I'm trying to compute the Radius of convergence of:
$$_2F_1(a,b;c;x) = 1+\frac{ab}{1!c}x+\frac{a(a+1)b(b+1)}{2!c(c+1)}x^2+\ldots+\frac{a(a+1)\cdots(a+n-1)b(b+1)\cdots(b+n-1)}{n!c(c+1)\cdots(c+n-1)}x^n+\ldots$$
Where $a$, $b$, and $c$ are real parameters, and $c$ is not a negative integer.
- First, I used the formula of the radius of convergence of a power series: $R= \lim\limits_{n \to \infty} \frac{|a_n|}{|a_{n+1}|}$; which gave me:
$ \lim\limits_{n \to \infty} |\frac{(a+n-1)(b+n-1)}{n!(c+n-1)}| ÷ |\frac{(a+(n+1)-1)(b+(n+1)-1)}{(n+1)!(c+(n+1)-1)}| $ (I'm not sure I have applied it correctly since the formula that's given is not in sigma notation...)
- After doing some simplification, I had:
$\lim\limits_{n \to \infty} \frac{(a+n-1)(b+n-1)(c+n)(n+1)}{(a+n)(b+n)(c+n-1)}$; which gave me:
- $\lim\limits_{n \to \infty} \frac{(a+n-1)(b+n-1)(c+n)(n+1)}{(a+n)(b+n)(c+n-1)} = \infty $
However, the answer of the problem is $R=1$ so... here I am, trying to understand why.
Any help would be appreciated! Thanks in advance :D
When the $n$th coefficient of the series is a fraction with a bunch of terms, including all of the terms that appears in the formula for the previous coefficient, then the ratio between coefficients is simply the fraction consisting of only the new terms that appear. For example, we have
$$ a_2=\frac{a(a+1)\,b(b+1)}{c(c+1)\,1(1+1)}, \quad a_3=\frac{a(a+1)(a+2)\,b(b+1)(b+2)}{c(c+1)(c+2)\,1(1+1)(1+2)}, \quad \frac{a_3}{a_2}=\frac{(a+2)(b+2)}{(c+2)(1+2)}. $$
More generally, the ratio is
$$ \frac{a_{n+1}}{a_n}=\frac{(a+n)(b+n)}{(c+n)(1+n)}. $$