TITLE: Prove that $\varlimsup _{n\rightarrow \infty}(u_n)^{\frac{1}{n}}=1$ where $u_{n+1}=\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}$ and $u_0=1$.
QUESTION [6 upvotes]: Prove that 
$$\varlimsup_{n\rightarrow \infty} (u_n)^{\frac{1}{n}}=1,$$ 
where $u_0=1$ and $$u_{n+1}=\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}\;.$$

REPLY [4 votes]: $u_n$ is clearly a non-negative sequence, and is decreasing since $$ \frac{u_{n+1} }{u_n} = \frac{2u^2_n + 2u_n +1 }{2u^2_n + 3u_n +1 } = 1 - \frac{u_n}{2u^2_n + 3u_n +1 }\leq 1.$$
Thus $u_n$ tends to some finite limit $L$ which must satisfy $\displaystyle L = \frac{2L^3+ 2L^2+ L}{2L^2 + 3L+1} .$ Thus $L=0$ and $$ \lim_{n\to\infty} \sqrt[n]{u_n} = \lim_{n\to\infty} \frac{u_{n+1}}{u_n} = 1. $$