Radius of Convergence of $(1-x)^{1/4}$ This is my first post so I am very new to MathJax formatting - i apologize in advance for the messy formatting
This equation becomes $$\sum_{n=2}^\infty \frac{(4n-5) x^n}{(4^n)n!}$$
The textbook says the ratio test would be:
$$\lim_{n\to \infty} \left\lvert\frac{(4n-1)(4n-5)(x^{n+1})(4^n)n!}{(4n-5)(x^n)(4^{n+1})(n+1)n!}\right\rvert$$
But I thought it would be:
$$\lim_{n\to \infty} \left\lvert\frac{(4n-1)(x^{n+1})(4^n)n!}{(4n-5)(x^n)(4^{n+1})(n+1)!}\right\rvert$$
So no extra $(4n-5)$ - but that would result in an addition $(n+1)$ in denominator (because of the expanded factorial) and the radius of convergence would now be infinite (and not $|x| < 1$ as above fraction suggests).
Thanks
 A: Your series as written obviously has an infinite radius of converges, since the denominator grows much faster than the numerator. However, this is because there is a mistake - the numerator should be $(4n-5)!.$
A: The generalized binomial series is
$$ (1 \ + \ x)^p \ \ = \ \ \sum_{n=0}^\infty \   \left( \begin{array}{c} p \\ n \end{array}   \right) \ x^n  \ \ , $$
which becomes $$ (1 \ - \ x)^p \ \ = \ \ \sum_{n=0}^\infty \  (-1)^n \ \left( \begin{array}{c} p \\ n \end{array}   \right) \ x^n  \ \  $$
for the base $ \ (1 \ - \ x) \ $ .  Writing out the first few terms of the series with the fractional binomial coefficients for $ \ p \ = \ \frac14 \ $ gives us
$$ (1 \ - \ x)^{1/4} \ \ = \ \  1 \ ·  \left( \begin{array}{c} \frac14 \\ 0 \end{array}   \right) \ x^0 \ \ + \ \  -1 \ ·  \left( \begin{array}{c} \frac14 \\ 1 \end{array}   \right) \ x^1 \ \ + \ \  1 \ ·  \left( \begin{array}{c} \frac14 \\ 2 \end{array}   \right) \ x^2 \ \ + \ \  -1 \ ·  \left( \begin{array}{c} \frac14 \\ 3 \end{array}   \right) \ x^3 \ \ + \  \ldots        $$
$$ = \ \ 1 \ · \frac{  1}{0!} \ x^0 \ \ - \ \ 1 \ · \frac{  \frac14   }{1!} \ x^1 \ \ + \ \ 1 \ · \frac{ \frac14 · \left(  \frac{-3}{4}   \right)}{2!} \ x^2 \ \ - \ \ 1 \ ·\frac{   \frac14 · \left(  \frac{-3}{4}   \right) ·  \left(  \frac{-7}{4}   \right)  }{3!} \ x^3 \ \ + \  \ldots        $$
$$ = \ \ 1  \ \ - \ \  \frac{  1   }{4 \ · \ 1!} \ x^1 \ \ + \ \  \frac{ 1 · (  -3)}{4^2 \ · \ 2!} \ x^2 \ \ - \ \ \frac{ 1 · (-3) · (-7) }{4^3 \ · \ 3!} \ x^3 \ \ + \  \ldots        $$
There are various ways that people choose for writing the general term; my guess is that this author(s) for your textbook decided to leave the first two terms outside of the general summation to start it at $ \ n \ = \ 2 \ $ (this has no effect on the interval of convergence):
$$ (1 \ - \ x)^{1/4} \ \ = \ \ 1  \ \ - \ \  \frac14 \ x \ \ - \ \ 
\sum_{n=2}^\infty \ \frac{3 \ · \ 7 \ · \ · \ldots \ · \ (4n-5) }{4^n \ · \ n!} \ x^n  \ \ .  $$
This is a product of factors in arithmetic progression in the numerator of the general term; that is what must be used in the Ratio Test.  Almost all of them will cancel in the calculation, but what will remain is
$$ \lim_{n\to \infty} \  \left\lvert \ \frac{(4[n+1]-5)(4n-5)(x^{n+1})}{4^{n+1}  \ · \ (n+1)!} \ · \ \frac{4^n \ · \ n!}{(4n-5)(x^n) } \ \right\rvert \ \ = \ \ \lim_{n\to \infty} \  \left\lvert \ \frac{(4n-1)(x)}{4  \ · \ (n+1)}  \ \right\rvert \ \ , $$
which will yield the radius of convergence $ \ R \ = \ 1 \ $ that obtains for the series representing $ \ (1 \ + \ x)^p \ $ .  [The generalized binomial series is exactly the same as you would get by deriving a Maclaurin series for these functions.]
