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, $$\sum_{j=1}^{\nu-1} \frac{(\nu-j)^n}{(\nu-1)^n} z^{-j} \to \sum_{j=1}^\infty z^{-j} = \frac{z^{-1}}{1-z^{-1}} = \frac{1}{z-1}$$