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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.

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  • $\begingroup$ It would be a great thing to have the same series in the title and in the question body. $\endgroup$ – Jack D'Aurizio Mar 15 '15 at 16:57
  • $\begingroup$ lol whoops, fixing that. $\endgroup$ – shinzou Mar 15 '15 at 16:58
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    $\begingroup$ I doubt it since $\frac{(-1)^{n+1}}{n^{1/n}}$ does not converge to zero. $\endgroup$ – Gregory Grant Mar 15 '15 at 16:59
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Mar 15 '15 at 17:10
  • $\begingroup$ @AndréNicolas I wanted to know for sure since it will probably come up more. Thanks. $\endgroup$ – shinzou Mar 15 '15 at 17:12
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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.

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  • $\begingroup$ Just to make sure, in this case the limit of term in the sum doesn't exist since it's either 1 or -1 right? $\endgroup$ – shinzou Mar 15 '15 at 18:48
  • $\begingroup$ You right, it doesn't exist, therefore it's not 0. $\endgroup$ – Surb Mar 15 '15 at 18:51
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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.

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

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