Convergence of $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{1/n}}$

Question

Does the series $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{1/n}}$ converge? converge in absolute value or conditionally?

It's easy to see that in absolute value the general term tends to $1$ so the series diverges in absolute value.

The term of the series doesn't tend to zero either so it's impossible to use the alternating series test. Can I conclude from the inability to use the alternating test that the series diverges?

Note: No integrals or Taylor's.

Answer 1

If the general terme doesn't converge to zero it can't converge. Indeed, if $S_n=\sum_{k=0}^n x_n$ converge, then $(S_{n})$ is a Cauchy Sequence and thus $$|S_{n+1}-S_n|=|x_{n+1}|<\varepsilon$$ if $n\geq N$ for a certain $N\in\mathbb N$ and all $\varepsilon>0$. Therefore $\lim_{n\to\infty }x_n=0$. Then, if $\lim_{n\to\infty }x_n\neq 0$, the series doesn't converge.

Answer 2

The terms must converge to zero for the sum to converge. That's true whether it is alternating or not. Since you apparently know already that the terms do not converge to zero, you are done.

Answer 3

$$\lim_{n\to\infty}n^{1/n}=1\not=0$$ therefore $$\lim_{n\to\infty}\dfrac{(-1)^n}{n^{1/n}} \text{does not exist}.$$