How to evaluate the series $1+frac34+frac{3cdot5}{4cdot8}+frac{3cdot5cdot7}{4cdot8cdot12}+cdots$

How can I evaluate the following series:

$$1+\frac{3}{4}+\frac{3\cdot 5}{4\cdot 8}+\frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12}+\frac{3\cdot 5\cdot 7\cdot 9}{4\cdot 8\cdot 12\cdot 16}+\cdots$$
In one book I saw this sum is equal to $\sqrt{8}$, but I don't know how to evaluate it.

Answer

$$a_n = \dfrac{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}{4 \cdot 8 \cdot 12 \cdots (4n)} = \dfrac{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}{4^n \cdot n!} = \dfrac{(2n+1)!}{2^n \cdot 4^n \cdot n! \cdot n!} = \dfrac1{8^n} \dfrac{(2n+1)!}{n! \cdot n!}$$
Now consider $f(x) = \dfrac1{(1-4x)^{3/2}}$. The Taylor series of $f(x)$ is
$$f(x) = \sum_{n=0}^{\infty} \dfrac{(2n+1)!}{n! \cdot n!} x^n$$ and is valid for $\vert x \vert < \dfrac14$. Hence, we get that
$$\sum_{n=0}^{\infty}\dfrac1{8^n} \dfrac{(2n+1)!}{n! \cdot n!} = f(1/8) = \dfrac1{(1-4\cdot 1/8)^{3/2}} = 2^{3/2} = \sqrt{8} = 2\sqrt{2}$$

EDIT

Below is a way on how to pick the appropriate function. First note that
$$\dfrac{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}{2^n} = (-1)^n \times \left(-\dfrac32\right) \times \left(-\dfrac32-1\right) \times \left(-\dfrac32-2\right) \times \cdots \times \left(-\dfrac32-n+1\right)$$
$$a_n = \dfrac1{2^n}\dfrac{(-1)^n \times \left(-\dfrac32\right) \times \left(-\dfrac32-1\right) \times \left(-\dfrac32-2\right) \times \cdots \times \left(-\dfrac32-n+1\right)}{n!}$$
Hence, $a_n = \left(\dfrac{-1}{2}\right)^n \dbinom{-3/2}{n}$. Hence, the idea is to consider $$g(x) = \sum_{n=0}^{\infty} \dbinom{-3/2}{n} x^n = (1+x)^{-3/2}$$
The motivation to choose such a $g(x)$ comes from the fact that $$(1+x)^{\alpha} = \sum_{n=0}^{\infty} \dbinom{\alpha}{n} x^n$$ where $\dbinom{\alpha}{n}$ is to be interpreted as $\dfrac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!}$ forall real $\alpha$.

Now set $x=-1/2$ to get that $$g(-1/2) = (1/2)^{-3/2} = 2^{3/2} = \sqrt{8} = 2\sqrt{2}$$

Once we have $g(x)$, we could either have it as such or we can play around with some nice scaling factor, to choose the function $f(x) = (1-4x)^{-3/2}$ to get the Taylor series $$\sum_{n=0}^{\infty} \dfrac{(2n+1)!}{n! \cdot n!}x^n$$