Prove that $1+\frac{1}{2!}+..+\frac{1}{n!}$ is a Cauchy sequence My attempt: 
We need to show that $|xn-xm|<e$  for all $m,n \geq H(e) $ By archimedian property we can always have an H such that He>1 and $(1/H)<e$ we chose n and m such that m=(n+p) where p is a natural number . 
$|1+\frac{1}{(2)!}+.. +\frac{1}{(n)}! -1-\frac{1}{(2)!}..-\frac{1}{(n+p)!}| = |\frac{1}{(n+1)!}+\frac{1}{(n+2)!}...+\frac{1}{(n+p)!}| < |\frac{n}{(n+1)!}|  = \frac{1}{((n-1)!(n+1))}<\frac{1}{(n+1)}<\frac{1}{n}$ now we know that $n\geq H$ so $1/n<1/H<e$ there fore the above criteria holds and it's a cauchy sequence  . Is my attempt correct?
 A: We know that the difference between any two terms in the sequence is $|\frac{1}{n!}-\frac{1}{m!}|$. Suppose without loss of generality that $n\leq m$, then since the sequence is monotonically decreasing, this suggests
$|\frac{1}{n!}-\frac{1}{m!} |\leq |\frac{1}{n!}-\frac{1}{(n+1)!}|= |\frac{1}{n!}-\frac{1}{n!(n+1)}| = |\frac{1}{n!} (1-\frac{1}{n+1})|$
Thus $\forall \epsilon>0,$ let $N(\epsilon)$ satisfy $|\frac{N!}{N+1}|\geq \frac{1}{\epsilon}$ s.t. if $n,m\geq N(\epsilon)$, then $|\frac{1}{n!}-\frac{1}{m!}| \leq \epsilon$ so that it is a Cauchy sequence. 
A: $m=n+p;$
$|S_m-S_n|= \sum_{k=n+1}^{n+p}1/(k!)\lt$
$\dfrac{1}{(n+1)!}(1+ \dfrac{1}{n+2} +$
$\dfrac{1}{(n+3)(n+2)} +$
$\dfrac{1}{(n+2)(n+3)....(n+p)})\lt $
$\dfrac{1}{(n+1)!}(1+\dfrac{1}{n+2}+$
$ \dfrac{1}{(n+2)^2}+..\dfrac{1}{(n+2)^{p-1}}) \lt$
$ \dfrac{1}{(n+1)!}(1+(1/2)+(1/2)^2 +..(1/2)^{p-1}) \lt 2\dfrac{1}{(n+1)!}\lt 2/n$.
Given $\epsilon >0$;
Archimedean principle:
There is a $n_0 \ge \ 2/\epsilon$ s.t.
for $m\ge n\ge n_0$ :
$|S_m-S_n| \lt 2/n \le 2/n_0 <\epsilon$.
Used: $1+(1/2)+(1/2)^2.......=2$(Infinite geometric series)
