Question
How can I find the sum of the series $$S=a^{\sqrt{1}}+a^{\sqrt{2} }+a^{\sqrt{3}}+a^{\sqrt{4}}+...$$ under the condition $0 \leq a < 1$
Short version
I think the sum of the series should be about $$S=\frac{2} {(\ln a)^2}+c$$ where c is a correction term. So the question is how should I find the value of $c$?
Long version
My approach so far:
The sum can be restated in terms of $y=f(x)$ where $$y_1=\sum_{n=1}^{\left \lfloor{x}\right \rfloor}a^{\sqrt{n}}$$ so the problem now is to find the limiting value of $y_1$ as $x \rightarrow \infty$. Instead of finding the limit of $y_1(x)$ as defined now, I tried to find another function with a similar growth rate: $$\frac{\mathrm{d} y_2}{\mathrm{d} x}=a^{\sqrt{x}}$$ $${\textrm{i.e. }} y_2=\int a^{\sqrt{x}}{\mathrm{d} x} $$ which on solving yields $$y_2=\frac{2 \sqrt{x} a^{\sqrt{x}}} {\ln a} - \frac{2a^{\sqrt{x}}} {(\ln a)^2}+\frac{2} {(\ln a)^2}$$ where the constant of integration is so chosen to make the curve pass through the origin.
Now plotting both these functions on Desmos
shows that $y_1$ and $y_2$ resemble each other closely. However $y_2$ has faster initial growth, and therefore it's limiting value is slightly larger than the limiting value of $y_1$ which is what I'm after. The limiting value of $y_2$ is
$$S_2=\lim_{x \rightarrow \infty}{y_2(x)} =\frac{2} {(\ln a)^2}$$
Therefore, I think the actual sum of the original series should be:
$$S=\frac{2} {(\ln a)^2}+c$$
where $c$ is a correction term, possibly depending on $a$.
I'm stuck on how to find the correction term $c$ and what it's form should be. Is my approach so far correct? Any help would be appreciated.