$\sum_{1}^{\infty}\int_{n}^{n+1} e^{-\sqrt{x}} dx,$ 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?

The function $x\mapsto e^{-\sqrt x}$ is decreasing then
$$0\le\int_n^{n+1}e^{-\sqrt x}dx\le e^{-\sqrt n}=:a_n$$
and the series $\sum a_n$ is convergent (we can prove it by several ways e.g. the ratio test or using $a_n=_\infty o\left(\frac1{n^2}\right)$, ...) so the given series is convergent by comparison.
