Taylor series expansion of $(1+x)^\frac{1}{n}$

Question

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

Answer 1

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:

$$ (1+x)^{\frac{1}{n} } = \sum_{k=0}^{\infty} (\frac{1}{n})^{ \underline{k} } \frac{x^k}{k!}$$

Answer 2

Just use $$\left((1+x)^{\frac{1}{n}}\right)^{(k)}_{x=0}=\frac{1}{n}\left(\frac{1}{n}-1\right)...\left(\frac{1}{n}-k+1\right).$$

Answer 3

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 $$(1+x)^{\frac{1}{n}}=\sum_{r=0}^{\infty}{\frac{1}{n}\choose r}x^{r}={\frac{1}{n}\choose 0}+{\frac{1}{n}\choose 1}x+{\frac{1}{n}\choose 2}x^{2}+...$$ $$=1+\frac{x}{n}+\frac{1}{n}(\frac{1}{n}-1)\frac{x^2}{2}+...$$