Limit of Integer to Reciprocal Power

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

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

Proof
First we show that $$\left \langle {n^{1/n}} \right \rangle$$ is decreasing for $$n \ge 3$$.

We want to show that $$\left({n + 1}\right)^{1/\left({n + 1}\right)} \le n^{1/n}$$.

This is the same as saying that $$\left({n + 1}\right)^n \le n^{n + 1}$$.

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But from One Plus Reciprocal to the Nth, $$\left({1 + \frac 1 n}\right)^n < 3$$.

Thus the reversible chain of implication can be invoked and we see that $$\left({n + 1}\right)^{1/\left({n + 1}\right)} \le n^{1/n}$$ when $$n \ge 3$$.

So $$\left \langle {n^{1/n}} \right \rangle$$ 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.

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

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

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

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

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

Thus $$n^{1/n} = n^{\frac {1}{2n}} \cdot n^{\frac {1}{2n}} \to l \cdot l = l^2$$ as $$n \to \infty$$.

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