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I need to show that $$ 1-x\leq e^{-x} (*) $$ Now we know that $e^x:=\sum_{k=0}^\infty \frac{x^k}{k!}$, which means $$ e^{-x}:=\sum_{k=0}^\infty \frac{(-x)^k}{k!}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}... $$ So after we subtract $1-x$ from both sides of the inequation (*) we have to show: $$ 0\leq \sum_{k=2}^\infty \frac{(-x)^k}{k!}=\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}... $$

1.) $x=0$ : $\sum_{k=2}^\infty \frac{(-x)^k}{k!} = 0$

2.)$x=1$: $\frac{1}{2n+1}\leq \frac{1}{2n}$ for all $n\in\mathbb{N}$

3.) $x<0$: This is obvious since ever summand is positive.

4.) $0<x<1$: This is also obvious (?) because for such $x$ it holds that $x^{2n}>x^{2n+1}$ for all $n\in\mathbb{N}$ and also $\frac{1}{(2n)!}>\frac{1}{(2n+1)!}$. So $\sum_{k=2}^\infty \frac{(-x)^k}{k!}=\sum_{n=1}^\infty \frac{x^{2n}}{(2n)!} - \frac{x^{2n+1}}{(2n+1)!}>0 $ since all summands are positive.

5.) $x>1$: Here is where I struggle and need help please. This is the a bit of the opposite of 4 maybe.

KingDingeling
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1 Answers1

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The derivative of $e^{-x}+x-1$ is $-e^{-x}+1 \geq 0$ because $e^{-x} \leq 1$. This function vanishes when $x=0$ and hence it is non-negative for $x \geq 0$. This gives $e^{-x} \geq 1-x$ for $x \geq 0$. For $x<0$ the inequality follows immediately from the series expansion (since all the terms are positive).

Kavi Rama Murthy
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