Evaluate $\lim_{n \to \infty} \prod_{k=n+1}^{2n} k^{1/k}$

Question

Problem: evaluate $\lim_{n \to \infty} \prod_{k=n+1}^{2n} k^{1/k}$.

My work: since $n+1 \le k \le 2n$, it is $\frac{1}{2n} \le \frac{1}{k} \le \frac{1}{n+1}$.

Since $k \ge 1$, the exponential with base $k$ is increasing and so, from $\frac{1}{2n} \le \frac{1}{k} \le \frac{1}{n+1}$, it follows that $k^{1/2n} \le k^{1/k} \le k^{1/(n+1)}$.

Since the $\alpha$-th ($\alpha$ real) power is increasing when its base is nonnegative and $k \ge 1$, it follows that $(n+1)^{1/2n} \le k^{1/2n}$ and $k^{1/(n+1)} \le (2n)^{1/(n+1)}$; in conclusion, it is $(n+1)^{1/2n} \le k^{1/k}\le(2n)^{1/(n+1)}$ for any $n+1 \le k \le 2n$ and for any $n\in\mathbb{N}$.

Hence $$\lim_{n \to \infty} \prod_{k=n+1}^{2n} k^{1/k} \ge \lim_{n \to \infty} \prod_{k=n+1}^{2n} (n+1)^{1/2n}=\lim_{n\to\infty}\sqrt{n+1}=\infty$$ $$\implies \lim_{n \to \infty} \prod_{k=n+1}^{2n} k^{1/k}=\infty$$ Is this correct? I'm not sure that what I've written about the estimation I get using the fact that the exponential and the real power are increasing.

Answer 1

This is right, yes. If you're worried about its rigour, you can take limits as late as possible, viz.$$k^{1/k}\ge k^{1/(2n)}\ge(n+1)^{1/(2n)}\implies\prod_{k=n+1}^{2n}k^{1/k}\ge\prod_{k=n+1}^{2n}(n+1)^{1/(2n)}=\sqrt{n+1}\stackrel{n\to\infty}{\rightarrow}\infty.$$