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I am looking for an example of a sequences such that $x_n $ converges to zero but $(x_n)^\frac{1}{n}$ diverges. Also please explain how to tackle such questions.

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  • $\begingroup$ If $x_n\to x$ and $x>0$ then $x_n^{1/n}\to 1$. For this, you can use the sandwich lemma. $\endgroup$ – amsmath Oct 4 '19 at 14:28
  • $\begingroup$ Others have shown various useful aspects of this. You could also note that $x_n$ can be negative when $n$ is odd and positive when $n$ is even. $\endgroup$ – Mark Bennet Oct 4 '19 at 14:42
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If $n$ is odd, let $x_n=2^{-n}$, and if $n$ is even let $x_n=3^{-n}$. Then the sequence converges to $0$ but $x_n^{\frac1n} $ alternates between two distinct values, hence diverges.

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  • $\begingroup$ Wow, Your answer kind of opened a new way of how I approach questions. Thank you very much. $\endgroup$ – Ankush Kothiyal Oct 4 '19 at 14:31
  • $\begingroup$ @Ank No problem. $\endgroup$ – Matt Samuel Oct 4 '19 at 14:35
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I would tackle the problem by flipping it on its back: Let $y_n := (x_n)^{1/n}$, then we're looking for a non-converging sequence $(y_n)$ such that $y_n^n \rightarrow 0$ (i.e. $0 < y_n < 1$). One solution would be $y_n = 1/2$ for $n$ even and $y_n = 1/4$ for $n$ odd.

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Note that if $x_n$ converges to $L > 0$, then it is NOT possible for $x_n$ to converge and $(x_n)^{\frac{1}{n}}$ to diverge. This is shown as follows:

Assume $x_n$ converges to $L > 0$. Then, by definition of convergence, we have

$\lim_{n \rightarrow \infty}x_n = L$ where $L \in \mathbb{R}$ s.t. $L>0$.

Now, lets find out if $(x_n)^{\frac{1}{n}}$ converges or diverges

$$\lim_{n \rightarrow \infty}(x_n)^{\frac{1}{n}}$$

$$\lim_{n \rightarrow \infty}e^{\ln (x_n)^{\frac{1}{n}}}$$

$$\lim_{n \rightarrow \infty}e^{\frac{1}{n} \ln (x_n)}$$

Now we take the limit of a composed function, and since the exponentiation function is continuous on $\mathbb{R}$, we have

$$\lim_{n \rightarrow \infty}e^{\frac{1}{n} \ln (x_n)} = e^{\lim_{n \rightarrow \infty} \frac{1}{n} \ln (x_n)}$$

Lets focus on the exponent

$$\lim_{n \rightarrow \infty} \frac{1}{n} \ln (x_n) = \lim_{n \rightarrow \infty} \frac{\ln (x_n)}{n} = \frac{\lim_{n \rightarrow \infty} \ln (x_n)}{\lim_{n \rightarrow \infty} n} = \frac{\ln \lim_{n \rightarrow \infty}(x_n)}{\lim_{n \rightarrow \infty} n} = \frac{\ln L}{\infty} = 0$$

Now lets take into the account the base of the exponential function...

$$e^{\lim_{n \rightarrow \infty} \frac{1}{n} \ln (x_n)} = e^0=1$$

So, in summary, we have $$\lim_{n \rightarrow \infty}(x_n)^{\frac{1}{n}}=1$$

and by definition of convergence, $(x_n)^{\frac{1}{n}}$ converges.

Thus, if $x_n$ converges to $L > 0$, then $(x_n)^{\frac{1}{n}}$ converges.

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