Prove $(x_n)=\sum_{k=0}^{n}\frac{x^k}{k!}$ is a Cauchy sequence If $(x_n)=\sum_{k=0}^{n}\frac{x^k}{k!}$, prove that $(x_n)$ is a Cauchy sequence.  Please give me a hint at how to proceed
 A: For a $n>N,m>M$ it must hold $|x_n-x_m| < \epsilon$ for a $\epsilon > 0$.
$|x_n-x_m| = |\sum_{k=m}^n \frac{x^k}{k!}|\leq \sum_{k=m}^n|\frac{x^k}{k!}|<\frac{(n-m)\max(x_n)^k}{m!}$ 
You can choose $\epsilon = \frac{(n-m)\max(x_n)^k}{m!} > 0$ that tends to Zero if $m>M$ becomes very very large.
A: If $x = 0$, the sequence is the zero sequence, which is Cauchy. So assume $x \neq 0$. By the Archimedean property, there exists a positive integer $N$ such that 
$$\frac{1}{N} < \frac{1}{2|x|}.$$
If $k > N$, then $\frac{1}{k} < \frac{1}{N} < \frac{1}{2|x|}$, which implies $\frac{|x|}{k} < \frac{1}{2}$. So for $k > N$,
$$\frac{|x|^k}{k!} = \frac{|x|}{k}\cdot\frac{|x|^{k-1}}{(k-1)!} < \frac{1}{2}\frac{|x|^{k-1}}{(k-1)!} < \frac{1}{2^2}\frac{|x|^{k-2}}{(k-2)!}< \cdots < \frac{1}{2^{k-N}}\frac{|x|^N}{N!}.$$
Therefore, if $m > n > N$,
$$|x_m - x_n| \le \sum_{k = n}^{m-1} |x_k - x_{k-1}| = \sum_{k = n}^{m-1} \frac{|x|^k}{k!} < \frac{|x|^N}{N!}\sum_{k = n}^{m-1} \frac{1}{2^{k-N}} \le \frac{|x|^N}{N!}\left(\frac{1}{2}\right)^{n-N-1}\tag{*}$$
Show that if $\varepsilon > 0$, there exists a positive integer $N'$ such that the rightmost expression of $(*)$ is less than $\varepsilon$ for all $n > N'$. Then if $m >  n > \max\{N,N'\}$, $|x_m - x_n| < \varepsilon$. This shows that $\{x_n\}$ is Cauchy.
