Absolute/Conditional Convergence or Divergence of $\sum\limits_{n=1}^{\infty}\frac{(-1)^{n}e^{1/n}}{n^3}$

Question

I'd just like to verify that my solution is right. This is for the series $$\sum\limits_{n=1}^{\infty}\frac{(-1)^{n}e^{1/n}}{n^3}\text{.}$$ Note $$\left|\dfrac{\left[\dfrac{(-1)^{n+1}e^{1/(n+1)}}{(n+1)^3}\right]}{\left[\dfrac{(-1)^{n}e^{1/n}}{n^3}\right]}\right| = \left|-e^{1/(n+1)}e^{-1/n}\left(\dfrac{n}{n+1}\right)^{3}\right| \to 1 \cdot 0 \cdot 1^{3} = 0$$ as $n \to \infty$. Hence by the ratio test, the series is absolutely convergent.

Answer 1

The solution is not correct. When one applies the Ratio Test, the limit is $1$. so the Ratio Test is inconclusive.

For a correct proof, note that the absolute value of the $n$-th term is $\le \frac{e}{n^3}$, so we have absolute convergence by Comparison.

Answer 2

A quick way to see that this converges which is applicable in a wide variety of cases is to consider how quickly the summand shrinks; in particular, using Big-O notation, we see that $e^{1/n}$ is $O(1)$, since it converges towards a limit as $n$ grows large. The denominator is $O(n^3)$, so the ratio, which is the summand, is clearly $O(n^{-3})$, so convergence of this series is the same as the convergence of $\sum_{n=1}^{\infty}n^{-3}$, which is convergent.