# Proof

Using the definition of the exponential series, the triangle inequality, the distributivity law for real numbers, the properties of the absolute value for real numbers, and the rules for calculating with inequalities or real numbers we can estimate the remainder term of the exponential series as follows

$\begin{array}{rcll} |r_{N + 1}(x)|&=& \sum_{n=N+1}^\infty\frac{x^n}{n! }&\text{definition of the exponential series}\\ &\le &\sum_{n=N+1}^\infty\frac{|x|^n}{n! }&\text{triangle inequality}\\ &=&\frac{|x|^{N+1}}{(N+1)!}\left(1 + \frac{|x|}{N+2} + \frac{|x|^2}{(N+2)(N+3)} + \ldots + \frac{|x|^k}{(N+2)\cdot\ldots\cdot(N+k+1)} + \ldots \right)&\text{properties of absolute value and distributivity law}\\ &\le&\frac{|x|^{N+1}}{(N+1)!}\left(1 + \frac{|x|}{N+2} + \left(\frac{|x|}{(N+2)}\right)^2 + \ldots + \left(\frac{|x|}{(N+2)}\right)^k + \ldots \right)&\text{rules for calculating with inequalities}\\ \end{array}$

If we require that $$|x|\le\frac{N+2}{2}$$ then we have $\frac{|x|}{N+2}\le \frac{\frac{N+2}{2}}{N+2}=\frac 12.$

Therefore, for $$|x|\le\frac{N+2}{2}$$ it follows with the formula for the infinite geometric series. $\begin{array}{rcl} |r_{N + 1}(x)|&\le & \frac{|x|^{N+1}}{(N+1)!}\left(1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^k} + \ldots \right)\\ &=&2\frac{|x|^{N+1}}{(N+1)!}. \end{array}$

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983