TITLE: How to evaluate the series $1+\frac34+\frac{3\cdot5}{4\cdot8}+\frac{3\cdot5\cdot7}{4\cdot8\cdot12}+\cdots$
QUESTION [13 upvotes]: 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.

REPLY [21 votes]: The series can be written as $$1+\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)+\frac{1}{2!}\left(-\frac{3}{2}\right)\left(-\frac{3}{2}-1\right)\left(-\frac{1}{2}\right)^2+\frac{1}{3!}\left(-\frac{3}{2}\right)\left(-\frac{3}{2}-1\right)\left(-\frac{3}{2}-2\right)\left(-\frac{1}{2}\right)^3+\cdots$$ which by binomial theorem is nothing but $(1-1/2)^{-3/2}=2^{3/2}=\sqrt{8}$