Find limit of recurrent sequence? Let $ \forall n \in \mathbf{N}, ~ u_{n+1} = \frac{1}{\tanh^2(u_n)} - \frac{1}{u^2_n} $ with $ u_0 = a > 0 $.
What is the limit of $(u_n)$ ?
I tried to find a fixed point of this sequence but the equation is impossible to solve algebraically.
 A: Since $f(x)=\coth^2(x)-\frac{1}{x^2}=1+\sum_{k\geq 1}\left[\frac{1}{(x-\pi k i)^2}+\frac{1}{(x+\pi k i)^2}\right]$ is an even function such that
$$ \left|f'(x)\right|=2\left|\frac{1}{x^3}-\frac{\cosh(x)}{\sinh^3(x)}\right|=2\left|\sum_{k\geq 1}\frac{1}{(x-\pi k i)^3}+\frac{1}{(x+\pi k i)^3}\right| $$
and the maximum of $\left|\frac{1}{(x-\pi k i)^3}+\frac{1}{(x+\pi k i)^3}\right|$ is achieved at $x=\pi k(\sqrt{2}-1)$, we have
$$ |f'(x)|\leq 2\sum_{k\geq 1}\frac{(\sqrt{2}+1)^2}{4\pi^3 k^3}=\frac{(\sqrt{2}+1)^2}{2\pi^3}\zeta(3)\leq 0.113 $$
and $f(x)$ is a contraction over $\mathbb{R}^+$. It follows that the limit of your sequence is the only positive solution of
$$ \frac{x^2}{\sinh^2(x)}=x^3-x^2+1 $$
which is clearly contained in $(0,1)$, and actually in $\left(\frac{2}{3},\frac{3}{4}\right)$, where the polynomial function is increasing and convex and $\frac{x^2}{\sinh^2(x)}$ is decreasing and concave. Few steps of Newton's method lead to the approximation
$$ \lim_{n\to +\infty} u_n \approx 0.6966915666. $$
