Convergence of $\int_{1}^{\infty}\frac{a^x}{\sqrt{x}}dx$ for $a>0$

Question

I have to study the following improper integral: $$\int_{1}^{\infty}\frac{a^x}{\sqrt{x}}dx,\,\, a>0$$ I have thought that $a^x$ converges to 0 when $x\to \infty$ if $|a|<1$ so $\forall \epsilon\,\, \exists \,M>0: \forall x>M: |a^x|<\epsilon$. So: $$\frac{a^x}{\sqrt{x}}<\frac{\epsilon}{\sqrt{x}}:=g(x)$$ Now the integral of g(x) diverges but this does not help me to conclude the divergence of the initial integral.
How can I proceed?