# 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 1

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:

$A_n = \dfrac {n - 1 + x} n = 1 + \dfrac{x - 1} n$
$x^{1/n} \le 1 + \dfrac{x - 1} n$

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:

$0 \le x^{1/n} - 1 \le \dfrac 1 n \paren {x - 1}$
$\dfrac 1 n \to 0$ as $n \to \infty$

From the Combination Theorem for Sequences:

$\dfrac 1 n \paren {x - 1} \to 0$ as $n \to \infty$

Thus by the Squeeze Theorem:

$x^{1/n} - 1 \to 0$ as $n \to \infty$

Hence, again from the Combination Theorem for Sequences:

$x^{1/n} \to 1$ as $n \to \infty$

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 $n \to \infty$

$\blacksquare$

## Proof 2

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

$x^{1/n} \ge 1$

Hence $\sequence {x^{1/n} }$ is bounded below by $1$.

Now consider $x^{1/n} / x^{1 / \paren {n + 1} }$:

 $\displaystyle \frac {x^{1/n} } {x^{\frac 1 {n + 1} } }$ $=$ $\displaystyle x^{\frac 1 n - \frac 1 {n + 1} }$ $\displaystyle$ $=$ $\displaystyle x^{\frac {n + 1 - n} {n \paren {n + 1} } }$ $\displaystyle$ $=$ $\displaystyle x^{\frac 1 {n \paren {n + 1} } }$ $\displaystyle$ $>$ $\displaystyle 1$ Root of Number Greater than One‎

So:

$x^{1/n} > x^{\frac 1 {n + 1} }$

and so $\sequence {x^{1 / n} }$ is strictly decreasing.

Hence from the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {x^{1 / n} }$ converges to a limit $l$ and that $l \ge 1$.

Now, since we know that $\sequence {x^{1 / n} }$ is convergent, we can apply Limit of Subsequence equals Limit of Real Sequence.

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

So we take the subsequence:

$\sequence {x^{1 / {2 n} } }$

From what has just been shown:

$x^{1 / {2 n} } \to l$ as $n \to \infty$

Using the Combination Theorem for Sequences, we have:

$x^{1 / n} = x^{1 / {2 n} } \cdot x^{1 / {2 n} } \to l \cdot l = l^2$ as $n \to \infty$

But a Convergent Real Sequence has Unique Limit, so $l^2 = l$ and so $l = 0$ or $l = 1$.

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

Hence the result.

$\blacksquare$