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

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
$x = 1$.
