# Integral Convergence test for Sum

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!

• The sum converges, hint : $e^{-\sqrt{n}}$ converges to $0$ faster than $n^k$ for all $k\geqslant 1$. May 1, 2022 at 16:44
• You likely entered the expression in incorrectly. (Your argument is correct.) May 1, 2022 at 16:45
• I dont know what online calculator you have used, but for example WolframAlpha says that it converges. Also you can try another convergence test to prove it May 1, 2022 at 16:47
• What online calculator did you use? May 1, 2022 at 16:54
• I've used SymboLab's online calculator for sum convergence. Probably a bug on their side, other calculators came to the conclusion that the sum is indeed convergent. May 1, 2022 at 16:57

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. Tell me I'm not blind, please.