What is the range of convergence of $\sum_{n=0}^{\infty} {(-1)}^n\binom{1/2}{n}\frac{1}{2n+3}.$

Question

I was fiddling with the integral $$\int_0^1 x^2\sqrt{1-x^2} \ dx $$ and I expanded the term under square root using a binomial series. Integrating, I got the result $$\sum_{n=0}^{\infty} {(-1)}^n\binom{1/2}{n}\frac{x^{2n+3}}{2n+3}\Biggr|_0^1.$$

I would like to know if evaluating this series at the upper limit 1 would make it converge, since binomial series has a convergence of $|x|<1$? Also if it does converge what is the range of convergence?

Answer 1

Since the coefficient of $x^{2n+3}$ is asymptotic to a constant times $n^{-3/2}$, and $\sum_n n^{-3/2}$ converges, this does converge for $|x| \le 1$. The answer, btw, is $\pi/16$.

Answer 2

Since

$\begin{array}\\ \dfrac{\binom{\frac12}{n+1}}{\binom{\frac12}{n}} &=\dfrac{\frac{\prod_{k=0}^{n+1}(\frac12-k)}{(n+1)!}}{\frac{\prod_{k=0}^{n}(\frac12-k)}{n!}}\\ &=\dfrac{\frac12-(n+1)}{(n+1)}\\ &=-(1-\dfrac{1}{2(n+1)})\\ \end{array} $

$(-1)^n\binom{\frac12}{n}$ decreases, as does $(-1)^n\binom{\frac12}{n}\dfrac1{2n+3} $.

Therefore the sum converges for $|x| < 1$ and conditionally converges for $x = -1$.

For $x=1$, we can do, for $n \ge 1$,

$\begin{array}\\ \binom{\frac12}{n} &=\dfrac{\prod_{k=0}^{n}(\frac12-k)}{n!}\\ &=\dfrac{\prod_{k=0}^{n}(1-2k)}{2^nn!}\\ &=\dfrac{\prod_{k=1}^{n}(1-2k)}{2^nn!}\\ &=\dfrac{(-1)^n\prod_{k=1}^{n}(2k-1)}{2^nn!}\\ &=\dfrac{(-1)^n\prod_{k=1}^{n}(2k-1)(2k)}{2^{2n}(n!)^2}\\ &=\dfrac{(-1)^n(2n)!}{2^{2n}(n!)^2}\\ &=\dfrac{(-1)^n}{2^{2n}}\binom{2n}{n}\\ &\approx\dfrac{(-1)^n}{2^{2n}}\dfrac{4^n}{\sqrt{\pi n}}\\ &=\dfrac{(-1)^n}{\sqrt{\pi n}}\\ \end{array} $

so $(-1)^n\binom{\frac12}{n}\dfrac{x^{2n+3}}{2n+3} \approx \dfrac{1}{\sqrt{\pi n}}\dfrac{x^{2n+3}}{2n+3} $

and the sum of these converges at $x = 1$.

Answer 3

From $\int_0^1 x^2\sqrt{1-x^2} \ dx$

Using the following substitution:

$t=x^2, \frac {dx}{dt}=\frac{1}{2}\frac{1}{\sqrt(t)}$ , at $x=0 \rightarrow t=0$ and at $x=1 \rightarrow t=1$

Applying the definition and identity of $\beta$ function:

$\frac{1}{2}\int\limits_0^1 t^\frac{1}{2}(1-t)^\frac{1}{2}dt= \frac{1}{2} \beta(\frac{3}{2},\frac{3}{2})=\frac{1}{2}\frac{\Gamma(\frac{3}{2})^2}{\Gamma(3)}= \frac{\pi}{16}$