# Limit of Integer to Reciprocal Power

## Theorem

Let $\sequence {x_n}$ be the real sequence defined as $x_n = n^{1/n}$, using exponentiation.

Then $\sequence {x_n}$ converges with a limit of $1$.

## Proof 1

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

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

Thus the subsequence of $\sequence {n^{1 / n} }$ consisting of all the terms of $\sequence {n^{1 / n} }$ where $n \ge 3$ is convergent by the Monotone Convergence Theorem (Real Analysis).

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 $\sequence {\paren {2 n}^{1 / {2 n} } }$.

By Limit of Subsequence equals Limit of Real Sequence, this will converge to $l$ also.

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

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

Thus:

- $n^{1 / n} = n^{1 / {2 n} } \cdot n^{1 / {2 n} } \to l \cdot l = l^2$

as $n \to \infty$.

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

$\blacksquare$

## Proof 2

We have the definition of the power to a real number:

- $\displaystyle n^{1/n} = \exp \left({\frac 1 n \ln n}\right)$.

From Powers Drown Logarithms, we have that:

- $\displaystyle \lim_{n \to \infty} \frac 1 n \ln n = 0$

Hence:

- $\displaystyle \lim_{n \to \infty} n^{1/n} = \exp 0 = 1$

and the result.

$\blacksquare$

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): Exercise $1.5: 10$