Used the ratio test. Having problems reducing it by canceling out factorials.

Question

$$\sum_{k=1}^{\infty} \frac{1..3...(2k-1)}{(2k)!}$$

Applying the ratio test:

$$\sum_{k=1}^{\infty} \frac{\frac{2(k+1)-1}{(2(k+1))!}}{\frac{2k-1}{2k!}}$$

which turns into:

$$\sum_{k=1}^{\infty} \frac{2k+1}{(2k+2)(2k+1)(2k)} \frac{2k!}{2k-1}$$

Im not sure how to deal with the factorials

Answer 1

Your $a_k$ is $$\frac{1..3...(2k-1)}{(2k)!}$$

Thus your $a_{k+1}$ is$$\frac{1..3...(2k-1)(2k+1)}{(2k+2)!}$$

Thus your $\frac {a_{k+1}}{a_k}$ simplifies to $$\frac {(2k+1)}{(2k+1)(2k+2)} = \frac {1}{2k+2}$$ which approaches to $0$ as $k$ goes to infinity.

Answer 2

We have that

$$\frac{a_{k+1}}{a_k}=\frac{1..3...(2k-1)(2k+1)}{(2k+2)!}\frac{(2k)!}{1..3...(2k-1)}=\frac{2k+1}{(2k+1)(2k+2)} \to 0<1$$

indeed

$$(2k+2)!=(2k+2)(2k+1)(2k)!$$

Answer 3

We have an embarrassment of riches here: The $k$th term is

$$\frac{1}{1\cdot 2}\frac{3}{3\cdot 4}\cdots \frac{2k-1}{(2k-1)\cdot 2k} = \frac{1}{2\cdot 4\cdot 6\cdot \cdot 2k} = \frac{1}{2^k}\frac{1}{k!}.$$

So life is easy.