Evaluation of $\Xi(z)=\sum_{t=1}^{\infty}\frac{t^z}{e^t}$ I would like to try and evaluate the following gamma function inspired sum.
$$\Xi(z)=\sum_{t=1}^{\infty}\frac{t^z}{e^t}$$
According to my computations, for large $z$, 
$$\Xi(z)\approx\Gamma (z+1)$$ 
and perhaps even 
$$\Xi(z) \sim \Gamma (z+1)$$
Does a closed form exist for this sum?
 A: Your sum is in the form of the Polylogarithm. In fact, it is equal to $\operatorname{Li}_{-z}(1/e).$ When $z$ is a negative integer, this sum is easily computed in closed form using the identity $\displaystyle  \frac{d}{dx} \operatorname{Li}_n (x) = x\operatorname{Li}_{n-1} (x).$
By applying the Abel-Plana summation to the Polylogarithm series, we get
$$\operatorname{Li}_s(z) = {z\over2} + {\Gamma(1 \!-\! s, -\ln z) \over (-\ln z)^{1-s}} + 2z \int_0^\infty \frac{\sin(s\arctan t \,- \,t\ln z)} {(1+t^2)^{s/2} \,(e^{2\pi t}-1)} \,\mathrm{d}t
$$
and so $$\operatorname{Li}_{-z}(1/e) = \frac{1}{2e} + \Gamma(z+1,1) + \frac{2}{e} \int^{\infty}_0 \frac{ (1+t^2)^{z/2} \sin(-z \tan^{-1}t+t)}{e^{2\pi t} -1} dt .$$
where $\Gamma(s,x)$ is the incomplete gamma function. Your asymptotic would be explained if you could show why the remaining integral is comparatively small. 
A: It  appears  that  we  can  obtain  a simple  proof  of  the  proposed
asymptotics using harmonic sum techniques when $\Re(z) = \sigma > 0$.
To see this, introduce the sum
$$S(x; z) = \sum_{n\ge 1} \frac{(nx)^z}{\exp(nx)}$$
so that $$\Xi(z) = S(1; z).$$
The sum term is harmonic and  may be evaluated by inverting its Mellin
transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad
g(x) = \frac{x^z}{\exp(x)}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty \frac{x^z}{\exp(x)} x^{s-1} dx
= \int_0^\infty e^{-x} x^{s+z-1} dx.$$
No  additional  computation  is  necessary  as this  integral  is  the
defining integral of the gamma function
$$\Gamma(s) = \int_0^\infty e^{-x} x^{s-1} dx$$
so that
$$g^*(s) = \Gamma(s+z)$$
with fundamental strip $\langle -\sigma, +\infty \rangle.$
It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x; z)$
is given by
$$Q(s) = \zeta(s) \Gamma(s+z)
\quad\text{because}\quad
\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \zeta(s).$$
The Mellin inversion integral here is
$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$
which we evaluate  by shifting it to the left  for an expansion about
zero.
The  line  for the  integral  is chosen  in  the  intersection of  the
fundamental  strip of $g^*(s)$  and the  half-plane of  convergence of
$\zeta(s)$, which is $\Re(s)>1.$
Collecting the residues from the poles we first obtain that
$$\mathrm{Res}\left(Q(s)/x^s; s=1\right) = \frac{1}{x} \Gamma(1+z).$$
The remaining poles are from  the gamma function. Keeping in mind that
$\Re(z) = \sigma > 0$ we find with $q\ge 0$
$$\mathrm{Res}\left(Q(s)/x^s; s=-z-q\right) = 
x^{z+q} \frac{(-1)^q}{q!} \zeta(-z-q).$$
We conclude that
$$\Xi(z) = S(1; z) \sim 
\Gamma(1+z) + \sum_{q\ge 0} \frac{(-1)^q}{q!} \zeta(-z-q).$$
For $z=m$ with $m$ a positive integer this simplifies to
$$\Xi(m) \sim m!
- \sum_{q\ge 0} \frac{(-1)^q}{q!} \frac{B_{m+q+1}}{m+q+1}.$$
We may  answer the  original question in  the affirmative, it  is true
that
$$\Xi(z) \sim \Gamma(1+z).$$
This MSE link points to another interesting Mellin transform computation.
Addendum. The convergence properties of the original sum as proposed above are quite interesting, there is an initial segment where it appears to diverge before it then begins to converge. This observation just in case someone decides to study the numerics of this problem.
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\color{#00f}{\large\Xi\pars{z}}& = \sum_{t = 1}^{\infty}{t^{z} \over \expo{t}}
=\sum_{t = 1}^{\infty}{\pars{\expo{-1}}^{t} \over t^{-z}}
= \color{#00f}{\large{\rm L_{i}}_{-z}\pars{1 \over e}}
\end{align}
where $\ds{{\rm L_{i}}_{s}\pars{z}}$ is the
Polylogarithm Function.

According to the 'link' given above, we can get the following relations:
  $$
\lim_{\Re\pars{z} \to -\infty}\Xi\pars{z} = {1 \over \expo{}}\,,\qquad\qquad
\Xi\pars{z} \sim \Gamma\pars{1 + z}\quad\mbox{when}\quad\Re\pars{z} \gg 1
$$

