Find $\lim_{n\to\infty} a_{n}$ when $a_{n+1}=\dfrac{a_{n}}{1+\frac{1}{n}+a_{n}+a_{n}^3}$ If $a_{n+1}=\dfrac{a_{n}}{1+\frac{1}{n}+a_{n}+a_{n}^3}$ for all positive integers $n$ and $a_{1}>0$, find the limit of  $$\lim_{n\to\infty} a_{n}$$
 A: I show that,
for $n \ge 3$,
$\dfrac{6a_3}{n(n-1)}
\lt a_n
\lt \dfrac1{n-1}
$.
I don't know which bound is 
close to the true value.
Since
$a_{n+1}
=\dfrac{a_{n}}{1+\frac{1}{n}+a_{n}+a_{n}^3}
$,
$a_{n+1}
\lt\dfrac{a_{n}}{1+a_{n}}
=\dfrac{1}{1+1/a_{n}}
$,
so
$\dfrac1{a_{n+1}}
\gt 1+\dfrac1{a_n}
$.
Let
$b_n = \dfrac1{a_n}
$.
Then
$b_{n+1} > 1+b_n
$
so,
since $b_1 > 0$,
$b_n > n-1
$
so
$a_n \lt \dfrac1{n-1}
\to 0$.
To get an upper bound,
$\begin{array}\\
a_{n+1}
&=\dfrac{a_{n}}{1+\frac{1}{n}+a_{n}+a_{n}^3}\\
&>\dfrac{a_{n}}{1+\frac{1}{n}+\frac1{n-1}+\frac1{(n-1)^3}}\\
&>\dfrac{a_{n}}{1+\frac{2}{n-1}}
\qquad\text{for } n \ge 3\\
\text{so}\\
\dfrac{a_{n+1}}{a_n}
&>\dfrac{1}{1+\frac{2}{n-1}}\\
&=\dfrac{n-1}{n+1}\\
\text{so}\\
\dfrac{a_{m}}{a_3}
&=\prod_{n=3}^{m-1}\dfrac{a_{n+1}}{a_n}\\
&>\prod_{n=3}^{m-1}\dfrac{n-1}{n+1}\\
&=\dfrac{\prod_{n=3}^{m-1}(n-1)}{\prod_{n=3}^{m-1}(n+1)}\\
&=\dfrac{\prod_{n=2}^{m-2}n}{\prod_{n=4}^{m}n}\\
&=\dfrac{2\cdot 3}{m(m-1)}\\
&=\dfrac{6}{m(m-1)}\\
\end{array}
$
A: $a_{n+1}$ is obtained from $a_n$ by dividing it by a number greater than $1$. Therefore, the sequence is monotonic and bounded (by $0$ and $a_1$), and the limit exists. Call it $L$.
We have $$L=\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{a_n}{1+\frac{1}{n}+a_n+a_n^3}=\frac{L}{1+L+L^3},$$ which implies $$L^2+L^4=0.$$ The only real solution to this equation is $\boxed{L=0}$, which must therefore be our limit.
