Integral Convergence test for Sum

Question

I'm trying to implement the integral test for sum convergence on the following series: $$ \sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n}\cdot e^{\sqrt{n}}} $$

I've proved that the function $f:[1,\infty]\rightarrow \mathbb{R} | f(x)=\dfrac{1}{\sqrt{x}\cdot e^{\sqrt{x}}}$ is monotonically descending and positive for all $x\ge 1$,

and also that the improper integral $\int_{1}^{\infty}\dfrac{1}{\sqrt{x}\cdot e^{\sqrt{x}}}dx$ converges, therefor the sum stated above is convergent.

However, according to an online calculator, the sum $\sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n}\cdot e^{\sqrt{n}}}$ is divergent, which makes no sense to me.

(said calculator only provides further information for money, as a student, I cannot afford such luxuries.)

Currently I am uncertain in my solution.

Any assistance/guidance is happily accepted!

Answer 1

For any $ X>1$,

$$F(X)=\int_1^X\frac{dx}{2\sqrt{x}e^{\sqrt{x}}}$$ $$=\Big[-e^{-\sqrt{x}}\Big]_1^X$$

$$=e^{-1}-e^{-\sqrt{X}}$$

$$\lim_{X\to\infty}F(X)=e^{-1}$$

the integrale is surely convergent.

Answer 2

Tell me I'm not blind, please.