Limit of Integer to Reciprocal Power/Proof 1

Theorem
Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as $x_n = n^{1/n}$.

Then $\left \langle {x_n} \right \rangle$ converges with a limit of $1$.

Proof
From Number to Reciprocal Power is Decreasing we have that the real sequence $\left \langle {n^{1/n}} \right \rangle$ is decreasing for $n \ge 3$.

Now, as $n^{1/n} > 0$ for all positive $n$, it follows that $\left \langle {n^{1/n}} \right \rangle$ is bounded below (by $0$, for a start).

Thus the subsequence of $\left \langle {n^{1/n}} \right \rangle$ consisting of all the elements of $\left \langle {n^{1/n}} \right \rangle$ where $n \ge 3$ is convergent by the Monotone Convergence Theorem.

Now we need to demonstrate that this limit is in fact $1$.

Let $n^{1/n} \to l$ as $n \to \infty$.

Having established this, we can investigate the subsequence $\left \langle {\left({2n}\right)^{1 / {2n}}} \right \rangle$.

By Limit of a Subsequence, this will converge to $l$ also.

From Limit of Root of Positive Real Number, we have that $2^{1 / {2n}} \to 1$ as $n \to \infty$.

So $n^{1 / {2n}} \to l$ as $n \to \infty$ by the Combination Theorem for Sequences.

Thus:
 * $n^{1/n} = n^{1 / {2n}} \cdot n^{1 / {2n}} \to l \cdot l = l^2$

as $n \to \infty$.

So $l^2 = l$, and as $l \ge 1$ the result follows.