Given $x_1 = a^{1\over k}$ and $x_{n+1} = \left({a\over x_n}\right)^{1\over k}$ prove that $x_n = a^{1-(-k)^{-n} \over k +1}$

Question

I'm solving various problems on sequences and here is one I'm stuck with:

Given $x_1 = a^{1\over k}$ where $a > 0$, and $x_{n+1} = \left({a\over x_n}\right)^{1\over k}$ where $n \in \mathbb N$. Prove that: $$ x_n = a^{1-(-k)^{-n} \over k +1} $$

Everything I've tried so far just gives me some identity. For example:

$$ \begin{align} x_{n+1} &= \left({a\over x_n}\right)^{1\over k} = \\ &= \frac{a^{1\over k}}{{x_n}^{1\over k}} = \\ &= \frac{x_1}{x_n^{1\over k}} \end{align} $$

Now try to express $x_n$:

$$ \begin{align} x_n^{1\over k} &= \frac{x_1}{x_{n+1}} \iff \\ \iff x_n &= \frac{x_1^k}{x_{n+1}^k} \end{align} $$

So this results in $x_n = x_n$ which doesn't show anything. I feel like there might be some smart substitution to translate the problem into some recurrence relation but don't see how.

Could someone please point me in the right direction?

Answer 1

$$x_1=a^{\frac{1}{k}}$$ now calculate $x_2$ using $x_{n+1} = \left({a\over x_n}\right)^{1\over k}$

thus $$x_2= (\frac{a}{x_1})^{\frac{1}{k}}$$ thus $$x_2=a^{\frac{1}{k}}*x_1^{\frac{-1}{k}}=a^{\frac{1}{k}-\frac{1}{k^2}}$$

now solve for $x_3$ using $x_2$ obtained above

you will get $$x_3=a^{\frac{1}{k}-\frac{1}{k^2}+\frac{1}{k^3}}$$ also $$x_4=a^{\frac{1}{k}-\frac{1}{k^2}+\frac{1}{k^3}-\frac{1}{k^4}}$$

hope you are seeing the pattern so we can write it for $x_n$

now the problem reduces to summing the series for n(even) $$S_n=\frac{1}{k}-\frac{1}{k^2}+\frac{1}{k^3}-\frac{1}{k^4}+.....-\frac{1}{k^n}$$

Answer 2

Hint. One may just rewrite the initial relation as $$ \left(-k\right)^{n+1}\ln (x_{n+1})-\left(-k\right)^{n}\ln (x_{n})=\ln \left[a^{\left(-k\right)^n}\right] $$ recognizing a telescoping sum then using logarithmic properties.