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

Question

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?

Answer 1

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

Answer 2

Hint: $\quad\displaystyle\sum_1^\infty\int_n^{n+1}e^{-\sqrt x}~dx~=~\int_1^\infty e^{-\sqrt x}~dx~\le~\int_0^\infty e^{-\sqrt x}~dx~=~2,~$ since, in general,

for $\color{blue}{n>0}$ we have $~\displaystyle\int_0^\infty e^{-\sqrt[n]x}~dx~=~n!~$ See $\Gamma$ function for more information.

Answer 3

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.