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?


  • $\begingroup$ You need it in $0$? $\endgroup$ – Michael Rozenberg Oct 1 '20 at 9:11
  • $\begingroup$ Yes I need the series around 0. $\endgroup$ – Iris---- Oct 1 '20 at 9:13

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:

$$ r^{ \underline{k} } = r (r-1)(r-2)..(r-k+1)$$


$$ k = 0$$

$$ r^{ \underline{0} } = 1$$

$$ k=2$$

$$ r^{ \underline{2} } = r(r-1)$$

Hence, the binomial series is:

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

  • $\begingroup$ This I got. I got confused trying to make a general summation formula for this as the powers of x are changing but the denominator 'n' is not so if I make a series that sums over n, it gets weird. $\endgroup$ – Iris---- Oct 1 '20 at 9:19
  • $\begingroup$ I have made an edit $\endgroup$ – Buraian Oct 1 '20 at 9:29
  • $\begingroup$ Thank you! I was not aware of the falling factorial notation. $\endgroup$ – Iris---- Oct 1 '20 at 9:33
  • $\begingroup$ You can accept the answer if you found it helpful $\endgroup$ – Buraian Oct 1 '20 at 9:34

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


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


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