Limit of Root of Positive Real Number/Proof 1

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 $a_1 = a_2 = \cdots = a_{n-1} = 1$ 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 $A_n = \dfrac {n - 1 + x} n = 1 + \dfrac{x - 1} n$.

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

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

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

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

From the Combination Theorem for Sequences it follows that $\dfrac 1 n \left({x - 1}\right) \to 0$ 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 $x = \dfrac 1 y$ where $y > 1$.

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

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