Limit and convergence of a sequence I have an exercise of sequences, I figured out half the solution but I'm stuck now, so please help.
This the question I can't figure out how to solve :
Let $a$ a positive real number that is strictly positive, and $\big(x_n\big)$ a sequence defined by :
$x_0=a$ and $\forall n\in \mathbb{N}, x_{n+1}=\frac{x_n}{1+(n+1)x_n^2}$
I proved that if $a=1$ then : $\big(\forall n\in\mathbb{N}^*\big)\ x_n=\frac{1}{n+1}$
Then I proved that if $a>1$ the $\big(x_n\big)$ is decreasing, bounded by $0$ and convergente,
but I couldn't find it's limit. I tried to turn $\frac{x_n}{1+(n+1)x_n^2}$ into a function,
so I wrote it this way : $f_n\big(x\big)=\frac{x}{1+(n+1)x^2}\, \forall x\in\mathbb{R^+}$, because I wanted to use the Intermediate value theorem
I proved that $f_n$ is injective and that $f_n\big(\big[0,+\infty\big[\big)=\big[0,+\infty\big[$, so I wanted to prove that $\exists!x_n\in\mathbb{R}_+^*/f_n\big(x_n\big)=\lambda$ when $\lambda$ is a number that exists in $\mathbb{R}_+^*$ that will help me to fin the limit, but I don't know how to choose that number.
I chose $1$ so it gives me :
$\frac{x_n}{1+(n+1)x_n^2}=1\quad\Rightarrow\quad x_n-\big(n+1\big)x_n^2=1$
$\Rightarrow\quad l-\lim\limits_{}(n+1)x_n^2=1$ where $l=\lim\limits_{}x_n$
but I can't solve this limit $\lim\limits_{}(n+1)x_n^2$.  
Please help!
 A: I think it's not necessary to use that Intermediate value theorem. Here is my proof:
First, we use mathematical induction to prove that $(x_n)$ is a monotonic decreasing sequence, and since $x_n>x_{n+1}=\dfrac{x_n}{1+(n+1)x_n^2}$, we obtain that $x_n>0$. So it must converge to some number. One might suspect $lim(x_n)=0$.
Change the equation: $x_{n+1} = \dfrac{x_n}{1+(n+1)x_n^2}$
we obtain: $x_{n+1}+(n+1)x_{n+1}x_n^2=x_n$
divide this by $(n+1)$: $\dfrac{x_{n+1}}{n+1}+x_{n+1}x_n^2=\dfrac{x_n}{n+1}$
Apply the fact that: $x = lim_n x_{n+1}$
$x^3=0\implies x=0$ as $n\to\infty$
So, $lim (x_n) = 0$
The proof written by Murphy is by contradition, my proof is really different from his, hope this gives you some ideas. I think your method may work but it's just very complicated.
A: 1) $0 <x_n$, $x_{n+1} <x_n$ , strictly decreasing.
Limit $L\ge 0$ exists.
Note :
$L<x_n$ , for all $n$.
2) $0\lt x_{n+1}< \dfrac{a}{1+(n+1)L^2}; $
Assume $L>0$, and take the limit.
Hence?
A: If $\lim x_n>0$ then $\frac {x_n}{1+(n+1)x_n^{2}} \to 0$ since the numerator has a finite limit and the denominator tends to $\infty$.  So we get a contradiction. Hence the limit is necessarily $0$. 
