(x)=\int_{1}^{\infty}  n^x\,\exp(-nx)\,dn=E_{-x}(x)$$ where appears the exponential integral function.
Since @Giulio R already provided some nice expansions and being myself unable to produce even decent plots,
the function varying quite fast, I give below a table of $x\,f(x)$ which is a quite nice function. I added the corresponding values generated by  @Giulio R expansion
$$\left(
\begin{array}{cc}
 x& x\,\text{Li}_{-x}\left(e^{-x}\right) & \text{approximation}\\
 0.00 & 1.00000 & 1.00000 \\
 0.01 & 1.03629 & 1.03627 \\
 0.02 & 1.05971 & 1.05965 \\
 0.03 & 1.07853 & 1.07838 \\
 0.04 & 1.09442 & 1.09417 \\
 0.05 & 1.10818 & 1.10780 \\
 0.06 & 1.12025 & 1.11974 \\
 0.07 & 1.13095 & 1.13030 \\
 0.08 & 1.14049 & 1.13971 \\
 0.09 & 1.14901 & 1.14812 \\
 0.10 & 1.15664 & 1.15567 \\
\end{array}
\right)$$
A: After your help,  I Think I found a way to prove it. Though I will use $\Gamma(x)$. If anyone has a another way not involving it. I am interested !
Due to $t\to t^x\exp(-tx)$ decreasing for $x>0$ and $t\ge 1$,  we know that : $$\forall n\in \mathbb N^*\backslash\{1\}, \int_{n-1}^{n}v^xe^{-vx}dv \ge n^xe^{-nx} \ge \int_{n}^{n+1}v^xe^{-vx}dv$$
So : $$\int_{1}^{+\infty}v^xe^{-vx}dv \ge \sum_{n=2}^{+\infty}n^xe^{-nx} \ge \int_{2}^{+\infty}v^xe^{-vx}dv$$
Yet $\int_{1}^{+\infty}t^x\exp(-tx)dt = x^{-x}\Gamma(x) - \int_{0}^{1}t^x\exp(-tx)dt$ and $\int_{2}^{+\infty}t^x\exp(-tx)dt = x^{-x}\Gamma(x) - \int_{0}^{2}t^x\exp(-tx)dt$
We conclude that : $$\forall x>0, x^{-x}\Gamma(x)- \int_{0}^{1}t^x\exp(-tx)dt \ge  \sum_{n=2}^{+\infty}n^xe^{-nx} \ge x^{-x}\Gamma(x) - \int_{0}^{2}t^x\exp(-tx)dt$$
Also $\forall x >0, \forall t\in[0,2], 0 \le t^x\exp(-tx) \le 1$.
Then : $$\forall x>0, x^{-x}\Gamma(x)+e^{-x} \ge  \sum_{n=1}^{+\infty}n^xe^{-nx} \ge x^{-x}\Gamma(x) - 2 + e^{-x}$$
And $x^{-x}\Gamma(x) \sim_{x\to0^+} x^{-1}$ and $x^{-1}\to_{x\to0^+} +\infty$
Finally :
$$\sum_{n=1}^{+\infty}n^xe^{-nx}\sim_{x\to0^+}x^{-1}$$
