$$\frac{1}{\sqrt{k}}＝\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-kq^{2}}dq$$ so
$$\sqrt{x}e^{-x}\sum_{k＝1}^{\infty}\frac{x^{k}}{k!\sqrt{k}}$$
$$＝\frac{2\sqrt{x}e^{-x}}{\sqrt{\pi}}\sum_{k＝1}^{\infty}\frac{x^{k}}{k!}\int_{0}^{\infty}e^{-kq^{2}}dq$$
$$＝\frac{2\sqrt{x}e^{-x}}{\sqrt{\pi}}\int_{0}^{\infty}(e^{xe^{-q^{2}}}-1)dq$$
The question is how to prove $$lim_{x\rightarrow \infty}\frac{2\sqrt{x}e^{-x}}{\sqrt{\pi}}\int_{0}^{\infty}(e^{xe^{-q^{2}}}-1)dq＝1$$
Thanks in advance