Be $$\sum\_{n=1}^\infty \int\_{n}^{n+1} \frac{1}{e^\sqrt x } dx$$ This is convergent or not?

So applying inequalities we have:
$$n\le x\le n+1$$
$$\sqrt n\le \sqrt x\le \sqrt {n+1}$$
$$e^\sqrt n\le e^\sqrt x\le e^\sqrt {n+1}$$
$$\frac{1}{e^\sqrt n}\ge \frac{1}{e^\sqrt x} \ge \frac{1}{e^\sqrt {n+1}}$$
Rearranging
$$\frac{1}{e^\sqrt {n+1}}\le \frac{1}{e^\sqrt x} \le \frac{1}{e^\sqrt n} $$
Applying integrals
$$\int\_{n}^{n+1}\frac{1}{e^\sqrt {n+1}}\le \int\_{n}^{n+1} \frac{1}{e^\sqrt x} \le \int\_{n}^{n+1}\frac{1}{e^\sqrt n} $$
but I get stuck here, I know that I have to apply $\sum\_{n=1}^\infty$ to both sides of the inequality and conclude by a comparison test but I do not how to proceed. So any help will be appreciated.