Find the limiting value of $S=a^{\sqrt{1}}+a^{\sqrt{2}}+a^{\sqrt{3}}+a^{\sqrt{4}}+...$ for $0 \leq a < 1$

We can keep going on the line of the comments for something that converges as close as necessary, starting from $$ S = \frac{2}{(\log a)^2} - \frac12 + \frac1\pi \int_0^\infty \frac{e^{-\beta\sqrt{t}}\sin(\beta\sqrt{t})}{e^t-1}dt \quad\text{ where }\quad \beta = \frac{-\log a}{2\sqrt{\pi}} $$ as given by @achille hui in the comments. We can try and get a form for the integral, $$ I(\beta) = \int_0^\infty \frac{e^{-\beta\sqrt{t}}\sin(\beta\sqrt{t})}{e^t-1}dt $$ by taking a Mellin transform of the integrand with respect to $\beta$ we get $$ \int_0^\infty \beta^{s-1}I(\beta)\; d\beta = \Gamma(s)\sin\left(\frac{\pi s}{4}\right)\int_0^\infty \frac{(2t)^{-s/2}}{e^t-1} \; dt $$ this is then $$ \int_0^\infty \beta^{s-1}I(\beta)\; d\beta = 2^{-s/2}\Gamma(s)\sin\left(\frac{\pi s}{4}\right)\Gamma\left(1-\frac{s}{2}\right)\text{Li}_{1-\frac{s}{2}}(1) $$ we want to take the inverse Mellin transform of both sides to get $I(\beta)$ but it's not trivial to do this for the RHS, so invoke the Ramanujan Master Theorem, that the Mellin transform of $I(\beta)$ is written $$ \int_0^\infty \beta^{s-1}I(\beta)\; d\beta = \Gamma(s)\phi(-s) $$ such that $$ I(\beta)=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\phi(n)\beta^n $$ then we have $$ \phi(s)= \sin\left(\frac{-\pi s}{4}\right)2^{s/2}\Gamma\left(1+\frac{s}{2}\right)\text{Li}_{1+\frac{s}{2}}(1) $$ and using $$ \text{Li}_{1+\frac{s}{2}}(1) = \zeta\left(1+\frac{s}{2}\right), \;\; s>0 $$ we can write $$ I(\beta)=\sum_{n=0}^\infty \frac{2^{n/2}}{n!}\sin\left(\frac{-\pi n}{4}\right)\Gamma\left(1+\frac{n}{2}\right)\zeta\left(1+\frac{n}{2}\right)\left(\frac{\log{a}}{2\sqrt{\pi}}\right)^n $$ finally $$ \sum_{n=1}^\infty a^{\sqrt{n}}=\frac{2}{(\log a)^2} - \frac12 -\frac{\log(a)\zeta\left(\frac{3}{2}\right)}{4\pi} -\frac{\log(a)^2\zeta\left(2\right)}{4\pi^2}-\frac{\log(a)^3\zeta\left(\frac{5}{2}\right)}{32\pi^2}+\frac{\log(a)^5\zeta\left(\frac{7}{2}\right)}{512\pi^3}+\cdots $$