For $s>0$, let $$\mathrm{Li}_{-s}(x) = \sum_{n=1}^\infty n^s x^n$$ be the polylogarithm function. It is convergent for $|x|<1$. As $x\to 1^-$, the series is divergent. What is the asymptotic behavior of $\mathrm{Li}_{-s}(x)$ as $x\to 1^-$?
Mathematica says that $$\mathrm{Li}_{-s}(x) \approx \frac{\Gamma(1+s)}{(1-x)^{1+s}}.$$ I want a proof of this result.
It is interesting that the Laplace transform of $t^{\alpha-1}$ has a similar form: $$\int_0^\infty t^{\alpha} e^{-\beta t} dt = \frac{\Gamma(1+\alpha)}{\beta^{1+\alpha}}.$$
To expand somewhat on the comment I provided, one has from here that
$$ \text{Li}_{-s}(z) = \Gamma(1+s)\left(\ln\left(\frac{1}{z} \right) \right)^{-s-1} + \sum_{n=0}^{\infty} \zeta(-s-n) \frac{\ln^n(z)}{n!}. $$
Around $z=1$, one then has that $\ln^n(z) \sim (z-1)^n$ and $\ln^{-s-1}(z^{-1}) \sim (1-z)^{-s-1}$ such that
$$ \text{Li}_{-s}(z) \sim \frac{\Gamma(1+s)}{(1-z)^{s+1}} + \sum_{n=0}^{\infty} \zeta(-s-n) \frac{(z-1)^n}{n!}. $$
For $\zeta(-s-n)$, one may use the reflection formula to write
$$ \zeta(-s-n) = \frac{2\Gamma(s+n+1)}{(2\pi)^{s+n+1}}\cos\left(\frac{\pi}{2}(s+n+1)\right)\zeta(s+n+1) $$
Hence,
$$ \text{Li}_{-s}(z) \sim \frac{\Gamma(1+s)}{(1-z)^{s+1}} + \frac{2}{(2\pi)^{s+1}}\sum_{n=0}^{\infty} \frac{(z-1)^n}{n!}\Gamma(s+n+1)\cos\left(\frac{\pi}{2}(s+n+1)\right)\zeta(s+n+1) $$
Each term in the sum decreases rapidly with increasing $n$ and to a first approximation, one arrives at
$$ \text{Li}_{-s}(z) \sim \frac{\Gamma(1+s)}{(1-z)^{s+1}}, \quad z\to 1^{-} $$
as provided by Mathematica. In contrast to the terms inside the sum, this term is divergent.