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.

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 $7x=x^3+6$.
