I would like to control the growth of the series $$ \sum_{n=1}^{\infty} \frac{r^n}{\sqrt{n}} $$ as $r\nearrow 1$; i.e. a lower bound of the form $\sum_{n=1}^{\infty} \frac{r^n}{\sqrt{n}} \geq C f(r)$ for some constant $C$. Simulations suggest the function $f$ might have the form $f(r)=\frac{1}{\sqrt{1-r}}$ or $ f(r) = \sqrt{\frac{1}{\log(2-r)}}$, but it could be something else.
My attempt: The easiest approximation is the integral $\int_1^{\infty}\frac{r^x}{\sqrt{x}}dx$ which yields $$\mathrm{Undefined}-\sqrt{\frac{\pi }{\ln \left(r\right)}}\text{erfi}\left(1\cdot \sqrt{\ln \left(r\right)}\right),$$
which seems to go in favour of my second suggestion of $f(r)$, but I'm not sure how to proceed or obtain this rigorously. Maybe there's a Taylor expansion I'm missing. Any help would be appreciated.
$$S(r)=\sum_{n=1}^\infty \frac{r^n}{\sqrt{n}}=\frac1{\sqrt\pi}\sum_{n=1}^\infty r^n\int_0^\infty t^{-\frac12}e^{-nt}dt$$ Changing the order of summation and integration and denoting $\epsilon=1-r$ $$=\frac r{\sqrt\pi}\int_0^\infty\frac1{e^t-r}\frac{dt}{\sqrt t}=\frac r{\sqrt\pi}\int_0^\infty\left(\frac1{e^t-1+\epsilon}-\frac1{t+\epsilon}+\frac1{t+\epsilon}\right)\frac{dt}{\sqrt t}$$ $$=\frac r{\sqrt\pi}\int_0^\infty\frac1{t+\epsilon}\frac{dt}{\sqrt t}-\frac r{\sqrt\pi}\int_0^\infty\frac{e^t-1-t}{(e^t-1+\epsilon)(t+\epsilon)}\frac{dt}{\sqrt t}$$ The second integral converges at $\epsilon=0$; therefore, $$S=\sqrt{\frac\pi\epsilon}\,r-\frac r{\sqrt\pi}\int_0^\infty t^{-\frac32}\frac{e^t-1-t}{e^t-1}dt+o(1)$$ $$S(r)=\sqrt{\frac{\pi}{1-r}}-\frac1{\sqrt\pi}\int_0^\infty t^{-\frac32}\frac{e^t-1-t}{e^t-1}dt+o(1)\tag{1}$$ Integrating the second integral by parts $$\int_0^\infty t^{-\frac32}\frac{e^t-1-t}{e^t-1}dt=-2t^{-\frac12}\left(1-\frac t{e^t-1}\right)\bigg|_{t=0}^\infty+2\int_0^\infty t^{-\frac12}\left(\frac{e^tt}{(e^t-1)^2}-\frac1{e^t-1}\right)dt$$ Considering analytical continuation (taking $\lambda>0$) and integrating by parts $$I(\lambda)=2\int_0^\infty t^\lambda\left(\frac{e^tt}{(e^t-1)^2}-\frac1{e^t-1}\right)dt$$ $$=-2\Gamma(1+\lambda)\zeta(1+\lambda)+2\int_0^\infty t^{\lambda+1}\frac{e^t}{(e^t-1)^2}dt$$ $$=-2\Gamma(1+\lambda)\zeta(1+\lambda)+2(1+\lambda)\int_0^\infty\frac{t^\lambda}{e^t-1}dt$$ $$I(\lambda)=2\lambda\Gamma(1+\lambda)\zeta(1+\lambda)$$ Choosing $\lambda=-\frac12$ and putting into (1) $$\boxed{\,\,S(r)=\sqrt{\frac\pi{1-r}}+\zeta\Big(\frac12\Big)+o(1)\,\,}$$ Numeric check (WA) confirms the answer.