Asymptotic behavior of $\sum \frac{n^K}{e^{n/T}}$ as $T\to \infty$ Consider the following series
$$f(T):= \sum_{n=1}^\infty \frac{n^K}{e^{n/T}}, \ \ \ K \in \mathbb N.$$
One can show that the series converges, using for example $e^x \ge x^{K+2}/(K+2)!$, then 
$$\sum_{n=1}^\infty \frac{n^K}{e^{n/T}} \le (K+2)! T^{K+2}\sum_{n=1}^\infty \frac{n^K}{n^{K+2}} =\frac{\pi^2}{6} (K+2)! T^{K+2}$$
It is also clear that when $T \to \infty$, the series tends to infinity. Indeed,
$$ \sum_{n=1}^\infty \frac{n^K}{e^{n/T}} \ge \sum_{n=1}^{[T]+1} \frac{n^K}{e^{n/T}}\ge \sum_{n=1}^{[T]+1} \frac{n^K}{e^2}\ge \frac{1}{e^2} T^K.$$
Question: I want to know the asymptotics of $f(T)$ as $T\to \infty$. In particular I want to know if 
$$\frac{ f(T)}{ T^\alpha} \to \text{something nonzero as } T\to \infty$$
for some $\alpha \in [K, K+2]$. Actually I guess $\alpha \in (K, K+2)$ as the above estimates are so weak. 
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\ds{\left.\sum_{n = 1}^{\infty}{n^{K} \over \expo{n/T}}
\right\vert_{\, K\ \in\ \mathbb{N}} \ =\
\sum_{n = 1}^{\infty}n^{K}\, z^{n}\,,\qquad z \equiv \expo{-1/T}}$.

\begin{align}
&\color{#f00}{\left.\sum_{n = 1}^{\infty}{n^{K} \over \expo{n/T}}
\right\vert_{\, K\ \in\ \mathbb{N}}} \ =\
\sum_{n = 1}^{\infty}n^{K}z^{n} =
\pars{z\,\partiald{}{z}}^{K}\sum_{n = 1}^{\infty}z^{n} =
\pars{z\,\partiald{}{z}}^{K}\pars{z \over 1 - z}
\end{align}
The right hand side is the PolyLogarithm Function $\ds{\Li{-K}\pars{z}}$.

$\ds{\Li{-K}\pars{z}}$ can be written as a series which involves the
Strirling Numbers of the Second Kind
$\ds{\braces{m \atop s}}$:
\begin{align}
&\color{#f00}{\left.\sum_{n = 1}^{\infty}{n^{K} \over \expo{n/T}}
\right\vert_{\, K\ \in\ \mathbb{N}}} \ =\
\Li{-K}\pars{z} = 
\sum_{\ell = 0}^{K}\ell!\braces{K + 1 \atop \ell + 1}
\pars{1 \over 1 - z}^{\ell + 1}\,,\qquad K = 0,1,2,\ldots
\end{align}

When $\ds{T \to \infty\,,\ z \to 1}$ such that the 'leading term' becomes:
\begin{align}
&\color{#f00}{\left.\sum_{n = 1}^{\infty}{n^{K} \over \expo{n/T}}
\right\vert_{\, K\ \in\ \mathbb{N}}} \ \sim\
K!\ \overbrace{\braces{K + 1 \atop K + 1}}^{\ds{=\ 1}}\
\pars{1 \over 1 - z}^{K + 1} =
\color{#f00}{K! \over \pars{1 - \expo{-1/T}}^{K + 1}}
\quad\mbox{as}\quad T \to \infty
\end{align}
A: As Felix Marin answered, almost from definition $$f(T)= \sum_{n=1}^\infty \frac{n^K}{e^{\frac n T}}=\text{Li}_{-K}\left(e^{-1/T}\right)$$ For large values of $T$, expansions of $f(T)$ lead to $$\left(\frac{1}{T}\right)^{-K} \left(\left(\frac{1}{T}\right)^K \left(\zeta
   (-K)-\frac{\zeta (-K-1)}{T}+\frac{\zeta (-K-2)}{2
   T^2}+O\left(\frac{1}{T^3}\right)\right)+\left(\Gamma (K+1)
   T+O\left(\frac{1}{T^3}\right)\right)\right)$$ that is to say $$f(T)=K!\, T^{K+1} +\frac{\zeta (-K-2)}{2 T^2}-\frac{\zeta (-K-1)}{T}+\zeta (-K)+O\left(\frac{1}{T^3}\right)$$
A: One can also approximate by the integral of $g(x) = x^K e^{-x/T}$, as suggested in the comment. Let me fill in the details.
Since $f(T)$ increasing in $T$, we just consider $T\in \mathbb N$. The function $g$ is increasing on $[0,KT)$ and decreasing at $(KT, \infty)$, thus 
$$\int_0^{KT}x^K e^{-x/T} dx + \int_{KT+1}^\infty x^K e^{-x/T} dx \le f(T) \le \int_1^{KT+1} x^K e^{-x/T} dx + \int_{KT}^\infty x^K e^{-x/T} dx$$
substitute $y = x/T$ and divide by $T^{K+1}$ gives 
$$\int_0^K y^K e^{-y} dy + \int_{(KT+1)/T}^\infty y^K e^{-y} dy \le \frac{f(T)}{T^{K+1}} \le \int_{1/T}^{(KT+1)/T} y^K e^{-y} dy + \int_{K}^\infty y^K e^{-y} dy .$$
taking $T\to \infty$ gives 
$$\lim_{T\to \infty} \frac{f(T)}{T^{K+1}} = \int_0^\infty y^K e^{-y} dy = K!.$$
