Find limit of recurrent sequence?

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 $$ \lim_{n\to +\infty} u_n \approx 0.6966915666. $$

Assuming that the limit $x$ exists, it is the solution of $$\coth ^2(x)-\frac{1}{x^2}-x=0$$ A quick look at the graph shows that the function is very linear and this is extremely good for any numerical method.

Using Taylor expansion around $x=0$ would give $$\coth ^2(x)-\frac{1}{x^2}-x=\frac{2}{3}-x+\frac{x^2}{15}+O\left(x^4\right)$$ Ignoring the higher order terms, an approximation of the root is $$x=\frac{15-\sqrt{185}}{2}\approx 0.699265$$ while, as already given in Jack d'Aurizio's answer, the solution is $0.696692$.

Using the approximation as $x_0$, Newton iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.69926474563227832749 \\ 1 & 0.69669127573131431017 \\ 2 & 0.69669156663855445854 \\ 3 & 0.69669156663855818578 \end{array} \right)$$

If, instead of Taylor series, we build the $[1,n]$ Padé approximants, the following sequence of rational numbers is generated $$\left\{\frac{2}{3},\frac{30}{43},\frac{86}{123},\frac{77490}{111217},\frac{222434}{31 9353},\frac{3193530}{4583941},\frac{9167882}{13158957},\frac{91191572010}{13089201 1709},\frac{261784023418}{375753454245}\right\}$$ For the fun of it, the $40^{\text{th}}$ term of the sequence is

$$\frac{6071119921652944530433232700069308841296441468886808607094226134858925518242} {8714214743470012630112071332705424737414514747250519055627479840496085107033}$$ The next would not fit in the page.

Inverse symbolic calculators do not find anything for this number.