# How to evaluate the series $1+\frac34+\frac{3\cdot5}{4\cdot8}+\frac{3\cdot5\cdot7}{4\cdot8\cdot12}+\cdots$

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.

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Since you indicate this is homework it would assist in giving you a good answer if you give some more context. Homework in which course? Which theorems were taught in relation to this exercise? – Ittay Weiss Dec 28 '12 at 9:02
possible duplicate math.stackexchange.com/questions/207062/sum-of-the-series/… – clark Dec 28 '12 at 9:30
This is a binomial series. – Lucian May 10 '14 at 15:18

$$a_n = \dfrac{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}{4 \cdot 8 \cdot 12 \cdots (4n)} = \dfrac{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}{4^n \cdot n!} = \dfrac{(2n+1)!}{2^n \cdot 4^n \cdot n! \cdot n!} = \dfrac1{8^n} \dfrac{(2n+1)!}{n! \cdot n!}$$ 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 $$\sum_{n=0}^{\infty} \dfrac{(2n+1)!}{n! \cdot n!}x^n$$

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It might help to have a comment on how you came to pick the particular function $f(x)$ ... since that would help to give insight into other similar problems – Mark Bennet Dec 28 '12 at 8:47
@MarkBennet Thanks. Have added it now. – user17762 Dec 28 '12 at 9:17
Very helpful - thanks. – Mark Bennet Dec 28 '12 at 9:19

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 $(1-1/2)^{-3/2}=2^{3/2}=\sqrt{8}$

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really smart solution! – 007resu Dec 28 '12 at 9:38
Brilliant solution. Concise and uses just binomial theorem – Aditya Sriram Dec 28 '12 at 13:31