Let $(X,d)$ be a complete metric space and $T:X\to X$ be a map such that for $x\in X$ there exists a sequence $(a_n(x))\in [0, \infty)$ such that
(A) $\lim _{n\to \infty} a_n(x)=a_{\infty}(x)<1$ and
(B) $d(T^n(x), T^n(y))\le a_n(x) d(x,y)$ , for all $y\in X$
show that $T$ has a unique fixed point and $\{T^n(x)\}$ converges to the unique fixed point for $x\in X$
I am trying to that the sequence is cauchy sequence but i am not get any idea
and i sir told me that use $\lim _{n\to \infty} \frac{a_{n+1}}{a_n}=l<1$ or $\lim _{n\to \infty} (a_n)^\frac{1}{n}=l<1$
what i have tried is
We have to prove that $\{T^nx\}$ is cauchy sequence
$d(T^mx,T^nx)\le \sum_{i=n}^{m-1}d(T^i(x),T^{i+1}(x))$ by triangle inequality
$ \le \sum_{i=n}^{m-1}a_i(x)d(x,Tx)$
if we show that this is less than $\infty$ then this is Cauchy but how to prove this