Note 21.6. Two things change when \(n\) is not a positive whole number.

The series is infinite. With \(n\) a positive integer the factor \(n-r\) eventually hits zero and every later term vanishes; with \(n=\frac {1}{2}\) the numerators \(\frac {1}{2},-\frac {1}{2},\frac {3}{2},\ldots \) never do.

And it requires \(\left |x\right |<1\). An infinite series only means something if it converges, and this one converges precisely on that interval. The condition is not decoration: putting \(x=4\) in the expansion of \((1+x)^{1/2}\) produces terms that grow without bound, and no amount of adding will reach \(\sqrt {5}\).

Note the form the theorem takes. It is stated for \((1+x)^{n}\), not \((a+b)^{n}\), so anything else must first be written that way. To expand \((4+x)^{1/2}\), take the \(4\) out: \[(4+x)^{1/2}=4^{1/2}\left (1+\tfrac {x}{4}\right )^{1/2} =2\left (1+\tfrac {x}{4}\right )^{1/2},\] valid now for \(\left |\frac {x}{4}\right |<1\), that is \(\left |x\right |<4\). The condition travels with the bracket, so it must be rewritten too.