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:
- $\ds n^{1/n} = \map \exp {\frac 1 n \ln n}$
From Powers Drown Logarithms, we have that:
- $\ds \lim_{n \mathop \to \infty} \frac 1 n \ln n = 0$
Hence:
- $\ds \lim_{n \mathop \to \infty} n^{1/n} = \exp 0 = 1$
and the result.
$\blacksquare$
Proof 3
Let $n^{1/n} = 1 + a_n$.
The strategy is to:
- $(1): \quad$ prove that $a_n > 0$ for $n > 1$
- $(2): \quad$ deduce that $n - 1 \ge \dfrac {n \paren {n - 1} } {2!} a_n^2$ for $n > 1$
and hence:
- $(3): \quad$ deduce that $0 \le a_n^2 \le \dfrac 2 n$
Let $n > 1$.
Then:
\(\ds n\) | \(=\) | \(\ds \paren {1 + a_n}^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + n a_n + \dfrac {n \paren {n - 1} } {2!} a_n^2\) |
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