Let $x_n$ denote a sequence, $n\in\Bbb N$: $$ \begin{cases} x_{n+1} = \sqrt[k]{5x_n}\\ x_1 = \sqrt[k]{5}\\ k\in\Bbb N \end{cases} $$ Find: $$ \lim_{n\to\infty} \sqrt[k]{x_n} $$
I would like to verify my approach on finding the limit, which doesn't seem very natural and potentially find a simpler one. I've started with writing down several first terms: $$ x_1 = \sqrt[k]{5}\\ x_2 = \sqrt[k]{5\sqrt[k]{5}}\\ x_3 = \sqrt[k]{5\sqrt[k]{5\sqrt[k]{5}}}\\ \dots $$
Rewrite as: $$ \begin{align} x_1 &= 5^{1\over k}\\ x_2 &= \left(5\cdot 5^{1\over k}\right)^{1\over k} = 5^{\frac{k+1}{k^2}}\\ &\cdots\\ x_n &= 5^{\frac{k^{n-1} + k^{n-2} + \cdots + 1}{k^n}} \end{align} $$
This is hardly readable using powers in mathjax. Rewrite: $$ \log_5 x_n = \frac{k^{n-1} + k^{n-2} + \cdots + 1}{k^n} $$
Nominator is a regular geometric series: $$ 1 + k + \cdots + k^{n-2} + k^{n-1} = \frac{k^n - 1}{k - 1} $$
Thus: $$ \frac{k^n - 1}{k - 1} \cdot \frac{1}{k^n} = \frac{k^n}{(k - 1)k^n} - \frac{1}{(k-1)k^n}\tag 1 $$
Since $a^x$ is continuous we may consider the limit of $(1)$: $$ \lim_{n\to\infty} \left(\frac{k^n}{(k - 1)k^n} - \frac{1}{(k-1)k^n}\right) = \frac{1}{k-1} $$ Which implies: $$ \lim_{n\to\infty}x_n = \sqrt[k-1]{5} $$
Is it the right way to approach this problem? Could it be simplified? Thank you!
We have $x_2 = \sqrt[k]{5\sqrt[k]{5}} \ge \sqrt[k]{5} = x_1$. If we assume that $x_{n} \ge x_{n-1}$, it follows $\sqrt[k]{5x_{n}} \ge \sqrt[k]{5x_{n-1}}$, or $x_{n+1} \ge x_n$. Induction implies that $(x_n)_n$ is increasing.
We have $x_1 = \sqrt[k]{5} \le \sqrt[k-1]{5}$. If we assume $x_n \le \sqrt[k-1]{5}$, we get $$x_{n+1} = \sqrt[k]{5x_n} \le \sqrt[k]{5\sqrt[k-1]{5}} = \left(5\cdot 5^{\frac1{k-1}}\right)^{1/k} = \sqrt[k-1]{5}$$ Induction implies that $x_n \le \sqrt[k-1]{5}, \forall n\in\mathbb{N}$.
The sequence $(x_n)_n$ is increasing and bounded from above so it converges to some $L = \lim_{n\to\infty} x_n$.
Letting $n\to\infty$ in the relation $x_{n+1} = \sqrt[k]{5x_n}$ gives $$L = \sqrt[k]{5L} \implies L(L^{k-1}-5) = 0$$
It cannot be $L = 0$ because $x_n \ge x_1 =\sqrt[k]{5}, \forall n\in\mathbb{N}$. Therefore $L = \sqrt[k-1]{5}$.