Proving that a recursive oscillating sequence is a cauchy-sequence Given $X_1 = 1$ and $X_n = 1 + \frac{3}{4 . X_{n-1}}$ for $n \geq 1$ Prove that this sequence is a Cauchy-sequence. I know that this sequence is converging to $1.5$ and I know that the sequence has two sub sequences $X_k$ where $k$ is an even number and $X_m$ where $m$ is an odd number. $X_k$ is falling and converging towards 1,5 $X_m$ is climbing and converging towards 1,5 I tried to prove that each sequence is converging using Induction but that didn't work. Any tips ? And how can I prove that the series is converging towards $1,5$ ?

First, we know that if $x_n$ has a limit $x$, it must satisfy $x = 1 + \frac3{4x}$, so that $x=\frac 32$ (the other root is negative, which we can disregard as we can show by induction that $x_n\geqslant1$ for all $n$).

Consider the subsequences $x_{2n}$ and $x_{2n-1}$, $n\geqslant 1$. For $n=1$, it is clear that $x_{2\cdot1-1}=x_1 = 1\leqslant \frac32$. Now assuming that $x_{2n-1}\leqslant\frac32$, we must show that $x_{2(n+1)-1}=x_{2n+1}\leqslant \frac 32$. We have \begin{align} x_{2n+1} &= 1 + \frac3{4x_{2n}}\\\ &= 1 + \frac3{4\left(1 + \frac3{4x_{2n-1}}\right)}\\\ &\leqslant 1 + \frac 3{4\left(1 + \frac3{4\cdot\frac32} \right)}\\\ &\leqslant 1 + \frac12\\\ &=\frac32. \end{align} I hope this provides enough guidance for you to finish the problem (use a similar argument to show that $x_{2n}\geqslant \frac32$, then show that $x_{2n-1}$ is increasing and that $x_{2n}$ is decreasing).