Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 3

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

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

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


 * $\paren {n + 1}^{1/n} = \map \exp {\dfrac 1 n \, \map \ln {n + 1} }$

For $n >= 1$ then $n + 1 \le 2 n$.

Hence:

By Powers Drown Logarithms:
 * $\displaystyle \lim_{n \mathop \to \infty} \frac 1 n \ln n = 0$

By Sequence of Reciprocals is Null Sequence:
 * $\displaystyle \lim_{n \mathop \to \infty} \frac 1 n = 0$

By Combined Sum Rule for Real Sequences:
 * $\displaystyle \lim_{n \mathop \to \infty} \paren {\frac {\ln 2} n + \frac 1 n \ln n} = \ln 2 \cdot 0 + 0 = 0$

By the Squeeze Theorem for Real Sequences:
 * $\displaystyle \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} = 0$

Hence:
 * $\displaystyle \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} = \exp 0 = 1$

and the result follows.