Numbers $a_i$, $b_i$ and $c_i$ satisfy $$a_i>0,\quad c_i>0,\quad b_i>a_i+c_i\,\,\,\text{for}\,\,\,i=1,2,\ldots,I-1$$ and $$u_i=\frac{c_i}{b_i-a_iu_{i-1}},\,\,\,i=1,2,\ldots,I-1\,\,\,\text{with}\,\,\,u_0=0.$$
Show by induction that $0<u_i<1$ for $i=1,2,\ldots,I-1$
I have worked through the following:
Take $i=1$: $$u_1 = \frac{c_1}{b_1 - a_1u_0} = \frac{c_1}{b_1}<\frac{c_1}{a_1+c_1}<1\,\,\text{ since $a_i>0$}.$$ A fraction of two positive numbers will always be greater than zero so $$0<u_1<1\,\,\text{ holds for $i=1$ case.}$$ Take $i=k$: Assume that $$0<u_k = \frac{c_k}{b_k - a_ku_{k-1}}<1\,\,\text{ is true.}$$ Take $i=k+1$: $$0<u_k = \frac{c_{k+1}}{b_{k+1} - a_{k+1}u_{k}}< \frac{c_{k+1}}{a_{k+1} + c_{k+1} - a_{k+1}u_{k}} = u_k = \frac{c_{k+1}}{a_{k+1}(1 - u_k) + c_{k+1}}$$ Now take the extreme cases: $$\LARGE 0<u_{k+1}=\begin{align*} \left\lbrace \begin{array}{r@{}l} \frac{c_{k+1}}{c_{k+1}} &\le 1\,\,\text{ if }\,\,u_k = 1 \\ \frac{c_{k+1}}{a_{k+1} + c_{k+1}} &< 1\,\,\text{ if }\,\,u_k = 0 \\ \end{array} \right. \end{align*}$$ So, since $0<u_k<1$ that then means that $u_{k+1}$ is strictly less that $1$ thus $$0<u_{k+1}<1\quad\text{QED.}$$
So is what I have done show the desired result properly? Can someone who is good at induction look this over for me?
The passage $k \rightarrow k+1$ is not very clear. Here is a better way to formulate it :
Suppose that $0 < u_k < 1$. Then, as $a_{k+1} > 0$, $$0 < a_{k+1} u_k < a_{k+1}$$ so $$b_{k+1} - a_{k+1} < b_{k+1} - a_{k+1}u_k < b_{k+1}$$ Moreover, $b_{k+1} - a_{k+1} > c_{k+1}$ so $$c_{k+1} < b_{k+1} - a_{k+1}u_k < b_{k+1}$$ You get finally, as $c_{k+1} > 0$ $$\frac{c_{k+1}}{b_{k+1}} < \frac{c_{k+1}}{b_{k+1} - a_{k+1}u_k} < \frac{c_{k+1}}{c_{k+1}}$$
so you have $$0 < u_{k+1} < 1$$