let $(u_n)$ be a sequence defined as :
$$ \begin{cases} u_{n+1}=3-\frac{10}{u_n+4} ,& 0<u_n<1\\ u_0=\frac{1}{4} \end{cases} $$ The above sequence is increasing by using the difference :$ u_{n+1}-u_n$ , really i have tried to look if the ratio $\frac{u_{n+1}}{u_n}$ greater than $1$, but i don't succeed ,Then is it possible to show that is increasing sequence using the ratio $\frac{u_{n+1}}{u_n}$ without using the difference:$ u_{n+1}-u_n$ ?
Using the inequality given: $$0<u_{n}<1$$ $$4< u_n + 4< 5$$ $$\frac{1}{5}< \frac{1}{u_{n} + 4}$$ $$\therefore 1< \frac{5}{u_{n} + 4}$$ Now using the given formula divide by $u_{n}$ and simplify: $$u_{n+1} = 3 - \frac{10}{u_n +4}$$ $$\frac{u_{n+1}}{u_n} = \frac{3u_n + 2}{u_n(u_n + 4)}$$ Working with inequalities again: $$0<u_{n}<1$$ $$2u_{n}<2$$ $$5u_n< 3u_n +2$$ $$\frac{5u_n}{u_n(u_n + 4)} < \frac{3u_n + 2}{u_n(u_n + 4)}$$ $$\frac{5}{u_n + 4} < \frac{3u_n + 2}{u_n(u_n + 4)}$$ $$1< \frac{3u_n + 2}{u_n(u_n + 4)} \:\:\: \because 1< \frac{5}{u_{n} + 4}$$ $$\therefore 1< \frac{u_{n+1}}{u_n}$$ This is a basic manipulation. This can be proved with other more efficient methods.