- #1
Crake
- 66
- 1
The Maclaurin series expansion for ##(1+z)^\alpha## is as follows:
$$(1+z)^\alpha = 1 + \sum_{n=0}^\infty \binom{\alpha}{n}z^n$$ with $$|z|<1$$
What I don't understand is why is ##|z|<1##?
$$(1+z)^\alpha = 1 + \sum_{n=0}^\infty \binom{\alpha}{n}z^n$$ with $$|z|<1$$
What I don't understand is why is ##|z|<1##?