How do you prove $\lim_{n\to\infty} 1/n^{1/n}$ using only basic limit theorems? How do you prove $\lim_{n\to\infty} \frac {1}{n^{1/n}}$ using only basic limit theorems? I thought it was $0$, but my book lists the solution as $1$. How come?

Method $1$: Clearly, $n^{1/n}>1$ for all $n \in \mathbb{N}$. Let $n^{1/n} = (1+a)$. We will prove that $a \to 0$ as $n \to \infty$. For $n \geq 2$, we have
$$n = (1+a)^n \geq \dfrac{n(n-1)}2a^2 \implies a^2 \leq \dfrac2{n-1}$$
Hence, as $n \to \infty$, we have $a \to 0$, which means that $n^{1/n} \to 1$.

Method $2$: From AM-GM, for $a > 1$, we have
$$\dfrac{a+a+\cdots+a+n}n \geq \sqrt[n]{na^{n-1}} \implies (1+a) - \dfrac{a}n \geq n^{1/n}a^{1-1/n}$$
Taking the limit as $n \to \infty$, we obtain
$$\lim_{n \to \infty} n^{1/n}a \leq (1+a) \implies \lim_{n \to \infty} n^{1/n} \leq 1+\dfrac1a$$
for all $a \in \mathbb{R}$. Taking the limit as $a \to \infty$, we obtain
$$n^{1/n} \leq 1$$
Further, $n^{1/n} \geq 1$ for all $n$. Hence, we have $\lim_{n \to \infty} n^{1/n}=1$.
