Comparing Infinite Series I want to prove that if $0 \le |x| < a$ then:
$$
\sum\limits_{n=1}^{\infty} \frac{|x|^{n-1}}{n!} < 
\sum\limits_{n=1}^{\infty} \frac{a^{n-1}}{n!}
$$
I initially attempted to use induction to prove that 
$\forall n > 1, \frac{|x|^{n-1}}{n!} < \frac{a^{n-1}}{n!}$,
but then learned that it is not the appropriate way of proving this.
Am I even correct to assume that the above statement can be proven by showing that $\frac{|x|^{n-1}}{n!} < \frac{a^{n-1}}{n!}$ holds $\forall n > 1$? What is the proper way of approaching this?
Does it suffice to show that $\lim\limits_{n \to \infty} \frac{|x|^{n-1}}{a^{n-1}} < 1$?
Thanks in advance.
 A: Remember that $<$ and $>$ become $≤$ and $≥$ (respectively) when passing to limits, that is the strictness of the inequality is lost. For example consider $1/n<1/2n$ for all $n$ yet they both tend to $0$. If you want to prove that your series are $≤$ then it is enough to know that $0≤|x|<a$. 
Hint: If you wanted to see that the inequality is strict you could note that your series are very close to the Taylor expansion for $e^x$, and that the exponential function is strictly increasing (and hence injective). 
A: first, it's easy to show that $\forall n>1, |x|^{n-1} < a^{n-1}$
You then have for all N,
$$\sum_{n=3}^N \frac{|x|^{n-1}}{n!} \leq \sum_{n=3}^N \frac{a^{n-1}}{n!}$$
Taking this to the limit, you get
$$\sum_{n=3}^{\infty} \frac{|x|^{n-1}}{n!} \leq \sum_{n=3}^{\infty} \frac{a^{n-1}}{n!}$$
Then, as $1+ \frac{|x|}{2} > 1+ \frac{a}{2}$, you get
$$1+ \frac{|x|}{2} + \sum_{n=3}^{\infty} \frac{|x|^{n-1}}{n!} < 1 + \frac{a}{2} + \sum_{n=3}^{\infty} \frac{a^{n-1}}{n!}$$
i.e.
$$\sum_{n=1}^{\infty} \frac{|x|^{n-1}}{n!} < \sum_{n=1}^{\infty} \frac{a^{n-1}}{n!}$$
