Let $B_n$ be a sequence such that $$\sum_{n=0}^\infty B_n$$ converges.
Let $(x_n)$ be a sequence in some metric space $(X, d)$ such that $d(x_n, x_{n+1})$ $\le$ $B_n$ for each $n$, then the sequence $(x_n)$ is Cauchy. If the sequence converges to $a \in X$ then is...? $$d(x_n, a) \le \sum_{k=n}^\infty B_k \textrm{ for every } n \in \Bbb N$$
This seems intuitive if we pick $n$ large since the tails of the series converges to zero, but what about the first few terms in the sequence? These could be large values and these large values need not be less than the sum of the series for small $n$. We would only need $d(x_n, a)$ to be less than the series at some value of $n$, at which point all the subsequent values will be less than the series for all $n \ge N$ (some $N \in \Bbb N$) and hence the sequence converges.
Supposedly the answer is for all $n \in \Bbb N$. What am I not seeing here?
Let $\ \epsilon>0.\ $ Then there exists natural $\ M\ >\ n\ $ such that $\ d(x_{_M}\ a) < \epsilon.\ $ Then
$$ d(x_n\ a)\ <\ d(x_n\ x_M) + \epsilon\ \le \ \sum_{k=n}^{M-1}\,B_k\ +\ \epsilon \ \le\ \sum_{k=n}^\infty\ B_k\ +\ \epsilon $$ hence $$ \forall_{\epsilon > 0}\quad d(x_n\ a)\ < \ \sum_{k=n}^\infty\ B_k\ +\ \epsilon $$ hence $$ d(x_n\ a)\ \le\ \sum_{k=n}^\infty\ B_k $$ Done.