Answer
Convergent
Work Step by Step
Let $f(x)=\frac{e^{1/x}}{x^2}$. This function is continuous and decreasing on $[1,\infty)$.
Evaluate $\int_1^\infty f(x)dx$:
$\int_1^\infty \frac{e^{1/x}}{x^2}dx=\int_1^\infty -e^{1/x}\cdot -\frac{1}{x^2}dx$ (Let $u=\frac{1}{x}\to du=-\frac{1}{x^2}dx$)
$=\int_1^0 -e^udu$
$=\int_0^1e^udu$
$=e^u]_0^1$
$=e^1-e^0$
$=e-1$
So, the integral is convergent.
Using the Integral Test for Series, $\sum_{n=1}^\infty \frac{e^{1/n}}{n^2}$ is convergent.
