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When $x<0$, the following inequality is true: $$e^x<1+x+\frac{x^2}{2!}+\dots+\frac{x^{2n}}{(2n)!}$$ for $n\in\mathbb{N}^+$.

My approach:

I think it is equivalent to $$\sum_{k=2n+1}^{\infty}\frac{x^k}{k!}<0$$ when $x<0$. If $0>x\geq-(2n+2)$, the summation can be written as:$$\sum_{k=n}^{\infty}\frac{x^{2k+1}}{(2k+1)!}\left(1+\frac{x}{2k+2}\right)$$then every term is negative. But for $x<-(2n+2)$, I can't figure out.

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2 Answers 2

up vote 2 down vote accepted

Just use the integral form for the remainder of the Taylor's series: if $x=-u$ and $u>0$, you have:

$$ e^x-\sum_{j=0}^{2n}\frac{x^j}{j!}=-\frac{e^x}{(2n)!}\int_{0}^{u}e^z\, z^{2n}\,dz < 0.$$

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Define $R_n(x)=\sum\limits_{k=n}^\infty\dfrac{x^k}{k!}$. Then for every $n\geq 1$, $R_n(0)=0$ and $R_n'=R_{n-1}$. Then by induction, you can prove that when $x<0$, $R_n(x)>0$ if $n$ is even and $R_n(x)< 0$ if $n$ is odd.

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1  
+1. Neat. $ $ $ $ –  Did Oct 30 '12 at 8:46

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