2

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.

Community
  • 1
Radix
  • 123

2 Answers2

1

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).

Reveillark
  • 13,044
1

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.

$$\sum_{n=1}^{\infty} \frac{|x|^{n-1}}{n!} < \sum_{n=1}^{\infty} \frac{a^{n-1}}{n!}$$

Tryss
  • 14,310