Mathematical Reflections 2012, Issue 5 - Problem U241

Problem:
Let $a>b$ be positive real numbers. Prove that $$c_n=\dfrac{\sqrt[n+1]{a^{n+1}-b^{n+1}}}{\sqrt[n]{a^n-b^n}}$$
is a decreasing sequence and find its limit.

Proposed by Ivan Borsenco.

Solution:
Let $c_n=\dfrac{\sqrt[n+1]{1-x^{n+1}}}{\sqrt[n]{1-x^n}}$, where $x=b/a < 1$. We first prove that $c_n > 1$. Let $b_n=c^{n(n+1)}_n$. Then $$b_n(x)=\dfrac{(1-x^{n+1})^n}{(1-x^n)^{n+1}}$$ and $$b'_n(x)=\dfrac{n(n+1)x^{n-1}(1-x^{n+1})^{n-1}(1-x^n)^n(1-x)}{(1-x^n)^{2(n+1)}}.$$ So, $b'_n(x) > 0$ for all $x \in (0,1)$, so $b_n(x)$ is increasing in $(0,1)$ and $b_n(x) > b_n(0)=1$, i.e. $c^{n(n+1)}_n > 1$, which yields $c_n > 1$. Now, we prove that $c_n$ is a decreasing sequence. We have $$\dfrac{b_{n-1}}{b_n}=\left(\dfrac{(1-x^n)^2}{(1-x^{n-1})(1-x^{n+1})}\right)^n > 1,$$ so $c_{n-1}^{(n-1)n}>c_n^{n(n+1)}=c_n^{(n-1)n}c_n^{2n}>c_n^{(n-1)n}$, i.e. $c_{n-1}>c_n$. Moreover,
$$\lim_{n \to \infty} \log c_n=\lim_{n \to \infty} \left[\dfrac{1}{n+1}\log (1-x^{n+1})-\dfrac{1}{n}\log(1-x^n)\right]=0,$$ so $$\lim_{n \to \infty} c_n= \lim_{n \to \infty} e^{\log c_n}=1.$$