Limit of Number to Reciprocal Power

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

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

	\(\displaystyle \)	\(\displaystyle \left({n + 1}\right)^{1/\left({n + 1}\right)}\)	\(\le\)	\(\displaystyle n^{1/n}\)	\(\displaystyle \)
	\(\displaystyle \iff\)	\(\displaystyle \left({n + 1}\right)^n\)	\(\le\)	\(\displaystyle n^{n + 1}\)	\(\displaystyle \)	raising both sides to the power of $n \left({n+1}\right)$
	\(\displaystyle \iff\)	\(\displaystyle \left({n \left({1 + \frac 1 n}\right)}\right)^n\)	\(\le\)	\(\displaystyle n^{n + 1}\)	\(\displaystyle \)
	\(\displaystyle \iff\)	\(\displaystyle \left({1 + \frac 1 n}\right)^n\)	\(\le\)	\(\displaystyle \frac {n^{n + 1} } {n^n} = n\)	\(\displaystyle \)

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

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.

$\Box$

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 $\left \langle {\left({2n}\right)^{\frac 1 {2n}}} \right \rangle$.

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.

$\blacksquare$

Alternatively, we can use 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

• W.A. Sutherland: Introduction to Metric and Topological Spaces (1975): Exercise $1.5: 10$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 4.20 \ (6)$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 5.7 \ (1)$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977): $\S 14.7 \ (4)$