Radius of Convergence of $(1-x)^{1/4}$

Question

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

Answer 1

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 $(4n-5)!.$

Answer 2

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.]