I'm trying to solve the following problem:
Given a non-constant sequence $a_n$ such that for all $n\in \mathbb{N}$, $\left|a_{n+1}-a_n\right|<q^n$ ($q>0$), prove that $a_n$ necessarily converges if and only if $q<1$
One direction I've proven successfully:
Let $0<q<1$. Using the given data, it can be shown that, for $m>n$: $$\:\left|a_m-a_n\right|\:=\:\left|\left(a_m-a_{m-1}\right)+\left(a_{m-1}+a_{m-2}\right)+...+\left(a_{n+1}-a_n\right)\right|\le \:\left|a_m-a_{m-1}\right|+\left|a_{m-1}+a_{m-2}\right|+...+\left|a_{n+1}-a_n\right|\le \:q^{m-1}+q^{m-2}+...+q^n=_{geometric\:progression\:sum}\frac{q^n\left(1-q^{m-n}\right)}{1-q}=\frac{q^n-q^m}{1-q}$$ Since $0<q<1$, it is easily verifiable that $\frac{q^n-q^m}{1-q}$ is always positive. Let $b_n=\frac{q^n}{1-q}$, so: $$\left|a_m-a_n\right|\le \frac{q^n-q^m}{1-q}=\left|\frac{q^n-q^m}{1-q}\right|=\left|b_n-b_m\right|$$ Since $0<q<1$, $\lim _{n\to \infty }\:q^n=0$, and simple arithmetic will yield that $\lim _{n\to \infty }\:b_n=0$ as well. $b_n$ is convergent, therefore by the Cauchy criterion: $$\forall ϵ>0,\:\exists > N\in \mathbb{R},\:\forall n,m>N,\:\left|b_n-b_m\right|<ϵ$$ And so: $$\left|a_m-a_n\right|\le \left|b_n-b_m\right|<ϵ$$ And $a_n$ is convergent by the Cauchy criterion.
The other direction (convergence yields $q<1$) is the one I'm still struggling with. I thought about trying contradiction - assuming $q>1$ and showing that $a_n$ diverges (with $q=1$ as a special case to be handled separately, as it renders impossible the use of the geometric sum formula I used earlier), but things don't develop in a satisfactory manner. Any help or guidance for that direction would be much appreciated.