Then show that $ \ \sqrt[n]{n^3} \to 1 \ \text{as} \ n \to \infty \ $

Question

We know that the sequence $ \ \sqrt[n]{n} \to 1 \ \text{as} \ n \to \infty \ $.

Then show that $ \ \sqrt[n]{n^3} \to 1 \ \text{as} \ n \to \infty \ $

Answer:

$ \sqrt[n]{n^3}=(n^{\frac{1}{n}})^3=(\sqrt[n]{n})^3 \to 1 \ \text{as } \ n \to \infty \ $

Is not it quite correct?

Answer 1

Let the limit be $L$, then $$\ln L = \frac{3\ln n }{n} \rightarrow 0$$ So $L \rightarrow e^0 = 1$

Answer 2

It's essentially correct. Maybe you still have to prove, that if a sequence $(a_n)$ converges to $1$, then the sequence $(a_n^3)$ also converges to one.

Answer 3

If $a_n \to 1$ then, for any $0 < c <1$, $a_n \ge 1-c$ for all large enough $n$.

Since $x^k-1 =(x-1)\sum_{j=0}^{k-1}x^j $, $a_n-1 =(a_n^{1/k})^k-1 =(a_n^{1/k}-1)\sum_{j=0}^{k-1}a_n^{j/k} $ so

$\begin{array}\\ |a_n-1| &=|a_n^{1/k}-1|\,|\sum_{j=0}^{k-1}a_n^{j/k}|\\ &\ge|a_n^{1/k}-1|\,|\sum_{j=0}^{k-1}(1-c)^{j/k}|\\ &\ge|a_n^{1/k}-1|k(1-c) \quad\text{since } x^r \ge x \text{ for } 0 < x , r < 1\\ \text{so}\\ |a_n^{1/k}-1| &\le \dfrac{|a_n-1|}{k(1-c)} \end{array} $

Therefore $|a_n-1|\to 0$ implies that $|a_n^{1/k}-1|\to 0$.

Answer 4

Notice that $$\lim_{n \to \infty} n^{\frac{1}{n}}=1.$$ Hence $$\lim_{n \to \infty} \sqrt[n]{n^3}=\lim_{n \to \infty} n^{\frac{3}{n}}=\lim_{n \to \infty} \left(n^{\frac{1}{n}}\right)^3=\left(\lim_{n \to \infty} n^{\frac{1}{n}}\right)^3=1^3=1.$$