Why do the endpoints of the Maclaurin series for arcsin converge?

Question

The series $$\sum_{n=0}^\infty {{-\frac {1} 2} \choose n} \frac{(-1)^n}{2n+1}$$ is an endpoint for the Maclaurin series for arcsin(x). (The other endpoint is just the negative of this one.) I played around with this a bit and turned it into three (potentially useful) forms :

$\sum_{n=0}^\infty \frac12\frac34\frac56\cdots\frac{2n-1}{2n} \frac{1}{2n+1}$

$\sum_{n=0}^\infty \left( 1-\frac 1 2 \right) \left( 1-\frac 1 4 \right)\cdots \left( 1-\frac 1 {2n} \right)\frac{1}{2n-1}$

$\sum_{n=0}^\infty \dfrac{(2n)!}{2^{2n}(n!)^2}\dfrac{1}{2n+1}$,

but I'm not sure where one could go from here.

Answer 1

Method 1:

First, we have $$ \frac{(-1)^n}{2n+1}\binom{-1/2}{n}=\frac{1}{2n+1}\binom{n-1/2}{n}\tag{1} $$ and $$ \binom{n-1/2}{n}=\frac{\Gamma(n+1/2)}{\Gamma(n+1)\Gamma(1/2)}=\frac{\Gamma(n+1/2)}{\Gamma(n+1)\sqrt{\pi}}\tag{2} $$ By Gautschi's Inequality, we have that $$ \frac{1}{\sqrt{n+1}}\le\frac{\Gamma(n+1/2)}{\Gamma(n+1)}\le\frac{1}{\sqrt{n}}\tag{3} $$ Therefore, we get that $$ \frac{1}{(2n+1)\sqrt{\pi}}\frac{1}{\sqrt{n+1}}\le\frac{(-1)^n}{2n+1}\binom{-1/2}{n}\le\frac{1}{(2n+1)\sqrt{\pi}}\frac{1}{\sqrt{n}}\tag{4} $$ By comparison to $$ \sum_{n=1}^\infty\frac{1}{n^{3/2}} $$ We get that the series converges.

---

Method 2:

As Gerry Myerson suggests, Stirling's Approximation gives $$ \binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag{5} $$ Thus the term in your third form is $$ \frac{(2n)!}{2^{2n}(n!)^2}\frac{1}{2n+1}\sim\frac{1}{\sqrt{\pi n}}\frac{1}{2n+1}\tag{6} $$ Again, comparison to $$ \sum_{n=1}^\infty\frac{1}{n^{3/2}} $$ yields the convergence of the series.

Answer 2

For $n \geq 1$ let $$f_n(x) = \frac{1}{2n - 1} + \frac{2n - 1}{2n} x $$ Then the $N$-th partial sum $S_N$ is $$ S_N = \sum_{n=0}^N\frac{1}{2} \frac{3}{4} \cdots \frac{2n - 1}{2n} \frac{1}{2n + 1} = f_1 \circ f_2 \circ \cdots \circ f_N \left( \frac{1}{2N+1} \right) $$

Each $f_n$ maps the interval $[0, 2]$ into itself and therefore all partial sums are bounded above by $2$.

Answer 3

Here's another approach. We write the shorthand $m!$ for $\Gamma(m+1)$, where $\Gamma$ is the gamma function.

Let $a_n = \displaystyle\frac{(-1)^n}{2n+1} {-\frac{1}{2} \choose n}$. Then $$\begin{eqnarray*} \frac{a_{n+1}}{a_n} &=& -\frac{2n+1}{2n+3} \, \frac{{-\frac{1}{2}\choose n+1}}{{-\frac{1}{2}\choose n}} \\ &=& -\frac{2n+1}{2n+3} \, \frac{(-\frac{1}{2})!}{(n+1)!(-\frac{1}{2}-n-1)!}\, \frac{n!(-\frac{1}{2}-n)!}{(-\frac{1}{2})!} \\ &=& -\frac{2n+1}{2n+3}\, \frac{-\frac{1}{2}-n}{n+1} \\ &=& \frac{(2n+1)^2}{2(n+1)(2n+3)} \\ &=& 1-\frac{3}{2n} +O\left(\frac{1}{n^2}\right). \end{eqnarray*}$$ Since $3/2>1$ the series converges by Raabe's test. (Note that the $a_n$s are positive, so we don't bother taking the absolute value.)

Answer 4

Another approach would be Stirling's closedly related Wallis' formula. It gives

$$\lim_{n \to \infty} \frac{(2n)!}{(2n-1)!}\frac{1}{\sqrt n}=\sqrt{\pi}$$

This means that your $n$th term is asymptotically equal to

$$ {\sqrt{\frac 1 {\pi n}}}\frac{1}{2n+1}$$

which explains why the series converges. In particular,

$$\lim_{n \to \infty} \frac{n^{3/2}}{\sqrt{n}(2n+1)}=\frac 1 2 $$ so the convergence of $$\sum_{n >0} n^{-3/2} $$ implies that of your series.