Find the limit of the sequence {$a_{n}$, given by

Question

Find the limit of the sequence {$a_{n}$}, given by$$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{n+1}=(1+a_{n}+a^{3}_{n-1}), \ for \ n \ > \ 1$$

My try:

$ a_{1}=0,a_{2}=\dfrac {1}{2},a_{3}=\dfrac {1}{2},a_{4}=0.54$ that is the sequence is incresing and each term is positive. Let the limit of the sequence be $x$. Then $ \lim _{n\rightarrow \infty }a_{n+1}=\lim _{n\rightarrow \infty }a_{n}=x$ $$ \lim _{n\rightarrow \infty }a_{n+1}= \lim _{n\rightarrow \infty }1+a_{n}+a^{3}_{n-1}$$

$\Rightarrow x=\dfrac {1}{3}( 1+x+x^3)$

$\Rightarrow x^3-2x+1=0$

and this equation has three roots $x=\dfrac {-1\pm \sqrt {5}}{2},1$

So the limit of the sequence is $\dfrac {-1 + \sqrt {5}}{2}$.

What is the other way to find the limit of the sequence?

Answer 1

From $a_{n+1}=1+a_{n}+a^{3}_{n-1}$ we get for the limit $x$:

$x=1+x+x^3$, hence $x=-1$.

Since $a_n \ge 0$ for all $n$, we have $x \ge 0$, a contradiction.

Consequence: $(a_n)$ is not convergent.

Answer 2

Each term is greater at least by $1$ than its predecessor: $$a_{n+1} \ge a_n+1$$ so the sequence grows not slower than the arithmetic progression $$(a_1 + (n-1))_{n>1}$$ hence it is divergent: $$\lim_{n\to\infty}a_n = \infty$$