Nicontio1

2022-01-14

Prove inequality ${e}^{x}>\frac{{x}^{n+1}}{n+1!}$

Durst37

Expert

Step 1
You can use Taylor series of the exponential, whose radius of convergence is infinite.
${e}^{x}=\sum _{k=0}^{+\mathrm{\infty }}\frac{{x}^{k}}{k!}$
Can you proceed?
Observe that due to the previous series, we can write the partial sum for k from 0 to $n+2$:
$1+x+\frac{{x}^{2}}{2!}+\cdots +\frac{{x}^{n}}{n!}+\frac{{x}^{n+1}}{\left(n+1\right)!}+\frac{{x}^{n+2}}{\left(n+2\right)!}+\cdots$
And this, being a sum of positive terms, is surely greater than $\frac{{x}^{n+1}}{\left(n+1\right)!}$

SlabydouluS62

Expert

Step 1
The inequality ${e}^{x}>\frac{{x}^{n+1}}{n+1!}$
can be proven by
${e}^{x}=\sum _{l=0}^{\mathrm{\infty }}\frac{{x}^{l}}{l!}\ge \sum _{l=n+1}^{\mathrm{\infty }}\frac{{x}^{l}}{l!}\ge \frac{{x}^{n+1}}{\left(n+1\right)!}$

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