> [!NOTE] Lemma > For all $x<1,$ $e^{x} \leq \frac{1}{1-x}$where $\exp(x)$ denotes the [[Real Exponential Function|real exponential function]]. **Proof**: If $0\leq x <1$ then by [[Geometric Series]], $\exp(x)= 1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\dots \leq 1+x+x^{2}+\dots= \frac{1}{1-x}$ Suppose $x=-u$ is negative. Then by [[Lower Bound for Real Exponential Function]], $\exp(u)\geq 1+u$ and hence $\exp(-x)\geq 1-x$ thus for all $x< 0,$ $\exp(x)\leq \frac{1}{1-x}.$