Taylor series expansion of $(1+x)^\frac{1}{n}$ I'm trying to find the formal Taylor series expansion of (1+x)1/n but I keep getting stuck. I tried the binomial expansion and while I managed it, I can't quite figure out how to change it to a general summation formula.
I tried the derivative method too but it got a bit complicated for me. Is there a shortcut for this. Maybe another Taylor series I can manipulate?
Thanks
 A: The binomial series is given as:
$$ (1+x)^j = 1 + jx + j(j-1) \frac{x^2}{2} + O(x^3)$$
Sub $j = \frac{1}{n}$
$$ ( 1+x)^{\frac{1}{n} }  = 1 + \frac{x}{n} + \frac{1}{n} ( \frac{1}{n} -1) \frac{x^2}{2}...$$

To write as sum you must use the falling factorial defined as follows:
$$ r^{ \underline{k} } = r (r-1)(r-2)..(r-k+1)$$
Examples:
$$ k = 0$$
$$ r^{ \underline{0} } = 1$$
$$ k=2$$
$$ r^{ \underline{2} } = r(r-1)$$
Hence, the binomial series is:
$$ (1+x)^{\frac{1}{n} }  = \sum_{k=0}^{\infty} (\frac{1}{n})^{ \underline{k} }  
 \frac{x^k}{k!}$$
A: Just use $$\left((1+x)^{\frac{1}{n}}\right)^{(k)}_{x=0}=\frac{1}{n}\left(\frac{1}{n}-1\right)...\left(\frac{1}{n}-k+1\right).$$
A: By the Binomial formula we have $$(1+x)^{n}=\sum_{k=0}^{\infty} {n\choose k}x^{k}$$
since we know that $${n\choose k}=\begin{cases} \frac{n!}{k!(n-k)!}\space\text{for $0\leq k \leq n$}\\\space\space\space\space0\space\space\space\space\space\space\space\space\text{otherwise}\end{cases}.$$
Thus we have $$(1+x)^{\frac{1}{n}}=\sum_{r=0}^{\infty}{\frac{1}{n}\choose r}x^{r}.$$
See this. Indeed  $$(1+x)^{\frac{1}{n}}=\sum_{r=0}^{\infty}{\frac{1}{n}\choose r}x^{r}={\frac{1}{n}\choose 0}+{\frac{1}{n}\choose 1}x+{\frac{1}{n}\choose 2}x^{2}+...$$
$$=1+\frac{x}{n}+\frac{1}{n}(\frac{1}{n}-1)\frac{x^2}{2}+...$$
