Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 7

Use the binomial series with exponent $1/2$ and argument $-3x$:Thus the first three nonzero terms are

The usual interior condition for the binomial series is $|3x|<1$, giving $-1/3<x<1/3$ as its open interval of convergence. In this particular square-root case the full series also converges absolutely at both endpoints, so the maximal real range on which its sum equals the original function isFor an endpoint check, write $\sqrt{1-u}=1-{\sum }_{n\ge 1}{b}_n{u}^n$, where $b_n=a_{n-1}-a_n>0$ and $a_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\to 0$. Then $\sum b_n=1$ by telescoping, establishing absolute convergence at $u=\pm 1$. The square-root binomial series at its endpoints proves continuity of the sum there. The endpoint values are $\sqrt{2}$ at $x=-1/3$ and $0$ at $x=1/3$. Outside this range the power-series terms do not tend to zero. A three-term truncation remains an approximation, rather than an exact formula throughout the interval.