Limit of Root of Positive Real Number

Theorem
Let $$x \in \R: x > 0$$ be a real number.

Let $$\left \langle {x_n} \right \rangle$$ be the sequence in $\R$ defined as $$x_n = x^{1/n}$$.

Then $$x_n \to 1$$ as $$n \to \infty$$.

Proof
Let us define $$1 = a_1 = a_2 = \cdots = a_{n-1}$$ and $$a_n = x$$.

Let $$G_n$$ be the geometric mean of $$a_1, \ldots, a_n$$.

Let $$A_n$$ be the arithmetic mean of $$a_1, \ldots, a_n$$.

From their definitions, $$G_n = x^{1/n}$$ and $$A_n = \frac {n - 1 + x} n = 1 + \frac{x - 1} n$$.

From Arithmetic Mean Never Less than Geometric Mean, $$x^{1/n} \le 1 + \frac{x - 1} n$$.

That is, $$x^{1/n} - 1 \le \frac{x - 1} n$$.

There are two cases to consider: $$x \ge 1$$ and $$0 < x < 1$$.


 * Let $$x \ge 1$$.

From Root of Number Greater than One‎, it follows that $$x^{1/n} \ge 1$$.

Thus $$0 \le x^{1/n} - 1 \le \frac 1 n \left({x - 1}\right)$$.

But from Power of Reciprocal, $$\frac 1 n \to 0$$ as $$n \to \infty$$.

From the Combination Theorem for Sequences it follows that $$\frac 1 n \left({x - 1}\right) \to 0$$ as $$n \to \infty$$.

Thus by the Squeeze Theorem, $$x^{1/n} - 1 \to 0$$ as $$n \to \infty$$.

Hence $$x^{1/n} \to 1$$ as $$n \to \infty$$, again from the Combination Theorem for Sequences.


 * Now let $$0 < x < 1$$.

Then $$x = \frac 1 y$$ where $$y > 1$$.

But from the above, $$y^{1/n} \to 1$$ as $$n \to \infty$$.

Hence by the Combination Theorem for Sequences, $$x^{1/n} = \frac 1 {y^{1/n}} \to \frac 1 1 = 1$$ as $$n \to \infty$$.

Alternative Proof
We consider the case where $$x \ge 1$$; when $$0 < x < 1$$ the proof can be completed as above.

From Root of Number Greater than One‎, we have $$x^{1/n} \ge 1$$.

Hence $$\left \langle {x^{1/n}} \right \rangle$$ is bounded below by $$1$$.

Now consider $$x^{1/n} / x^{1/\left({n+1}\right)}$$.

$$ $$ $$ $$

So $$x^{1/n} > x^{\frac 1 {n+1}}$$ and so $$\left \langle {x^{1/n}} \right \rangle$$ is decreasing.

Hence from the Monotone Convergence Theorem, it follows that $$\left \langle {x^{1/n}} \right \rangle$$ converges to a limit $$l$$ and that $$l \ge 1$$.

Now, since we know that $$\left \langle {x^{1/n}} \right \rangle$$ is convergent, we can apply Limit of a Subsequence.

That is, any subsequence of $$\left \langle {x^{1/n}} \right \rangle$$ must also converge to $$l$$.

So we take the subsequence $$\left \langle {x^{1/{2n}}} \right \rangle$$.

From what we've just shown, $$x^{1/{2n}} \to l$$ as $$n \to \infty$$.

Using the Combination Theorem for Sequences, we have $$x^{1/n} = x^{1/{2n}} \cdot x^{1/{2n}} \to l \cdot l = l^2$$ as $$n \to \infty$$.

But a Sequence has One Limit at Most, so $$l^2 = l$$ and so $$l = 0$$ or $$l = 1$$.

But $$l \ge 1$$ and so $$l = 1$$.

Hence the result.