Let be the following sequence $x_{n}$, $n\geq0$, $0
Having given this, i'm required to calculate: $\lim_{n\rightarrow\infty} n x_{n} $
Let be the following sequence $x_{n}$, $n\geq0$, $0
Having given this, i'm required to calculate: $\lim_{n\rightarrow\infty} n x_{n} $
The recurrence yields $ \frac{x_{n+1} }{x_n} = (1-x_n) (1+x_n^2 + x_n^4) .$
For $z\in (0,1)$ we have $ 0< (1-z)(1+z^2+z^4) < 1$ so $x_n$ is montonically decreasing and non-negative. Also note that since $x_n$ is a non-negative sequence we have $0\leq x_n \leq 1$ for all $n\in \mathbb{N}$ (for if $x_k>1$ then $x_{k+1}<0$). Thus $x_n\to x$ where $x\in[0,1].$ It must satisfy $x=x(1-x)(1+x^2+x^4)$ which can be written $ x\left((1-x)(1+x^2+x^4) -1\right)=0.$ Thus $\displaystyle \lim_{n\to \infty} x_n=0$ and consequentially, $\displaystyle \lim_{n\to\infty} \frac{x_{n+1}}{x_n} = 1.$
Now we calculate $ \frac{(n+1) - n}{\frac{1}{x_{n+1}} - \frac{1}{x_n} }= \frac{x_{n+1} x_n}{x_n - x_{n+1}}= \frac{x_{n+1}x_n}{x_n^2-x_n^3 +x_n^4 -x_n^5 +x_n^6} $
$= \frac{x_{n+1}/x_n }{1-x_n +x_n^2 - x_n^3 + x_n^4}\to 1.$
Hence, by the Stolz–Cesàro theorem we conclude $\displaystyle \lim_{n\to\infty} nx_n =1.$