Integral test for convergence: $\sum _1^\infty \frac{e^{1/n}}{n^2}$ Integral test for convergence: $$\sum _1^\infty \frac{e^{1/n}}{n^2}$$
I tried approaching this as an IBP but I haven't been able to sort the solution. Can this be made into a improper integral? and if so could someone show me the process?

$$
\int_1^\infty e^{1/x} \left(\frac{dx}{x^2}\right) = \int_1^0 e^u \, (-du)
$$
Or by comparison:
$$
\frac{e^{1/n}}{n^2} \le \frac{3}{n^2}
$$
and $\sum_n (3/n^2)$ converges.  But of course that is also usually shown via an integral test.
