$\sum_{1}^{\infty}\int_{n}^{n+1} e^{-\sqrt{x}} dx,$ converge or diverge? Since
$$D^{-1} e^{-\sqrt{x}} \big|_{x := u^{2}} = D^{-1} e^{-u} Du^{2} = 2D^{-1} e^{-u} u = -2(u+1)e^{-u} + C
= -2(\sqrt{x} + 1)e^{-\sqrt{x}} + C,$$
we have
$$\int_{n}^{n+1} e^{-\sqrt{x}} dx = 2(\sqrt{n} + 1)e^{-\sqrt{n}} - 2(\sqrt{n+1} + 1)e^{-\sqrt{n+1}}.$$
Then I am not sure how to proceed to conclude the convergence or divergence?

$$\sum_{1}^{\infty}\int_{n}^{n+1} e^{-\sqrt{x}} dx = \int_{1}^{\infty} e^{-\sqrt{x}} dx =2\int_{1}^{\infty} e^{-t} t dt = \left[-2(t+1)e^{-t}\right]_1^\infty =\frac{4}{e} $$
Since the integral has a finite value $\frac{4}{e}$ , the sum is convergent.
