Convergence of the following hypergeometric power series

Question

Let $$a_n = \frac{1.3...(2n-1)}{2.4...(2n)}$$ It is required to discuss the convergence of the following series $$ \sum a_nx^n$$ where $ x \in \Bbb R$. My claim is this series converges absolutely for all values of $x$. First notice that

$$ |a_nx^n| =\frac{|x|^n}{2^{2n}n!} \le \frac{|x^n|}{n!}$$

Since $\sum \frac{|x^n|}{n!}$ converges for every $x$, by the comparison test,$\sum|a_nx^n|$ converges for every $x$. Hence $\sum a_nx^n$ converges absolutely for every $x$. Is there any mistake in my reasoning?

Answer 1

It is not true that$$a_n=\frac1{2^{2n}n!}.$$

On the other hand,$$\left\lvert\frac{a_{n+1}}{a_n}\right\rvert=\frac{2n+1}{2n+2}\to1$$and therfore the radius of convergence of the series $\sum_{n=1}^\infty a_nx^n$ is $1$.

Answer 2

Note that

$$a_n = \frac{1.3...(2n-1)}{2.4...(2n)}=\frac{(2n)!}{4^{n}(n!)^2}$$

and

$$\frac{(2n)!}{(n!)^2}=\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}$$

Refer also to

• To show for following sequence $\lim_{n \to \infty} a_n = 0$ where $a_n$ = $1.3.5 ... (2n-1)\over 2.4.6...(2n)$