I encountered recently the following partial sum $\sum_{k=1}^n a^k k^{-1/2}$ with $a$ a constant approximately equal to $2.955$.
I was wondering if there were any clever way to find an asymptotic to this partial sum. I recognized the partial sum of a (divergent) polylogarithm but was not able to find any further literature on this question.
Any help would be welcome !
Fist of all, there is a exact result. For $a>1$ $$S_n=\sum_{k=1}^n \frac {a^k}{\sqrt k}=\text{Li}_{\frac{1}{2}}(a)-a^{n+1} \Phi \left(a,\frac{1}{2},n+1\right)$$ where $\Phi(.)$ is the Hurwitz-Lerch transcendent function.
Using Euler-MacLaurin summation $$S_n=A_0+ a^n \,\frac {A_1}{ n^\frac 12}+ a^n \,\frac {A_2}{ n^\frac 32}+\cdots$$ $$A_0=E_{\frac{1}{2}}(-\log (a)) + a A_3$$ $$A_1=\frac{1}{x}+\frac{1}{2}+\frac{x}{12}-\frac{x^3}{720}+\frac{x^5}{30240}-\frac{x^7}{1209600}+\frac{x^9}{47900160}$$ $$A_2=\frac{1}{2 x^2}-\frac{1}{24}+\frac{x^2}{480}-\frac{x^4}{12096}+\frac{x^6}{345600}-\frac{x^8}{10644480}$$
For this level of expansion $$A_3=\frac{2125613}{3932160}-\frac{482719 x}{5898240}-\frac{6103 x^2}{5160960}+\frac{2699 x^3}{2580480}+\frac{1807 x^4}{17031168}-\frac{7943 x^5}{212889600}+\frac{x^6}{2534400}+\frac{x^7}{3801600}+\frac{x^8}{10644480}-\frac{x^9}{47900160}$$ where $\color{red}{x=\log(a)}$.
For $a=3$ and $n=10^4$, this gives $\color{red}{2.4470864}56\color{red}{\times 10^{4769}}$ while the exact value is $\color{red}{2.447086463}\color{red}{\times 10^{4769}}$
@Svyatoslav's much simpler formula
$$\frac{a^{n+1}}{a-1}\frac1{\sqrt n}\left(1+\frac1{2(a-1)}\frac1n+O\left(\frac1{n^2}\right)\right)$$ would give for the same case $\color{red}{2.4470864}54\color{red}{\times 10^{4769}}$