How can I test the convergence of the following sequence with odd products over even ones?

Question

The sequence is:

$$a_n = \frac {2^{2n} \cdot1\cdot3\cdot5\cdot...\cdot(2n+1)} {(2n!)\cdot2\cdot4\cdot6\cdot...\cdot(2n)} $$

Answer 1

Investigating the ratio of consecutive terms, as suggested by André Nicolas, is probably easiest. Alternatively, note that

$$2\cdot 4\cdot 6\cdot\ldots\cdot(2n)=2^nn!\;,$$

and

$$1\cdot3\cdot5\cdot\ldots\cdot(2n+1)\le 2\cdot4\cdot6\cdot\ldots\cdot 2n\cdot(2n+2)=2^{n+1}(n+1)!\;,$$

so

$$0<a_n\le\frac{2^{3n+1}(n+1)!}{2^nn!^2}=\frac{2^{2n+1}(n+1)}{n!}<\frac{4^{n+1}(n+1)}{n!}<\frac{4^{n+2}}{(n-1)!}\;,$$

and consider what you know about

$$\lim_{n\to\infty}\frac{4^n}{n!}\;.$$

Answer 2

A direct approach is to try to show that is the same as $a_n=(2n+1)!/(2(n!)^{3})$, which goes to zero as $n\rightarrow\infty$. Alternatively, use the ratio test as suggested in the comments above.

Answer 3

In the same spirit as Brian M. Scott's answer, using $$2\cdot 4\cdot 6\cdot\ldots\cdot(2n)=2^nn!$$ and $${1\cdot 3\cdot 5\cdots (2n+1)}=\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdots (2n+1)}{2\cdot4\cdot6\cdots(2n)}=\frac{(2n+1)!}{2^n n! }$$ All of this makes $$a_n=\frac{2^{2n}\frac{(2n+1)!}{2^n n!}}{2^n n! (2n)!}=\frac{2n+1}{ (n!)^2}$$ which, following André Nicolas's suggestion, seems to make $$\frac{a_{n+1}}{a_n}=\frac{2 n+3}{(n+1)^2 (2 n+1)}$$