Prove that $1+\frac{1}{2!}+..+\frac{1}{n!}$ is a Cauchy sequence

Question

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?

Answer 1

$m=n+p;$

$|S_m-S_n|= \sum_{k=n+1}^{n+p}1/(k!)\lt$

$\dfrac{1}{(n+1)!}(1+ \dfrac{1}{n+2} +$

$\dfrac{1}{(n+3)(n+2)} +$

$\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)

Answer 2

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.