Taylor series for $\sqrt[3]{x+1}$ I'm trying to find the Taylor series for $\sqrt[3]{x}$, but since the n-th derivative of the function $$f(x)=\sqrt[3]{x}$$
not definded at $x=0$, I've swithched to $f(x)=\sqrt[3]{x+1}$, This is my Attempt:
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)x^n}{n!}$$
$$\sqrt[3]{x+1}=1+\frac{x}{3}-\frac{2x^2}{9\cdot2!}+\frac{10x^3}{27\cdot 3!}-\frac{80x^4}{81\cdot4!}+...$$
but when i plug $x=125 \iff \sqrt[3]{x+1}=\sqrt[3]{126}$ i get something way bigger than the acttual value which is about $5.01...$, you can check it here.
But I can't find the mistake in the equation.
 A: We can write the $n$th term exactly as
$$ R_n(x) = \left(\frac{1}{3}\right)_{n}\frac{x^n}{n!} $$
where $( a )_{n} = \prod\limits_{k=0}^{n-1} a-k$ is a falling factorial. The ratio
$$ \biggr|\frac{R_{n}(x)}{R_{n-1}(x)}\biggr| = |x|\biggr|\frac{4}{3n} - 1\biggr| \to |x| $$
so the series definitely does not converge for $|x| > 1$ by the ratio test.

If you want a method of computing cube roots effectively, the paper A Way of Approximation of a Cube
Root, 2018 by By Mohammad H. Poursaeed has your back:
$$\sqrt[3]{x} \approx \frac{k}{2} + \sqrt{\frac{4x-k^3}{12k}}$$
where $k$ is chosen such that $k^3 < x < (k+1)^3$.
In your example, $x = 126$, and so $k = 5$, giving
$$\sqrt[3]{126} \approx \frac{5}{2} + \sqrt{\frac{4(126)-5^3}{12(5)}} = 5.013297966\cdots$$
while $\sqrt[3]{126} = 5.013297935\cdots$ in actuality. It's quite spectacularly close, and very fast (as long as you have a fast method for computing square roots).
A: For $x\gt 1$, expand in terms of $\frac1x<1$ to ensure convergence
$$\sqrt[3]{x+1}= \sqrt[3]{x} \sqrt[3]{1+\frac1x}
= \sqrt[3]{x}  (1+\frac{1}{3x}-\frac{2}{9\cdot2!}\frac1{x^2}+\frac{10}{27\cdot 3!}\frac1{x^3}-\frac{80}{81\cdot4!}\frac1{x^4}+...)
$$
Then, set $ x=125$ to obtain the approximation
$$\sqrt[3]{126} =5(1+\frac1{3\cdot 125}-\cdots )=5.0133$$
A: The binomial series for $(a+x)^\alpha$ is only valid if the absolute value of x is smaller than one, unless $\alpha$ is a non-negative integer. Your value is x=125>>1 and $\alpha$ is not integer, thus the divergence. To get a good approximation you would have to expand around $x_0=125$, which of course requires knowledge of the third root of 126.
