I have to prove from definition that the following sequence is Cauchy:
${1+\frac{1}{1!}+\frac{1}{2!}...+\frac{1}{n!}}$.
Definition of Cauchy sequence: A sequence $(a_n)$ is said to be a Cauchy sequence if given $\epsilon>0$, however small, there exists a $m\in N$ such that $|a_{m+p}-a_p|$<$\epsilon$ for all $n\geq m$ and for $p=1,2,3...$
Hint given in textbook is : $(n+1)!\geq 2^n$
So should I take $p=1?$ That would give me ${1+\frac{1}{1!}+\frac{1}{2!}...+\frac{1}{n!}+\frac{1}{(n+1)!}}$.
Then,
$|u_{n+1}-u_n|=\frac{1}{{(n+1)}!}\leq \frac{1}{2^n}$
You need to prove it for any $p$, not just $p=1$, but you're on the right track.
$$|u_{n+p} - u_n| = \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots + \frac{1}{(n+p)!} \le \frac{1}{2^n} + \frac{1}{2^{n+1}} + \cdots + \frac{1}{2^{n+p-1}} = \frac{1}{2^n} \frac{1-2^{-p}}{1-2^{-1}} \le \frac{1}{2^{n-1}}.$$
So, given $\epsilon$, can you choose $m$ such that $|u_{n+p} - u_n| < \epsilon$ for any $n \ge m$ and any $p$?