Proving the sequence is increasing Suppose that a sequence $a_n$ of real numbers satisfies $7a_{n+1 }= a_n^3+6$
for $n ≥ 1$. If $a_1 =\frac{1}{2}$
, prove
that the sequence increases and find its limit.
Now I calculated the difference of two successive terms and factorized it. I got $$a_{n+1}-a_{n}=\frac{(a_n-1)(a_n-2)(a_n+3)}{7}$$ But I am unable to proceed further.
 A: Use induction you can show $a_n$ is always less than $1$ so the product is positive.
$a_{n+1} = \frac{a_n^3+6}{7}$, so when $a_n<1$, $a_n^3<1$, which implies $a_{n+1} = \frac{a_n^3+6}{7}<1$. Base case is $a_1=\frac{1}{2}$. 
So $(a_n-1)(a_n-2)(a_n+3)>0$. Thus $a_{n+1}-a_n>0$.
Now the limit can be found by investigating the solutions to $7x=x^3+6$.
A: A slightly quicker answer is to note that the function $f(x) = x^3 + 6$ has $f'(x) = 3x^2 \ge 0$, which means the function is increasing, and this in turn implies the sequence above is increasing also.The limit is the solution of the equation $x^3 + 6 = 7x, x \ge \dfrac{1}{2}$ which can be found by synthetic division in elementary algebra.
A: It is quite obviously that all memberes of sequnce are positive and less than 1 (see previous posts). Then by $AM-GM$ we have:
$$ a_{n+1} = {a_n^3+1+1+1+1+1+1\over 7} \geq \sqrt[7]{a_n^3} > a_n $$ 
A: Note that by induction it follows that $a_n\ge0$ $\forall n\in\mathbb N$.
Now $$a_{n+1}=\frac{a_n^3+6}{7}$$ and $$a_{n}=\frac{a_{n-1}^3+6}{7}$$ so $$a_{n+1}-a_n=\frac{(a_n-a_{n-1})(a_n^2+a_{n-1}^2+a_na_{n-1})}{7}$$
So, $ a_{n+1}\ge or\le a_n$ according as $a_n\ge or\leq a_{n-1} $ respectively.But we see that $a_1=\frac{1}{2}\le\frac{49}{56}=a_2$. Hence the sequence is a increasing sequence.
Again by induction it follows that $a_n\le 1$ $\forall n\in\mathbb N.$ So the sequence converges(monotonically increasing sequence bounded above).
Let $\lim_{n\to \infty}a_n=l$ then $$7l=l^3+6$$ the roots are $l=-3,1,2$ 
As $0\leq a_n\leq 1$ the only case is $l=1$
