Use any test to determine convergence of the series $sum_{n=1}^infty frac{(-1)^ne^{frac{1}{n}}}{n^3}$

I am asked to find whether the series $\sum_{n=1}^\infty \frac{(-1)^ne^{\frac{1}{n}}}{n^3}$ is absolutely convergent, conditionally convergent or divergent. So far i have determined that the series is convergent for positive values but i am stuck on determining the absolute convergence.

I began by using the alternating series test:

let $b_n = \frac{e^{\frac{1}{n}}}{n^3}$, it can be seen that $b_n = \frac{e^{\frac{1}{n}}}{n^3} \geq 0$ and that $\frac{e^{\frac{1}{n+1}}}{(n+1)^3} \lt \frac{e^{\frac{1}{n}}}{n^3} \implies$ $b_n$ is decreasing for $n \gt1$. With both conditions for the alternating series test met I can take the limit as $n$ approaches infinity. $\lim_{n \to \infty} \frac{e^{\frac{1}{n}}}{n^3} = 0$ and so the series is convergent for positive terms of the series.

I now want to check if the series is absolutely convergent - if for negative terms and positive terms the series diverges then i will know that the series is conditionally convergent otherwise it will be absolutely convergent.

I don't think using a root test or a ratio test will be particularly helpful here so instead I will take the absolute value of the series:

$$\sum_{n=1}^\infty|\frac{(-1)^ne^{\frac{1}{n}}} {n^3}|=\sum_{n=1}^\infty\frac{e^{\frac{1}{n}}}{n^3}$$

$\sum_{n=1}^\infty\frac{1}{n^3}$ is something i know - it's a convergent p-series, however the above series is bigger than this one and so it doesn't really help to do a direct comparison test. At this point I am stuck.

Answer

Comparison test goes through: $e^{1/n}\leq 3$ so $\left|(-1)^{n}\dfrac{e^{1/n}}{n^{3}}\right|\leq\dfrac{3}{n^{3}}$.