Analysis Challenge Question using Cauchy sequences in $\mathbb{R}^p$. Suppose that $S=(\bar{x}_n)$ is a sequence in $\mathbb{R}^p$. Assume that $$ \left\|\bar{x}_{n+1}-\bar{x}_n\right\|_2 \leq \frac{1}{n!}\left\|\bar{x}_n-\bar{x}_{n-1}\right\|_2$$ for every natural number n $\geq$ 2. Prove that there exists a vector $\bar{a}\ \epsilon\ \mathbb{R}^p$ so that S converges to $\bar{a}$.

We are expected to first show that this is a Cauchy sequence and use the theorem to then say that is converges to some $\bar{a}\ \epsilon\ \mathbb{R}$. My professor also told us that we have to use the fact that $\sum_{n=1}^{\infty}\frac{1}{n!} = e-1$.

I've tried starting off by considering some $m,n\ \epsilon\ \mathbb{N}$ with $m,n\geq2$ and w.l.o.g. say that $m\geq n$.
Then consider $\left\|\bar{x}_m-\bar{x}_n\right\|_2 = \left\|\bar{x}_m-\bar{x}_{m-1}+\bar{x}_{m-1}-...-\bar{x}_{n+1}+\bar{x}_{n+1}-\bar{x}_n\right\|_2$
$$\leq \left\|\bar{x}_m-\bar{x}_{m-1}\right\|_2+\left\|\bar{x}_{m-1}-\bar{x}_{m-2}\right\|_2+...+\left\|\bar{x}_{n+2}-\bar{x}_{n+1}\right\|_2+\left\|\bar{x}_{n+1}-\bar{x}_n\right\|_2$$.
I then tried using the inequality above that relates the next to terms to the previous ones so that all the Euclidean norms above are reduced to the sum of the same Euclidean norm but with different coefficients:$$\leq \frac{1}{(m-1)!(m-2)!...(n+2)!(n+1)!}\left\|\bar{x}_{n+1}-\bar{x}_n\right\|_2+\frac{1}{(m-2)!(m-3)!...(n+2)!(n+1)!}\left\|\bar{x}_{n+1}-\bar{x}_n\right\|_2+...+\frac{1}{(n+1)!}\left\|\bar{x}_{n+1}-\bar{x}_n\right\|_2+\left\|\bar{x}_{n+1}-\bar{x}_n\right\|_2$$.
But this is where I'm not sure how to proceed, specifically with using the given convergent series to show that the sequence is Cauchy.
 A: If $k>1$ and $\|x_{n+1}-x_n\|\le (1/k)\|x_n-x_{n-1}\|$ whenever $n\ge 2$ then by induction on $n\ge 2$ we have $$\|x_{n+1}-x_n\|\le (1/k^{n-1})\|x_2-x_1\|$$ whenever $n\ge 2.$
Now if $2\le a<b$ then $$\|x_a-x_b\|\le\sum_{j=0}^{b-a-1}\|x_{a+j}-x_{a+j+1}\|\le$$ $$\le \sum_{j=0}^{b-a-1}(1/k^{a+j-1})\|x_2-x_1\|\le $$ $$\le\sum_{j=0}^{\infty}(1/k^{a+j-1})\|x_2-x_1\|=$$ $$=(1/k^{a-1})\frac {\|x_2-x_1\|}{k-1}.$$ So if $\epsilon >0$ and $n\ge 2$ is large enough that $(1/k^{n-1})\frac {\|x_2-x_1\|}{k-1}<\epsilon$ then $$n\le a<b\implies \|x_a-x_b\|\le (1/k^{a-1})\frac {\|x_2-x_1\|}{k-1}\le$$ $$\le (1/k^{n-1})\frac {\|x_2-x_1\|}{k-1}<\epsilon.$$
A: As mentioned by Kavi Rama Murthy in the comment section, if a series $\sum a_n$ converges, then $\sum\limits_{k=m}^{n} a_k \to 0$ as $m,n\to \infty$. Since
$$\sum\limits_{n>0}\frac{1}{n!} = e-1, $$
the series
$\frac{1}{(m-1)!}+\frac{1}{m!}+\dots+\frac{1}{n!}$ $\to 0$ as $m,n\to \infty$.
Now, as you have little bit noted down the iterarion process, can you prove the following?
$$ ||\overline{x_{n+1}} - \overline{x_n}||_2 \leq \frac{1}{n!(n-1)!\dots 2!}||\overline{x_2}-\overline{x_1}||_2\leq \frac{1}{n!}||\overline{x_2}-\overline{x_1}||_2.$$
It is now easy for us to show that the given sequence is Cauchy. Just observe that for $m>n$,
$$||\overline{x_m} - \overline{x_n}||_2 \leq \left(\sum\limits_{k=n}^{m-1} \frac{1}{k!}\right)||\overline{x_2} - \overline{x_1}||_2.$$
