Deducing an inequality from a Taylor series expansion

Question

Consider the inequality
$$
1-\frac{x}{2}-\frac{x^2}{2} \le \sqrt{1-x} < 1-\frac{x}{2}
$$
for $0 < x < 1$. The upper bound can be read off the Taylor expansion for $\sqrt{1-x}$ around $0$,
$$
\sqrt{1-x} = 1 – \frac{x}{2} – \frac{x^2}{8} – \frac{x^3}{16} – \dots
$$
by noting that all the non linear terms are negative. Can the left side inequality be read-off the expansion by a similar reasoning? Please do not try to prove the left side inequality by other means (such as minimizing $\sqrt{1-x} – 1 + \frac{x}{2} + \frac{x^2}{2}$ using derivatives).

Answer 1

We have a function $$f(x) =\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \frac{5 x^4}{128 }- \frac{7 x^5}{256}- \cdots = \\=1 - a_1 x - a_2 x^2 -a_3 x^3 \cdots $$ The series converges for every $x\in [0,1]$, so we have $$a_1 + a_2 + \cdots = 1$$

Therefore we can write $$1 - \frac{x}{2} - \frac{x^2}{2} = 1 - a_1 x - (a_2 + a_3 + \cdots ) x^2< \\ <1 - a_1 x - a_2 x^2 - a_3 x^3 -\cdots = f(x)$$ for $x\in (0,1)$.