Taylor Series — Question 9

Question 9 – Solution

Step 1: Use the binomial pattern.

$$(1+x)^{\alpha}=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n,\qquad \binom{\alpha}{n}=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}.$$ With $\alpha=1/2$, $$\binom{1/2}{0}=1,\quad\binom{1/2}{1}=\frac12,\quad\binom{1/2}{2}=-\frac18,\quad\binom{1/2}{3}=\frac1{16}.$$ Therefore $$\boxed{\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\cdots}.$$

Step 2: State radius and endpoints.

The binomial series has $R=1$. Its coefficients have magnitude asymptotic to a constant times $n^{-3/2}$, so it converges absolutely at both $x=1$ and $x=-1$.

Conclusion.

The interval is $\boxed{[-1,1]}$, with sums $\sqrt2$ and $0$ at the endpoints.