Question: Is there any way I can bound this series for fixed $x \in \mathbb{R}$?

Sequence is $a\_n = \sum\_{k=0}^n \frac{x^k}{k!}$ and I assumed here WLOG $n<m$ and let $m=n+k$ for some $k$.

$|\sum\_{i=n+1}^{n+k} \frac{x^i}{i!}| \leq |\frac{x^{n+1}}{(n+1)!}|+...+|\frac{x^{n+k}}{(n+k)!}| \leq ...$ 

Does anyone have an idea? I'm trying to show that it's Cauchy so I want to show that series is smaller than for any $\epsilon >0$ by choosing $N>0$.

I have no idea how to bound this by. Please help!! thanks!!!

The series itself kind of looks like the exponential for Taylor expansion but I'm not sure if that'll come in help here.. :(

Answer: 
Choose an integer $N$ such that $N > (2|x|)^{2}$. For $k >2(N-1)$ we have $\frac {|x|^{k}} {k!} < \frac {|x|^{k}} {N^{k-N+1}}$ because $k! >(1)(1)...(1)(N)(N)...(N)$ (where there are $(N-1)$ $1$'s). I leave it to you to show that $\frac {|x|^{k}} {N^{k-N+1}}< \frac 1 {2^{k}}$ for all $k >2(N-1)$. Finally use the fact that partial sums of $\sum \frac 1 {2^{k}}$ form a Cauchy sequence. Hint: $1-\frac N k +\frac 1 k > \frac 1 2$.
Starting with some fairly small $i$, we have $2^i < i!$, form then on, and forever:
$$
2^3 > 3!, \quad \mbox{ but } \quad
2^4 = 16, \; 4! = 24 \quad
2^5 = 32, \; 5! = 120 \quad, \ldots.
$$
Therefore, for $N \geq M \geq 4$,
$$
\sum\_{i = M}^{N}{|x|^i \over i!} \leq \sum\_{i = M}^{N}{|x|^i \over 2^i},
$$
and the latter series can be bounded by using the geometric series formula.