Sign permanence of locally Lipschitz functions calculated on a sequence

Suppose I have a sequence $a_k(m)>0$ with $m\in\mathbb{N}$ such that, given $k\in\mathbb{N}$ and $p\geq 1$, I can show that $$|a_{k+p}(m)-a_k(m) |\leq \frac{m^2}{k^2}$$ with $a_{k+p}<a_{k}$. I also know that $$\lim_{k\to\infty} a_{k}(m) = 0$$Wich implies that $a_{k}(m) $ is a Cauchy sequence.

Now, suppose I have a locally Lipschitz (continuous) function $f_n(a_k(m))\equiv f_n(k,m)$ such that:$$|f_n(k+p,m)-f_n(k,m)|\leq \Lambda_n\, |a_{k+p}(m)-a_k(m)|,\qquad \Lambda_n>0$$

Suppose I know that $0<f_n(0,m)\leq 1$ for every $n,m\in\mathbb{N}$ and that $$\lim_{k\to\infty} f_{n}(k,m) = 0,\qquad \forall n,m\in\mathbb{N}$$I would like to use continuity and induction to prove that $0<f_n(k,m)\leq 1$ for every $k$.

I assume it is true for $f_n(k,m)$ for some $k>0$ and should try to prove it for $k+1$.Since $a_{k}(m)$ is Cauchy, there exists a $K\in\mathbb{N}$ such that, for $k>K$:$$|a_{k+1}(m)-a_k(m) |\leq \frac{m^2}{k^2}\leq \frac{m^2}{K^2}\leq \delta$$So I can alway find a $K =\frac{m}{\sqrt{\delta}}$.When this is true I can write:$$|f_n(k+1,m)-f_n(k,m)|\leq \Lambda_n\delta\leq \epsilon$$If I choose $\epsilon = \frac{f_n(k,m)}{2}<1$ I could prove that$$f_n(k+1,m)>0$$and if I choose $\epsilon = \frac{1-f_n(k,m)}{2}<1$ (which i can do by inductive hyptohesis) I can prove that$$f_n(k+1,m)<1$$the question is, can I do it?

I mean, I think $\delta$ must be such that$$\delta\leq \frac{\epsilon}{\Lambda_n} \leq\frac{1}{2\Lambda_n}$$but this translates to:$$\frac{m^2}{K_n^2}\leq\frac{1}{2\Lambda_n}\to K_n\geq\sqrt{2\Lambda_n}m$$is it enough to prove that I can always find a $K_n$ for every $n$?