If a sequence $( x_ n )_{ n ∈ \mathbb{N}} \in X$ satisfies $d ( x _n ,x_{n +1 }) < α_ n$ , for all $n \in N$ , then the sequence is convergent.

Question

Let $(X,d)$ be a complete metric space and $(α_n)_{n \in > \mathbb{N}}$ a sequence of positive real numbers such that $\sum_{n=0}^{\infty} \alpha_n$ converges. Show that if a sequence $(x_n)_{n \in N} \in X$ satisfies $d ( x _n ,x_ {n +1} ) < α_ n$ , for all $n \in N$ , then the sequence is convergent.

Answer 1

Note that $\sum_{n=0}^\infty \alpha_n$ converges if and only if $\forall \varepsilon > 0\;\exists N\in\mathbb{N}\;\forall n\geq m\geq N: \sum_{k=m}^n \alpha_n < \varepsilon$, and that $d(x_{n+m},x_n) \leq \sum_{k=n}^{m-1} d(x_{k+1},x_k)$.