Is there an alternative proof for $\lim a_n^{\frac{1}{q}}=(\lim a_n)^{\frac1q}$?

Question

Let $\left(a_n\right)_n$ be a sequence convergent to $a$.

I want to see that if $a_n \geq 0$, for $n \in \mathbb{N}$, then $\lim a_n^{\frac{1}{q}} = (\lim a_n)^{\frac{1}{q}} = a^{\frac{1}{q}}$ for all $q \in \mathbb{N}$.

The proof written by my teacher is the following.

Suppose there exists $\varepsilon > 0$ such that $$M_\varepsilon = \left\{n \in \mathbb{N} \colon a_n^{\frac{1}{q}} \not\in \left(a^{\frac{1}{q}} - \varepsilon, a^{\frac{1}{q}} + \varepsilon\right)\right\}$$ is infinite. So if $n \in M_\varepsilon$ then $a_n^{\frac{1}{q}} \not\in \left(a^{\frac{1}{q}} - \varepsilon, a^{\frac{1}{q}} + \varepsilon\right)$, which means that $a_n^{\frac{1}{q}} \leq a^{\frac{1}{q}} - \varepsilon$ or $a_n^{\frac{1}{q}} \geq a^{\frac{1}{q}} + \varepsilon > 0$.

If $a_n^{\frac{1}{q}} \leq a^{\frac{1}{q}} - \varepsilon$ then $a^{\frac{1}{q}} \geq a_n^{\frac{1}{q}} + \varepsilon > 0$. Therefore, $$a \geq \left(a_n^{\frac{1}{q}} + \varepsilon\right)^q \geq a_n + \varepsilon^q.$$ Thus $a_n \leq a - \varepsilon^q$.

If $a_n^{\frac{1}{q}} \geq a^{\frac{1}{q}} + \varepsilon > 0$ then $$a_n \geq \left(a^{\frac{1}{q}} + \varepsilon\right)^q \geq a + \varepsilon^q.$$ Hence $a_n \geq a + \varepsilon^q$.

Consequently, if $n \in M_\varepsilon$ then $a_n \not\in \left(a - \varepsilon^q, a + \varepsilon^q\right)$, which means that $a$ cannot be the limit of the sequence $\left(a_n\right)_n$.

This proof is generally confusing to me, and that's why I was wondering if you could please provide me with another one if possible. Thank you very much!

Answer 1

Let $b_n=a_n^{\frac1q}\geq0$ and $b=a^{\frac1q}\geq0$, then $a_n=b_n^q$ and $a=b^q$.

If $b>0$, since $$a_n-a=b_n^q-b^q=(b_n-b)(b_n^{q-1}+b_n^{q-2}b+\cdots+b_nb^{q-2}+b^{q-1}),$$ we have $|a_n-a|\geq |b_n-b|b^{q-1}$ and thus $$|b_n-b|\leq b^{1-q}|a_n-a|,$$ which implies $\lim_{n\to\infty} b_n=b$, recalling that $\lim_{n\to\infty} a_n=a$.

If $b=0$, then for any $\varepsilon>0$, since $\lim_{n\to\infty} a_n=0$, there exists $N$ such that $0\leq a_n<\varepsilon^q$ for all $n>N$; hence $0\leq b_n<\varepsilon$ for all $n>N$, which implies that $\lim_{n\to\infty} b_n=0=b$.

Answer 2

The function $\exp: \mathbb{R} \to \mathbb{R}^{>0}$ and its inverse $\log : \mathbb{R}^{>0} \to \mathbb{R}$ are continuous. Clearly $x \to x/q$ is also continuous on $\mathbb{R}^{>0}$. Thus $f(x) = x^{1/q} = \exp (\frac{1}{q} \log x)$ is also continuous on $\mathbb{R}^{>0}$. Thus for any $a_n \to a$ in $\mathbb{R}^{>0}$, we have $f(a_n) \to f(a)$. We're done if $a > 0$ (and thus $a_n > 0$ for sufficiently large $n$); if $a = 0$, note that if $|a_n| < \epsilon$, then $|a_n^{1/q}| = |a_n|^{1/q} < \epsilon^{1/q}$.