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I´m having trouble analyzing the convergence of another series:

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

If you could just give me a hint about which test for convergence should I use, it would be great.

Thanks so much!

Lessa121
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3 Answers3

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Hint: We have $n\lt 2^n$, so $n^{1/n}\lt (2^n)^{1/n}=2$.

André Nicolas
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  • So I use that as a lower bound to prove it does not converge to zero? I´ve tried to use L'Hôpital's rule instead to prove that $\frac 1 n log(n) $ converges to zero... Therefore $$n^{\frac 1 n}=e^{\frac 1 n log(n)}$$ converges to 1, as the function $e^x$ is continuos. Is it okay to do that? – Lessa121 Feb 13 '14 at 18:56
  • Yes, the hint tells you quickly that the $n$-th term has absolute value $\gt \frac{1}{2}$. A necessary (but not sufficient) condition for convergence is that the terms go to $0$. Since the function $e^x$ is continuous, you can indeed use L'Hospital's Rule to show that $\lim_{x\to\infty} x^{1/x}=1$ and therefore $\lim_{n\to\infty} n^{1/n}=1$. But the hint I gave uses much less machinery. – André Nicolas Feb 13 '14 at 19:02
  • You are welcome. – André Nicolas Feb 13 '14 at 20:50
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An essential limit to know is that:

$$\sqrt[n]{n}\to 1$$

Which is a consequence of the following also essential limit:

$$(\forall a>1)\quad \frac{a^n}{n}\to0$$

The latter can be proven by looking at ratios of successive terms, you'll find that it's almost like a geometric sequence, for large enough $n$. In particular, then, we have eventually $n<a^n$, and so $\sqrt[n]{n}<a$, and this is for all $a>1$. Since $\sqrt[n]{n}$ is clearly eventually greater than $1$, we have $\sqrt[n]{n}\to1$. Incidentally, by shuffling exponents around and taking the limit of a product, this means that $\sqrt[n]{n^k}\to1$, too, and without too much more work, even that $\sqrt[n]{P(n)}\to1$ for any polynomial $P$.

Anyway, this almost immediately implies that the general term of your sequence converges to $1$ (in absolute value), and thus that the series does not converge.

Jack M
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The series diverges because $$ \lim_n\frac{1}{n^{1/n}}=\lim_n\exp(-\frac{\ln n}{n})=1 \ne 0. $$

HorizonsMaths
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