# Example of sequence such that $x_n$ converges to zero but $(x_n)^\frac{1}{n}$ diverges?

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.

• If $x_n\to x$ and $x>0$ then $x_n^{1/n}\to 1$. For this, you can use the sandwich lemma. Oct 4, 2019 at 14:28
• 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. Oct 4, 2019 at 14:42

## 3 Answers

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.

• Wow, Your answer kind of opened a new way of how I approach questions. Thank you very much. Oct 4, 2019 at 14:31
• @Ank No problem. Oct 4, 2019 at 14:35

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.

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.