Consider the sequence $\left\{\dfrac{1}{2}, \dfrac{1}{2+\frac{1}{2}}, \dfrac{1}{2+\dfrac{1}{2+\frac{1}{2}}}, ...\right\}$
This is the continued fraction for $\sqrt{2} - 1$.
I was trying to prove that $x_n>x_{n+1}$ ie decreasing.
So I came up with recursive formula $x_{n+1}= \frac{1}{2+x_n} \implies x_n=\frac{1}{x_{n+1}}-2$
And here is my attempt to prove that it is decreasing:
for n=1, $x_1$=$\frac{1}{2}$, $x_2$=$\frac{2}{5}$ so this checks out
Now suppose its true for "n"
so if $x_n$ > $x_{n+1}$
$\implies$ $\frac{1}{x_{n+1}}$-2>$x_{n+1}$ from the recursive formula.
$\implies$$\frac{1}{x_{n+1}}$>$x_{n+1}$+2
$\implies$$\frac{1}{x_{n+1}+2}$>$x_{n+1}$
$\implies$$x_{n+2}$>$x_{n+1}$
Which is pretty much the opposite of what I was trying to prove...
Hints would be very appreciated so that I can retain my own sanity!
Actually, $\langle x_n\rangle$ is neither decreasing nor increasing but it's "especial" subsequences i.e. $\langle x_{2n}\rangle$ and $\langle x_{2n+1}\rangle$ both are decreasing and have the same limit $\sqrt{2}-1$. Therefore , $\langle x_n\rangle $ converges to $\sqrt{2}-1$.