# 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.

• There are two different things: (1) to prove that the sequence converge to a unique value and (2) to find that value. I imagine that your are asked to work on (1). The solution $l >0$ is satisfying the equation $f(x)=x$ where $f(x)= 1/\tanh^2 x - 1/x^2$. Sep 27, 2019 at 7:47

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.

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