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 $A_n = \dfrac {n - 1 + x} n = 1 + \dfrac{x - 1} n$.

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 $n \to \infty$.

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

	\(\displaystyle \)	\(\displaystyle \frac {x^{1/n} } {x^{\frac 1 {n+1} } }\)	\(=\)	\(\displaystyle x^{\frac 1 n - \frac 1 {n+1} }\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x^{\frac {n+1-n} {n \left({n+1}\right)} }\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x^{\frac 1 {n \left({n+1}\right)} }\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(>\)	\(\displaystyle 1\)	\(\displaystyle \)	from Root of Number Greater than One‎

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.

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 4.14$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 5.4$