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$$\sum\limits_{n=1}^\infty\dfrac{(-1)^{n}}{n^{2/n}}$$

I applied alternating series test. I tried to check if it fulfills the condition that it goes to zero. $\frac{2}{n}$ goes to $0$ . However $n^0$ goes to $1$ despite $n$ being infinity. Am I doing something wrong here?

Raymond
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  • You're wrong in the manner of calculus: $$n^{2/n}=e^{\frac2n\log n}$$ but the series still divergent. –  Mar 23 '14 at 07:40
  • okay based on the above and apply l'hopital rule i would get 1, which means alternating series test fails right? But if it's 1, the test of divergence also fails so how do i actually proof it? – Raymond Mar 23 '14 at 07:49
  • Since the sequence $$\left(\dfrac{(-1)^{n}}{n^{2/n}}\right)$$ doesn't converge to $0$ then the series is divergent. –  Mar 23 '14 at 07:55
  • For the limit of the denominator, see here: http://math.stackexchange.com/questions/115822/how-to-show-that-lim-n-to-infty-n-frac1n-1 and http://math.stackexchange.com/questions/28348/proof-that-lim-n-rightarrow-infty-sqrtnn-1/ – Martin Sleziak Mar 23 '14 at 09:48

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Note that $$\lim_{n \to \infty} \frac{1}{n^{2/n}} = \lim_{n \to \infty} e^{\ln n^{-2/n}} = \lim_{n \to \infty} e^{(-2/n)\ln n} = 1$$ because $$\lim_{n \to \infty} \frac{\ln n}{n} = 0.$$ But $$\lim_{n \to \infty} \frac{(-1)^n}{n^{2/n}} = \lim_{n \to \infty} (-1)^n e^{(-2/n)\ln n}$$ does not converge to zero. In fact, it does not converge at all. Therefore $$\sum_{n = 1}^\infty \frac{1}{n^{2/n}}$$ diverges.

glebovg
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