Limit of Root of Positive Real Number/Proof 2

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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

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

\(\displaystyle \frac {x^{1/n} } {x^{\frac 1 {n + 1} } }\) \(=\) \(\displaystyle x^{\frac 1 n - \frac 1 {n + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle x^{\frac {n + 1 - n} {n \paren {n + 1} } }\)
\(\displaystyle \) \(=\) \(\displaystyle x^{\frac 1 {n \paren {n + 1} } }\)
\(\displaystyle \) \(>\) \(\displaystyle 1\) 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.

$\blacksquare$


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