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

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.

• It would be a great thing to have the same series in the title and in the question body. – Jack D'Aurizio Mar 15 '15 at 16:57
• lol whoops, fixing that. – shinzou Mar 15 '15 at 16:58
• I doubt it since $\frac{(-1)^{n+1}}{n^{1/n}}$ does not converge to zero. – Gregory Grant Mar 15 '15 at 16:59
• Being unable to use the Alternating Series Test tells us nothing. Being unable to use it because the terms do not have limit $0$ tells us the series does not converge. But there is no need to drag in the irrelevant Alternating Series Test. Since the terms do not have limit $0$, the series does not converge. – André Nicolas Mar 15 '15 at 17:10
• @AndréNicolas I wanted to know for sure since it will probably come up more. Thanks. – shinzou Mar 15 '15 at 17:12

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