How do I prove $B_n$ is a Cauchy sequence given $$|B_n - B_{n-1}| < \frac{1}{n!}$$ I can prove that that the above approaches $0$ as $n$ becomes large since $n! > 2^n$ and thus setting $N = \log_2\frac{1}{\epsilon}$ but don't know how to relate it to the definition of a Cauchy sequence:
$\forall \epsilon > 0, \exists N \in \mathbb{N}$ such that $n,m> N$ $\Rightarrow |B_n - B_m| ≤ \epsilon$
Thanks
Note that if $m>n\geq 1$ then $$|B_m - B_{n}| \leq |B_m - B_{m-1}| +\dots+|B_{n+1} - B_{n}| <\sum_{k=n+1}^{m}\frac{1}{k!}\leq \sum_{k=n+1}^{m}\frac{1}{2^{k-1}}<\sum_{k=n+1}^{\infty}\frac{1}{2^{k-1}}$$ where $2^{k-1}\leq k!$ for $k\geq 2$. Can you evaluate the last (geometric) series and finish the job?