Represent the function $f(x) =\sqrt{x+1}$ as a Maclaurin series, and find its interval of convergence.

Solution Write $\sqrt{x+1}=(1+x) ^{1/2}$ and use the binomial series with $m=\frac{1}{2}$. The result is $$\begin{eqnarray*} (1+x)^{1/2} &=& 1+\frac{1}{2}x+\frac{\left( \frac{1}{2}\right) \left( -\frac{1 }{2}\right) }{2!}\,x^{2}+\frac{\frac{1}{2}\left( -\frac{1}{2}\right) \left( -\frac{3}{2}\right) }{3!}x^{3}+\cdots \\[12pt] &=& 1+\frac{1}{2}x-\frac{1}{8} x^{2}+\frac{1}{16}x^{3} - \cdots \end{eqnarray*}$$

Since $m=\frac{1}{2}>0$ and $m$ is not an integer, the series converges on the closed interval $[-1,\,1]$.