Asymptotic rate of growth of a sum Consider $$\Phi_0(x) = \sum_{i=0}^{\infty} (1-x)^i,$$ where $x \in (0,1)$. As $x \rightarrow 0$, $\Phi_0(x)$ blows up as $\Theta(1/x)$. Similarly, consider $$ \Phi_1(x) = \sum_{i=0}^{\infty} i (1-x)^i.$$ As $x \rightarrow 0$, a direct calculation shows that $\Phi_1(x)$ blows up as $\Theta(1/x^2)$. What I am interested in is the sum
$$ \Phi_{1/2}(x) = \sum_{i=0}^{\infty} \sqrt{i} (1-x)^i.$$ My question is: how fast does $\Phi_{1/2}(x)$ blow up as $x \rightarrow 0$?
I suspect this is routine and easily solvable with some trick. In the case of $\Phi_0$ and $\Phi_1$, explicit expressions make the asymptotic rate of blowup easily calculable. I don't know a simple expression for $\Phi_{1/2}$.
 A: One trick that I like is to estimate the sum via integrals. The following is a rough informal outline, but you can formalize it via upper and lower bounds.
The basic idea is that roughly $\sum_{n=0}^{\infty} f(n)(1-x)^n \sim \int_0^{\infty} f(n)\exp(-n(-\log(1-x)))dn \sim \int_0^{\infty} f(n)\exp(-nx)dn$ for small $x$. One particularly nice aspect of this method is that you can use it for any nice enough $f(n)$.
In particular, for $f(n) = n^{\alpha},\,\alpha \geq 0$, the integral is $\frac{\Gamma(1+\alpha)}{x^{1+\alpha}}$.
A: The Polylogarithm function is defined as
$$\text{Li}_s(z) = \sum_{k=1}^{\infty} \dfrac{z^k}{k^s}$$
Hence, $$\text{Li}_{-1/2}(z) = \sum_{k=1}^{\infty} \sqrt{k} z^k$$
Hence, the series you have is $\text{Li}_{-1/2}(1-x)$. And you also have that
$$\lim_{x \to 0} x^{3/2} \text{Li}_{-1/2}(1-x) = \dfrac{\sqrt{\pi}}2$$ i.e. as $x \to 0$ $$\text{Li}_{-1/2}(1-x) \sim \dfrac{\sqrt{\pi}}{2 x^{3/2}}$$
In general as $x \to 0$, we have
$$\text{Li}_{s}(1-x) \sim \dfrac{\Gamma(1-s)}{x^{1-s}}$$
