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?

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. 
