trinhquockhanh wrote: Consider the sequence $(a_n)$ satisfying $a_1=\dfrac{1}{2},a_{n+1}=\sqrt[3]{3a_{n+1}-a_n}$ and $0\le a_n\le 1,\forall n\ge 1.$ a. Prove that the sequence $(a_n)$ is determined uniquely and has finite limit. $a_{n+1}$ is a root of $x^3-3x+a_n=0$ belonging to $[0,1]$ Over $[0,1]$, the cubic is strictly decreasing from $a_n\ge 0$ to $a_n-2<0$ and so such a root exists and is unique. $a_n^3-3a_n+a_n\le 0$ and so the root is $\le a_n$ So $a_{n+1}\le a_n$ and, nonincreasing and nonnegative, sequence $a_n$ has a limit $l\in[0,1]$ which matches equaltion $l^3-3l+l=0$ And so $\boxed{\lim_{n\to+\infty}a_n=0}$

a. We have $a_{n+1}=\sqrt[3]{3a_{n+1}-a_n}\Leftrightarrow 3a_{n+1}-a_{n+1}^3=a_n$ Firstly, we will prove that the equation $3x-x^3=m$ has only one real root in $[0,1],\forall m\in [0,1]$ (Lemma 1) Consider $f(x)=x^3-3x+m=0\Rightarrow f(0).f(1)<0\Rightarrow \exists a\in [0,1]\; \text{such that}\; f(a)=0$ $\text{Let}\; a=2\sin(t), t\in \left[0,\dfrac{\pi}{6} \right]\cup \left[\dfrac{5\pi}{6},\pi \right] \Rightarrow 4 \sin ^3(t)-3\sin(t)+\dfrac{m}{2}=0\Rightarrow \sin(3t)=\dfrac{m}{2}\Rightarrow t=\dfrac{\arcsin \left( \dfrac{m}{2} \right)}{3}+2k\pi$ With $k\in \mathbb{Z},$ note that $t\in \left[0,\dfrac{\pi}{6} \right]\cup \left[\dfrac{5\pi}{6},\pi \right]\Rightarrow t$ can get only one value $\Rightarrow a$ can get only one value Hence the Lemma 1 is proven, implying the sequence $(a_n)$ is determined uniquely. By induction, we can prove that $a_n=2\sin \left(\dfrac{\theta}{3^{n-1}} \right),$ with $\theta=\arcsin \left(\dfrac{1}{4}\right)\Rightarrow (a_n)$ has finite limit. b. It's easy to prove that $(b_n)$ is increasing, so we need to prove that $(b_n)$ is bounded We have $\ln (b_n)= \ln (1+2a_1) +...+\ln (1+2^na_n),$ note that $\ln(1+x)\le x,\forall x >0$ $\Rightarrow \ln (b_n)\le 2a_1+...+2^na_n=2^2.\sin (\theta)+...2^{n+1}\sin \left(\dfrac{\theta}{3^{n-1}} \right) =\sum\limits_{i = 1}^n {{2^{i + 1}}\sin \left( {\frac{\theta }{{{3^{i - 1}}}}} \right)}$ $= 4\theta .\sum\limits_{i = 1}^n {{{\left( {\dfrac{2}{3}} \right)}^{i - 1}}\frac{{\sin \left( {\dfrac{\theta }{{{3^{i - 1}}}}} \right)}}{{\dfrac{\theta }{{{3^{i - 1}}}}}}} \le 4\theta .\sum\limits_{i = 1}^n {{{\left( {\dfrac{2}{3}} \right)}^{i - 1}} = } 4\theta .\dfrac{{{{\left( {\dfrac{2}{3}} \right)}^n} - 1}}{{\dfrac{2}{3} - 1}} = 4\theta .\dfrac{{1 - {{\left( {\dfrac{2}{3}} \right)}^n}}}{{\dfrac{1}{3}}} = 4\theta .\left( {3 - 3.{{\left( {\dfrac{2}{3}} \right)}^n}} \right) < 12\theta $ Hence $\ln (b_n)$ is always bounded $\Rightarrow (b_n)$ is bounded, this means $(b_n)$ has finite limit.

Actually we can shorten part b by denote that $$( 1+2a_{1})\left( 1+2^{2} a_{2}\right) ...\left( 1+2^{n} a_{n}\right) < e^{2a_{1} +2^{2} a_{2} +...+2^{n} a_{n}}$$using the famous inequality $\displaystyle 1+x\leqslant e^{x} ,\forall x$ Thus we just need to prove that $\displaystyle \sum _{i=1}^{\infty } 2^{i} a_{i}$ is a convergent series. This is true since $$\displaystyle \frac{2^{n+1} a_{n+1}}{2^{n} a_{n}} =2\frac{a_{n+1}}{a_{n}} =2\frac{a_{n+1}}{3a_{n+1} -a_{n+1}^{3}} =\frac{2}{3-a_{n+1}^{2}} < 1$$hold by Ratio' sTest