12
$\begingroup$

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.

$\endgroup$
26
$\begingroup$

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

$\endgroup$
  • 4
    $\begingroup$ 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 $\endgroup$ – Mark Bennet Dec 28 '12 at 8:47
  • 1
    $\begingroup$ @MarkBennet Thanks. Have added it now. $\endgroup$ – user17762 Dec 28 '12 at 9:17
  • $\begingroup$ Very helpful - thanks. $\endgroup$ – Mark Bennet Dec 28 '12 at 9:19
20
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.