Use the Banach fixed point theorem to show that
the following sequence converges. What is the limit of this
sequence?
$$\left(\frac{1}{3},
\frac{1}{3+\frac{1}{3}},
\frac{1}{3+\frac{1}{3+\frac{1}{3}}},
\dots\right)$$
I inferred the closed form of the sequence
$$\begin{align}
x_0 &= \frac{1}{3} \\
x_n &= \frac{1}{3 + x_{n-1}}
\end{align}$$
and then the function that would, I assume, be that which would be applied repeatedly to converge the sequence (per the Banach fixed point theorem)
$$f(x)=\frac{1}{3+x}$$
That's about as far as I know to go for sure, but I assumed then that I needed to prove $f(x)$ was a contraction mapping (as the hypothesis of the Banach fixed point theorem dictates), that is
$$d(f(x),f(y)) \leq r d(x,y)$$
where $r \in [0, 1)$, and so
$$ \begin{align}
\frac{1}{3+x} – \frac{1}{3+y} &\leq r(x – y) \\
\frac{1}{(x – y)(3 + x)} – \frac{1}{(x – y)(3 + y)} &\leq r \\
\frac{-1}{(x + 3)(y + 3)} &\leq r
\end{align}$$
Although I don't think the above proves anything useful at all. (Unless $(x+3)(y+3) < -1$, where it is shown that $\exists r \in [0,1) : \dots$.)
Best Answer
You're on the right track (though you forgot the rather crucial absolute value signs in your inequality). Try showing that $f$ is a contraction map on some subinterval of $\Bbb R$. (It isn't even defined at $-3$, so we have to consider a proper subinterval, anyway.) Since we're taking our initial point to be $\frac13$, we might as well consider $[0,\infty)$. We could instead use $[\alpha,\infty)$ for some $-3<\alpha<0$, if we liked, but $0$ works just fine.
Show that if $x\in[0,\infty)$ then $f(x)\in[0,\infty).$ Then for all $x,y\in[0,\infty)$, you can see that $$\left|\frac1{3+x}-\frac1{3+y}\right|=\left|\frac{y-x}{(3+x)(3+y)}\right|=\frac1{(3+x)(3+y)}|x-y|\le r|x-y|,$$ where $r=???$