Geometric sum with square root What is the result of
$$\sum_{t=0}^\infty \sqrt t a^t ?$$
where $a \in (0,1)?$
My effort: The geometric sum writes
$$\forall a\in (0,1) \qquad \sum_{t=0}^\infty a^t = \frac{1}{1-a}$$
Informally, we can make the derivative of both sides having
$$\forall a\in (0,1) \qquad \sum_{t=0}^\infty (t+1)a^t = \frac{1}{(1-a)^2}$$
which entails $\sum_{t=0}^\infty ta^t = \frac{1}{(1-a)^2}-\frac{1}{1-a}$. Of course, this trick only works for integer powers, not for $\sqrt t$, but I suspect that
$\sum_{t=0}^\infty \sqrt t a^t \approx \frac{1}{(1-a)^{3/2}}$
 A: As requested. We have
$$
f(a)=\sum\limits_{t=0}^\infty t^{1/2}a^t=\operatorname{Li}_{-1/2}(a)
$$
Where $\operatorname{Li}$ is the polylogarithm. First, I plotted the ratio of $f(a)/(1-a)^{-3/2}$, since $(1-a)^{-3/2}$ was the conjectured behavior. As $a \to 1$ the result approached a finite number- great! Then I consulted the asymptotic expansion wiki page and the DLMF. The useful expression is
$$\tag{ 25.12.12}
\operatorname{Li}_{s}(z)=\Gamma(1-s)(-\ln z)^{s-1}+\sum\limits_{n=0}^\infty\zeta(s-n)\frac{(\ln z)^n}{n!}
$$
Where $\Gamma$ is the gamma function, and $\zeta$ is the Riemann zeta function. As $z\to 1$, $\ln z \to 0$, which means that  as long as $s-1<0$, the factor of $(-\ln z)^{s-1}$ will be large (going to inifnity) compared to any $(\ln z)^n$ (going to zero). The leading order behavior is then
$$
\operatorname{Li}_{-1/2}(a)\sim\Gamma(3/2)(-\ln a)^{-3/2} \qquad,\qquad a\to1
$$
Where I've set $z=a$ and $s=-1/2$. Then with $\Gamma(3/2)=\sqrt{\pi}/2$, and
$$
-\ln a\sim 1-a \qquad,\qquad a\to 1
$$
We have directly
$$
f(a)\sim \frac{\sqrt{\pi}}{2}(1-a)^{-3/2} \qquad,\qquad a\to 1
$$
Eq (25.12.12) furnishes the whole asymptotic series for $a\to 1$. For example, the three term series is
$$
f(a)\sim \frac{\sqrt{\pi}}{2}(1-a)^{-3/2} +\zeta(-1/2)+\zeta(-3/2)(a-1)\qquad,\qquad a\to 1
$$
Note that all terms after the second vanish when $a= 1$.
Aside: we may take the fractional derivative
$$
D^{1/2}g(x)=\frac{1}{\Gamma(1/2)}\frac{d}{dx}\int_0^x dx' \frac{g(x')}{\sqrt{x-x'}}
$$
of $g(x)=\sum\limits_{t=0}^\infty x^t=(1-x)^{-1}$. Formally, this yields
$$\tag{*}
\sqrt{\pi x} D^{1/2}g(x)=\frac{1}{(1-x)}+\frac{\sqrt{x}\sin^{-1}(\sqrt{x})}{(1-x)^{3/2}}
$$
Which is not $\operatorname{Li}_{-1/2}$.The reason is that the fractional half derivative (in the sense above) doesn't pull out the $t^{1/2}$, in fact we have
$$
D^{1/2}x^t=\left[\frac{\Gamma(t+1)}{\Gamma(t+1/2)}\right]x^{t-1/2}\neq t^{1/2}x^{t-1/2}
$$
Consequently, eq (*) has not computed the sum $\sum\limits_t t^{1/2}x^t$, but another, similar sum. Interestingly, eq (*) does provide the same leading order behavior as $x\to 1$ as $\operatorname{Li}_{-1/2}$, which is; I suppose, because  $\frac{\Gamma(t+1)}{\Gamma(t+1/2)} \sim \sqrt{t}$ as $t\to \infty$, in fact the difference for all $t$ is very small.
