what is Binomial series of $(1+z)^{\frac{1}{n}}=1+\frac{1}{n}z+O(z^2)$

Question

what is Binomial series of : $(1+z)^{\frac{1}{n}}$ as $z\to 0$

I know that: $$(1+z)^{\frac{1}{n}}=1+\frac{1}{n}z+O(z^2)$$

But I want :

$$(1+z)^{\frac{1}{n}}=1+\frac{1}{n}z+...+O(z^4)$$

Now how is it?

Answer 1

The series representation of the binomial power is $${(1 + z)}^{\frac1n} = \sum_{k = 0}^\infty \binom{1/n}{k} z^k$$ where we extend the standard binomial coefficient to work with arbitrary complex numbers $\alpha$: $$\binom{\alpha}{k} = \frac{\alpha(\alpha - 1)\cdots(\alpha - k + 1)}{k(k - 1)\cdots1},\qquad\text{for }\alpha \in \mathbb C \text{ and } k \in \mathbb N$$

If you expand the series up to the third power you get: $$\begin{align*} {(1 + z)}^{\frac1n} &= 1 + \frac1n x + \left(-\frac1{2n} + \frac1{2n^2}\right)x^2 + \left(\frac1{3n} - \frac1{2n^2} + \frac1{6n^3}\right)x^3 + O(x^4) =\\ &= 1 + \frac1n x + \frac{1 - n}{2n^2}x^2 + \frac{2n^2 - 3n + 1}{6n^3}x^3 + O(x^4) \end{align*}$$