# Asymptotic approximations for $\Phi(z,-n,0)_\nu=\sum_{k=0}^{\nu-1}k^nz^k$ as $\nu\to\infty$

I am interested in asymptotic approximations of $$\Phi(z,-n,0)_\nu:=\sum_{k=0}^{\nu-1}k^nz^k=\Phi(z,-n,0)-z^\nu\Phi(z,-n,\nu),\quad n\in\Bbb N$$ for large $$\nu$$ where $$\Phi(z,s,a)$$ is the Lerch Transcendent. I tried using the Euler–Maclaurin formula which yields $$\Phi(z,-n,0)_\nu\sim\int_0^{\nu-1}x^nz^x\,\mathrm dx+\frac{1}{2}(\nu-1)^nz^{\nu-1}.$$ The integral is evaluated by substituting $$y=-x\log z$$ to get $$\Phi(z,-n,0)_\nu\sim(-1)^{n+1}\frac{\gamma(n+1,(1-\nu)\log z)}{\log^{n+1}z}+\frac{1}{2}(\nu-1)^nz^{\nu-1},$$ with $$\gamma(s,z)$$ being the lower incomplete gamma function. To deal with the removeable singularity at $$z=1$$ I further used the confluent hypergeometric function $${_1F_1}(a;b;z)$$ to write $$\Phi(z,-n,0)_\nu\sim(\nu-1)^nz^{\nu-1}\left(\frac{\nu-1}{n+1}{_1F_1}(1;n+2;(1-\nu)\log z)+\frac 12\right).$$ This asymptotic approximation works quite well. Plots for various $$n$$ on $$z\in(0,2)$$ show that the two functions become nearly indistinguishable for $$\nu>2$$. My problem is that this asymptotic approximation becomes difficult to evaluate for large $$\nu$$ which seems to defeat the point of this exercise.

What can I do to this asymptotic expression to make it easier to compute for large $$\nu$$? Alternatively, is there a better approach (i.e. not the EM-formula) for deriving asymptotic expressions that will yield more computationally friendly results?

If $$|z| < 1$$, the infinite series $$\sum_{k=0}^\infty k^n z^k$$ converges to $$\text{polylog}(-n,z)$$.
If $$|z| > 1$$, we can write $$\Phi(z,-n,0)_\nu = (\nu-1)^n z^{\nu} \sum_{j=1}^{\nu-1} \frac{(\nu-j)^n}{(\nu-1)^n} z^{-j}$$ where using Dominated Convergence, $$\sum_{j=1}^{\nu-1} \frac{(\nu-j)^n}{(\nu-1)^n} z^{-j} \to \sum_{j=1}^\infty z^{-j} = \frac{z^{-1}}{1-z^{-1}} = \frac{1}{z-1}$$