Let $(X,d)$ be a metric space and let $(\beta_n)_{n\in\mathbb{N}}$ be a sequence in $\mathbb{R}_{\geq 0}$ such that $\sum\limits_{n=0}^{\infty}\beta_n$ converges. Let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $X$. Prove that if $d(x_n,x_{n+1})\leq \beta_n$ for all $n\in\mathbb{N}$ then $(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence.
How could I prove this? I tried using the fact that the tails of the infinite series converge to $0$ i.e. $$\lim\limits_{n\rightarrow\infty}\sum\limits_{k=n}^\infty\beta_k=0$$ and somehow relate that to the sequence $(x_n)_{n\in\mathbb{N}}$ but I am still a little confused on how to proceed from there.
$d(x_n,x_{n+m}) \leq d(x_n,x_{n+1})+d(x_{n+1},x_{n+2})+...+d(x_{n+m-1},x_{n+m})\leq \beta_n+\beta_{n+1}+...\beta_{n+m-1}$. Since $\sum \beta_n$ converges $\beta_n+\beta_{n+1}+...\to 0 $.