# double factorial series problem

In the problem the sum is given as $$\sum\limits_{n=1}^\infty \frac{(2n-1)!!}{(2n)!!}$$ and then when I try to solve it using Gauss's test I get $$\frac{a_{n}}{a_{n+1}}=\frac{2n+2}{2n+1}$$ but in the solution there is given: $$\frac{a_{n}}{a_{n+1}}=\frac{2n}{2n-1}$$

my reasoning was:

$$\frac{a_{n}}{a_{n+1}}=\frac{\frac{(2n-1)!!}{(2n)!!}}{\frac{(2n+1)!!}{(2n+2)!!}}=\frac{1\cdot3\cdot...\cdot(2n-1) \times 2\cdot4\cdot...\cdot2n\cdot(2n+2)}{2\cdot4\cdot...\cdot2n \times 1\cdot3\cdot(2n-1)\cdot(2n+1)}=\frac{2n+2}{2n+1}$$ I believe that I made a mistake, but I don't know where?

• What is it you are trying to do? Nov 3, 2015 at 17:55
• Your calculation of the ratio $\frac{a_n}{a_{n+1}}$ is correct. The one you quote seems to have an index off by one issue. The ratio is unfortunately not directly useful in dealing with the issue of convergence. Nov 3, 2015 at 17:58
• To see is my reasoning true or not, because from that point I can solve the rest of the problem... Nov 3, 2015 at 17:58
• As an aside, $\displaystyle\sum_{n=0}^\infty \bigg[\frac{(2n-3)!!}{(2n)!!}\bigg]^2~=~\frac4\pi$ Nov 3, 2015 at 20:41

$$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{(2n)!!^2} = \frac{1}{4^n}\binom{2n}{n}$$ but the RHS behaves like $\frac{1}{\sqrt{\pi n}}$, hence the series is diverging. Also without the exact asymptotics:
$$\left(\frac{(2n-1)!!}{(2n)!!}\right)^2 = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4}\prod_{k=1}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right) \geq \frac{1}{4n}.$$