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 $u_0=1$. Prove that 
$$\varlimsup_{n\rightarrow \infty} (u_n)^{\frac{1}{n}}=1,$$ 
where $u_0=1$ and $$u_{n+1}=\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}\;.$$
 A: We can do even better: $\lim\limits_{n\to\infty} (u_n)^{1/n}=1$. Note that for $x>0$ we have
$$\frac{d}{dx}\frac{2x^3+2x^2+x}{2x^2+3x+1}=\frac{4x^4+12x^3+10x^2+4x+1}{(2x^2+3x+1)^2}>0$$
thus since clearly each $u_n>0$, $\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}$ is an increasing function of $u_n$. This allows us to show that $u_n\geq \frac{1}{n+1}$ for all $n\in\mathbb N$, as $u_0=\frac{1}{1}$ and if $u_n\geq \frac{1}{n+1}$ then
$$\begin{eqnarray}
u_{n+1}&=&\frac{2u_n^3+2u_n^2+u_n}{2u_n^2+3u_n+1}\\
&\geq& \frac{2/(n+1)^3+2/(n+1)^2+1/(n+1)}{2/(n+1)^2+3/(n+1)+1}\\
&\geq&\frac{1}{n+1}-\frac{1}{(n+1)^2+3(n+1)+2}\\
&\geq& \frac{1}{n+1}-\frac{1}{(n+1)^2+(n+1)}\\
&\geq&\frac{1}{n+2}\end{eqnarray}$$
and so by induction $u_n\geq \frac{1}{n+1}$ for all $n$. Since it is clear that $(u_{n})^{1/n}\leq 1$ for all $n$, we need only show that $\lim\limits_{x\to\infty}\left(\frac{1}{x+1}\right)^{1/x}=1$, or equivalently (since $\ln x$ is continuous) that $$\lim\limits_{x\to\infty}\ln\left(\left(\frac{1}{x+1}\right)^{1/x}\right)=\ln 1=0$$
which we can do using L'Hospital's rule
$$\lim\limits_{x\to\infty}\ln\left(\left(\frac{1}{x+1}\right)^{1/x}\right)=\lim\limits_{x\to\infty}\frac{1}{x}\ln\left(\frac{1}{x+1}\right)=\lim\limits_{x\to\infty}\frac{-1}{x+1}=0$$
and so desired result follows from squeezing $\lim\limits_{n\to\infty}(u_n)^{1/n}$ between $\lim\limits_{n\to\infty} \left(\frac{1}{n+1}\right)^{1/n}$ and $\lim\limits_{n\to\infty} 1$.
A: $u_n$ 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 $$ \lim_{n\to\infty} \sqrt[n]{u_n} = \lim_{n\to\infty} \frac{u_{n+1}}{u_n} = 1. $$
