Application of Banach Fixed Point Theorem to sequences

Question

I have the following question

Show that there is unique real bounded sequence $(a_n: n \in \mathbb{N})$ such that $$a_n = \frac{n+1}{n}+\sum^\infty_{m=1}\frac{\sqrt{3a^2_{m+n}+1}}{4m^2}$$ for all $n \in \mathbb{N}$.

I know I need to use the Contraction/Fixed Point Theorem, so I first defined $\Phi : l_{\infty}(\mathbb{N}) \to l_{\infty}(\mathbb{N})$ by

$$\Phi(a)_n=\frac{n+1}{n}+\sum^\infty_{m=1}\frac{\sqrt{3a^2_{m+n}+1}}{4m^2}$$ where $a\in l_{\infty}(\mathbb{N})$ and $n \in \mathbb{N}$.

But I got stuck in my attempt to show $\Phi$ is a contraction. My last line of work is

$$|\Phi(a)_n - \Phi(b)_n| \leq \cdots \leq \left|3 \sum^{\infty}_{m=1} \frac{a^2_{n+m}-b^2_{n+m}}{4m^2\left(\sqrt{3a^2 _{n+m}+1}+\sqrt{3b^2_{n+m}+1}\right)}\right|$$ Thanks in advance

Answer 1

Just to finalize John Ma's idea.

Note the sequence of inequalities $$ \begin{align} |\Phi(a)_n - \Phi(b)_n| &\leq \left|3 \sum^{\infty}_{m=1} \frac{a^2_{n+m}-b^2_{n+m}}{4m^2\left(\sqrt{3a^2 _{n+m}+1}+\sqrt{3b^2_{n+m}+1}\right)}\right|\\ &\leq \left|3 \sum^{\infty}_{m=1} \frac{a^2_{n+m}-b^2_{n+m}}{4m^2\left(\sqrt{3a^2 _{n+m}}+\sqrt{3b^2_{n+m}}\right)}\right|\\ &\leq \left|3 \sum^{\infty}_{m=1} \frac{(a^2_{n+m}-b^2_{n+m})}{4m^2\sqrt{3}(|a_{n+m}|+|b_{n+m}|)}\right|\\ &\leq \left| \sum^{\infty}_{m=1} \frac{\sqrt{3}}{4m^2}(|a_{n+m}|-|b_{n+m}|)\right|\\ &\leq \sup_m|a_{n+m}-b_{n+m}|\sum^{\infty}_{m=1} \frac{\sqrt{3}}{4m^2}\\ &\leq \frac{\pi^2\sqrt{3}}{24}\sup_m|a_{n+m}-b_{n+m}| \\ \end{align} $$ Then $$ \Vert \Phi(a) - \Phi(b)\Vert \leq\frac{\pi^2\sqrt{3}}{24}\sup_n\sup_m|a_{n+m}-b_{n+m}| \leq\frac{\pi^2\sqrt{3}}{24}\Vert a-b\Vert $$ Since $\alpha=\frac{\pi^2\sqrt{3}}{24}<1$ we may apply Banach fixed point theorem.