Binomial Series — Question 5

*Question 5 For $(1+x)^{1/3}=\sum_{n=0}^{\infty}c_nx^n$, derive a recurrence for $c_{n+1}$ from $c_n$. Use it to compute through $x^4$, then state the convergence interval.

*Question 5 – Solution

Since $c_n=\binom{1/3}{n}$, $$\frac{c_{n+1}}{c_n}=\frac{(1/3)-n}{n+1},\qquad \boxed{c_{n+1}=c_n\frac{1/3-n}{n+1}},\quad c_0=1.$$

$$c_1=\frac13,\qquad c_2=-\frac19,\qquad c_3=\frac5{81},\qquad c_4=-\frac{10}{243}.$$ Therefore $$\boxed{(1+x)^{1/3}=1+\frac{x}{3}-\frac{x^2}{9}+\frac{5x^3}{81}-\frac{10x^4}{243}+\cdots}.$$

The radius is $1$. Since the coefficient magnitudes are order $n^{-4/3}$, both endpoint series converge absolutely.

The interval is $\boxed{[-1,1]}$.