Show that lim of the sequence is $\ell$

Question

Given the sequence defined as :

$\displaystyle{u_{n + 1} = {u_{n}^{2} \over 1 - 2u_{n}^{2}}}\quad\forall\ n \in \mathbb{N}$

And $u_{0} =a $ such that $a \in \left(0,{1 \over 4}\right)$

Let's take $s_{n} = \sum_{k = 0}^{n}\,\left(-1\right)^{k}\,u_{k}$

$w_{n} = s_{2n + 1}$

$v_{n} = s_{2n}$

Let $\lim v_{n} = \lim w_{n} = \ell$

How can one prove that $\lim s_{n} = \ell$

I see that we need to majorate the expression of $s_{n}$ by $v_{n}$ or $w_{n}$ to find that $\lim s_{n} = \ell$ but it seems difficult since i don't see a direct relation between $s_{n}$ and $s_{2n}$ and $s_{2n + 1}$.

Answer 1

Notice that if $0 < x < \frac{1}{4}$, then

$$ \frac{x^2}{1-2x^2} < \frac{\frac{1}{4}x}{1-\frac{1}{8}} = \frac{2}{7}x. $$

Then by a simple induction, we find that $0 < u_n < (2/7)^n u_0$. This proves that $ \sum_{n=0}^{\infty} (-1)^n u_n$ converges absolutely, and if we denote the value of this series by $\ell$, then both $v_n$ and $w_n$ also converge to $\ell$ since they are subsequences of the convergent sequence $(s_n)$.