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!
$\endgroup$ 72 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$Tell me I'm not blind, please.
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