# Taylor series expansion and radius of convergence

The problem is:
Expand the given function using Taylor's expansion around $a=1$. $f(x)=(5x-4)^{-7/3}$, and then find the radius of convergence of the obtained series. Hint : Write the nth derivative in the summation notation.

I've found a few derivatives but I don't know how to write the nth one, especially in the summation notation. Could you please help me solve this problem, I would be very grateful.

• $5x-4=5(x-1)+1$ could help. Define $y=5(x-1)$ and use binomial expansion. – Claude Leibovici Sep 10 '15 at 13:24

The problem is equivalent to finding the Taylor series (and the radius of convergence of such a series centered at $z=0$) of $$g(z) = (1+5z)^{-7/3}. \tag{1}$$ The Taylor series is given by the extended binomial theorem: $$\begin{eqnarray*} g(z)=\sum_{k\geq 0}\binom{-7/3}{k} 5^k z^k &=& \sum_{k\geq 0}\frac{\Gamma(-4/3)}{\Gamma(k+1)\Gamma(-4/3-k)}\,5^k z^k\\&=&\sum_{k\geq 0}\frac{(-1)^k 9\sqrt{3}\,\Gamma(2/3)}{8\pi}\cdot\frac{\Gamma(k+7/3)}{\Gamma(k+1)}\cdot 5^k z^k\tag{2}\end{eqnarray*}$$ and the term $\frac{\Gamma(k+7/3)}{\Gamma(k+1)}$ behaves like $k^{4/3}$ by Gautchi's inequality, hence the radius of convergence is $\large\color{red}{\rho=\frac{1}{5}}$, as expected, since $z=-\frac{1}{5}$ is the closest singularity to the origin for $g(z)$.