Prove that series stays on an interval using induction Hello fellow mathematicians,
I am currently working on induction and fell on this example :
$U_0 = 0$
$U_{n+1} = \frac {3+2U_n}{5+2U_n}$
Knowing that $n\,\epsilon\ R$ prove that $\frac 35 \le U_n \le \frac57$
How should I do it ? I tried it but didn't manage to get very far. I think I am missing how to start the $P_n \Rightarrow P_{n+1}$ process. Of course I already initialized the induction but that is the trivial part.
 A: Write
$$
U_{n+1} = 1 - \frac {2}{5+2U_n} = f(U_n)
$$
where $f(t) = 1 - \frac {2}{5+2t}$ is an increasing function for $t > 0$. Therefore
$$
 \frac 35 \le U_n \le \frac 57 \implies f(\frac 35) \le U_{n+1} \le f(\frac 57)
$$
and it remains to show that $f(\frac 35) \ge \frac 35$ and $f(\frac 57) \le \frac 57$.
A: I assume you want to see a valid induction step.
As you already mentioned, the induction start is easy:
$$\frac{3}{5} \leq U_1 = \frac{3}{5} \leq \frac{5}{7} \quad\mbox{true!}$$
Induction step assuming the induction hypothesis 


*

*(IH): $\frac{3}{5} \leq U_n  \leq \frac{5}{7}$ for an $n\in \mathbb{N}$.


To show is
\begin{eqnarray*}
&\frac{3}{5} \leq \frac{3+2U_n}{5+2U_n}  \leq \frac{5}{7}& \\
& \Leftrightarrow & \\
&15 + 6U_n \leq 15 + 10U_n  \mbox{ and }  21 + 14U_n \leq 25 + 10U_n &\\
& \Leftrightarrow & \\
&0 \leq U_n \mbox{ and } U_n \leq 1 \quad\mbox{which is true according to IH}&
\end{eqnarray*}
A: Hint: Compute $$\frac{5}{7}-\frac{3+2U_n}{5+2U_n}=\frac{4}{7}\times \frac{1-U_n}{5+2U_n}$$ and $$U_n-\frac{3}{5}=\frac{4}{5}\times \frac{U_n}{5+2U_n}$$
