How did we come to this inequality $|\frac{e^x-1}{x}-1|\leq|\sum_{n=1}^{\infty}{\frac{|x|^n}{(n+1)!}}|$

Question

How did we come to this inequality $|\frac{e^x-1}{x}-1|\leq|\sum_{n=1}^{\infty}{\frac{|x|^n}{(n+1)!}}|$.

Using $e^x=\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$.

I have this inequality in the proof of $\lim_{x\to 0}{\frac{e^x-1}{x}}=1$, I undestand the proof except that one inequality.

I have tried to rewrite $e^x$ with the sum but I couldn't come to that right side in the inequality.

Answer 1

It is mostly index manipulation $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ $$e^x-1=\sum_{n=1}^\infty\frac{x^n}{n!}$$ $$\frac{e^x-1}{x}=\sum_{n=1}^\infty\frac{x^{n-1}}{n!} =1+\sum_{n=2}^\infty\frac{x^{n-1}}{n!} =1+\sum_{n=1}^\infty\frac{x^{n}}{(n+1)!}$$ $$\frac{e^x-1}{x}-1=\sum_{n=1}^\infty\frac{x^{n}}{(n+1)!}$$ Now use the triangle inequality:

$$\left|\frac{e^x-1}{x}-1\right| =\left|\sum_{n=1}^\infty\frac{x^{n}}{(n+1)!}\right| \le\sum_{n=1}^\infty\frac{\left|x\right|^{n}}{(n+1)!} $$

Answer 2

$$ \frac{e^x-1}{x}-1=\sum_{n=1}^\infty\frac{x^{n-1}}{n!}-1=\sum_{n=2}^\infty\frac{x^{n-1}}{n!}=\sum_{n=1}^\infty\frac{x^{n}}{(n+1)!}. $$

Answer 3

Before expanding it is convenient to rearrange the expression (assuming $x\neq0$): $$ \left|\frac{e^{x}-1}{x}-1\right|=\left|\frac{e^{x}-1-x}{x}\right|=\left|\frac{\sum_{2}^{\infty}\left(x^{n}/n!\right)}{x}\right|=\left|\sum_{2}^{\infty}\frac{x^{n-1}}{n!}\right|\leq\sum_{1}^{\infty}\frac{\left|x\right|^{n}}{(n+1)!} $$ where the last follows from $\sum_{1}^{\infty}\frac{\left|x^{n}\right|}{(n+1)!}<\infty $