Limit of Root of Positive Real Number
Theorem
Let $x \in \R: x > 0$ be a real number.
Let $\sequence {x_n}$ 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:
- $A_n = \dfrac {n - 1 + x} n = 1 + \dfrac{x - 1} n$
From Arithmetic Mean is 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:
- $0 \le x^{1/n} - 1 \le \dfrac 1 n \paren {x - 1}$
But from Sequence of Powers of Reciprocals is Null Sequence:
- $\dfrac 1 n \to 0$ 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.
From Root of Number Greater than One:
- $x^{1/n} \ge 1$
Hence $\sequence {x^{1/n} }$ is bounded below by $1$.
Now consider $x^{1/n} / x^{1 / \paren {n + 1} }$:
\(\ds \frac {x^{1/n} } {x^{\frac 1 {n + 1} } }\) | \(=\) | \(\ds x^{\frac 1 n - \frac 1 {n + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^{\frac {n + 1 - n} {n \paren {n + 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^{\frac 1 {n \paren {n + 1} } }\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 1\) | 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.
$\blacksquare$