Convergence of $\sum_{n=1}^{\infty} \int_n^{n+1} e^{- \sqrt x} dx$

Question

Test the convergence of $$\sum_{n=1}^{\infty} \int_n^{n+1} e^{- \sqrt x} dx$$

Attempt: For sufficiently large $x$, we have $e^{-\sqrt x} > e ^{- x}$.

I also tried solving the integral by By Parts rule, which resulted in a more hodgepodge.

Unfortunately, these methods did not prove to be of much help.

Please guide me on how to go about this problem.

Thank you very much for your help.

Answer 1

The $N$th partial sum of the series is $$\sum_{n = 1}^N \int_n^{n + 1} e^{-\sqrt{x}} \,dx = \int_1^{N + 1} e^{-\sqrt{x}} \,dx.$$ But in the limit $N \to \infty$ the right-hand side converges, because $\int_1^{\infty} e^{-\sqrt{x}} \,dx$ does: One can also evaluate it directly using the substitution $x = u^2$ followed by i.b.p.

(We don't need both directions here, but the convergence of the sum and integral are equivalent because the integrand is positive.)

Answer 2

One more way is to notice that for every $n \int_{n}^{n+1} e^{-\sqrt{x}}dx \leq e^{-\sqrt{n}}$ because the integrand is a monotone decreasing function. So the sum is upperbounded by $\sum_{n=1}^{\infty} e^{-\sqrt{n}}$ which again converges by integral test: $\int_{1}^{\infty} e^{-\sqrt{x}}dx = 2 \int_{1}^{\infty}t e^{-t}dt$