Vtyjkyuk

Let $(a_n)$ be the sequence defined by $a_1 = 1$ and $a_n+1 = frac13+a_n$ for all $n in mathbb N.$ Show that $(a_n)$ converges and find its limit.

I have no idea on how to handle this though so I would definitely appreciate some help for a rigorous proof.

First we note that $a_nin (0,1/3)$ for $n>1$.

Since
$$a_n+1-a_n=frac13+a_n-frac13+a_n-1=fraca_n-1-a_n(3+a_n)(3+a_n-1),$$
and $(3+a_n)(3+a_n-1)>9$, we see that
$$|a_n+1-a_n|<frac19 |a_n-a_n-1|.$$
This shows that $a_n$ is Cauchy, so it converges to some number $ain[0,1/3]$. To find $a$, take limits on both sides of
$$a_n+1=frac13+a_n.$$

Using contraction mapping technique. We will look at the function $f(x)=frac13+x$ s.t. $a_n+1=f(a_n)$.

1. $forall x in left[0,1right] => f(x) in left[0,1right]$. Indeed
   $$0leq xleq 1 Rightarrow 3leq 3+xleq 4 Rightarrow frac13geq frac13+xgeq frac14>0$$ or $$1geq f(x)geq 0$$


2. $f(x)$ is a contraction mapping on $left[0,1right]$, from MVT $forall x,y in left[0,1right], exists c$ in between them s.t.
   $$|f(x)-f(y)|=|f'(c)||x-y|=left|-frac1(3+c)^2right||x-y|leqfrac19|x-y|$$

Since $a_1 in left[0,1right]$, from Banach fixed-point theorem, the sequence has a limit on $left[0,1right]$ which you can find from $L=frac13+L$.