Convergence (or Divergence) of [(-1)^(n-1)*e^(1/n)]/n

Question

so I'm trying to find out whether the following sum converges ot not

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}e^{1/n}}{n}$$

I tried to prove it via the Liebnitz criterion, and got:

$$\lim_{n\to\infty} \frac{(-1)^{n-1}e^{1/n}}{n}=0$$

but now when i have to show that

$$ |a_{n+1}|<|a_n| $$

that is: $$ \frac{e^{1/(n+1)}}{n+1}<\frac{e^{1/n}}{n} $$

but i only get to the point:

$$ \frac{|n|}{|n+1|}<e^{\frac{1}{n(n+1)}} $$ or $$ ln(\frac{n}{n+1})<\frac{1}{n(n+1)} $$

does anyone know how to prove or disprove one of these inequalities, or an alternative way of testing the convergence (or divergence) of this series?

Answer 1

Write it as $\sum_{n \geq 1}(-1)^{n-1}\dfrac{(e^{1/n} - 1)}{n} + \sum_{n \geq 1}\dfrac{(-1)^{n-1}}{n}$.

$\dfrac{e^{1/n}-1}{n}$ is $\mathcal{O}(1/n^2)$ so the first term converges absolutely, and the second term is known to converge.