Proving the inequality $e^{-2x}\leq 1-x$
For example, $(1-x)(1+2x)=1+x-2x^2\geq 1$ so $$ e^{2x}\geq 1+2x \geq \frac{1}{1-x} $$
A slightly ugly but standard beginning calculus approach is to let $$f(x)=1-x-e^{-2x}.$$ We want to show that $f(x)\ge 0$ in the interval $[0,1/2]$.
A first experimental step might be to use software to graph $y=f(x)$ as $x$ ranges over our interval. If we have a high degree of trust in the graphing software, the picture tells us that $f(x)$ is quite likely to be $\ge 0$ in our interval.
Certainty is better. Use the derivative to study the behaviour of $f(x)$. Note that $f(0)=0$, and $$f'(x)=2e^{-2x} -1.$$ The derivative is positive at $x=0$, and is clearly decreasing. It reaches $0$ at $x=(\ln 2)/2\approx 0.3465$. So $f(x)$ is increasing in the interval $[0,(\ln 2)/2]$, and decreasing from $(\ln 2)/2$ on. By $x=1/2$, $f(x)$ is about $0.13212$, and in particular still positive.
Thus $f(x)$ is $\ge 0$ for $0 \le x\le (\ln 2)/2$, and $f(x) >0$ for $(\ln 2)/2 \le x\le 1/2$. It follows that $f(x)\ge 0$ on the whole interval $[0,1/2]$ (and somewhat beyond $1/2$).
Comment: There are far better ways to prove the inequality. But let's stick to calculusy approaches. To cut down on the negatives, note that equivalently we want to show that $e^{2x}(1-x) \ge 1$ on our interval (we multiplied both sides by the positive number $e^{2x}$). Let $g(x)=e^{2x}(1-x)-1$. Then $g(0)=0$. Also, $g'(x)=(1-2x)e^{2x}$, so $g$ is increasing on the interval $[0,1/2]$, and we are finished.
Let $f(x)=e^{-2x}-(1-x)$. We have:
- $f''(x)=4e^{-2x}$, which is always positive. Hence, $f$ is convex.
- $f(0)=0$
- $f\left(\dfrac{1}{2}\right)=\dfrac{1}{e}-\dfrac{1}{2}\leq0$.
Since $f$ is convex, its curve is always below $0$ on $\left[0,\dfrac{1}{2}\right]$. This proves that $e^{-2x}\leq 1-x$.
By Taylor expansion, we have $$\ln \frac{1}{1-x} = \sum_{n=1}^{\infty} \frac{x^n}{n},$$ whose convergence radius is $R=1$.
The equation above can also be achieved by integrating both sides of $$\frac{1}{1-x} = \sum_{n=0}^{\infty} {x^n} .$$
For $0\leq x\leq 1/2$ we have the following upper bound $$ \sum_{n=1}^{\infty }\frac{x^{n}}{n}\leq \sum_{n=1}^{\infty }\frac{1}{ n\left( 2^{n}\right) }=\ln 2 \leq 2. $$ Therefore $$ -\ln \left( 1-x\right) =\sum_{n=1}^{\infty }\frac{x^{n}}{n}\leq 2x. $$ The given inequality follows.