Prove that $\sqrt{8}=1+\dfrac34+\dfrac{3\cdot5}{4\cdot8}+\dfrac{3\cdot5\cdot7}{4\cdot8\cdot12}+\ldots$

Question

Prove that $\sqrt{8}=1+\dfrac34+\dfrac{3\cdot5}{4\cdot8}+\dfrac{3\cdot5\cdot7}{4\cdot8\cdot12}+\ldots$

My work:
$\sqrt8=\bigg(1-\dfrac12\bigg)^{-\frac32}$

Now, I suppose there is some "binomial expansion with rational co-efficients" or something similar for this, which I do not know. Please help.

N.B.: Any other solution is also acceptable, I do not have any restriction regarding the solution.

Answer 1

The binomial series is "just" the Taylor series of $(1+x)^{\alpha}$ at $x=0$.

Start deriving $f(x)=(1+x)^{\alpha}$ by $x$ and you get $\alpha(1+x)^{\alpha-1}$, then $\alpha(\alpha-1)(1+x)^{\alpha-2}$ for the first derivative etc. The $n$th derivative is

$$\frac{d^n}{dx^n} (1+x)^{\alpha} = \alpha(\alpha-1)\cdots(\alpha-n+1)(1+x)^{\alpha-n}$$

So the Taylor series is

$$\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n= \sum_{n=0}^\infty \frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!} x^n$$

If $\alpha$ is a non-negative integer this series becomes a finite sum and the $n$th coefficient is simply ${\alpha\choose n}$, so it makes sense to call it a "generalized" binomial coefficient and denote it by the same symbol, also if $\alpha$ is not a non-negative integer.

Now notice that the Taylor series has convergence radius $1$ and that $(1+x)^{\alpha}$ is real-analytic, therefore

$$(1+x)^{\alpha} = \sum_{n=0}^\infty {\alpha\choose n} x^n$$

for all $|x|<1$, in particular $x=-\frac{1}{2}$ and $\alpha=-3/2$. Now notice that $${-3/2\choose n}=\frac{(-1)^n}{n!2^n} 3\cdot 5\cdots (2n+1)$$

Then,

$$\begin{align} \sqrt{8} &=(1-1/2)^{-3/2} = \sum_{n=0}^\infty \frac{(-1)^n}{n!2^n} 3\cdot 5\cdots (2n+1) \frac{1}{(-2)^n} = \sum_{n=0}^\infty \frac{3\cdot 5\cdots (2n+1)}{n! 4^n}\\ &= 1 + \frac{3}{4} + \frac{3\cdot 5}{4\cdot 8} + \frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12} + \cdots \end{align} $$

Answer 2

It can be shown that $\displaystyle (1 + x)^{\alpha} = \sum_{n = 0}^{\infty} \binom{\alpha}{n}x^n$ when $|x| < 1$. This is not really precalculus though...

EDIT: I'll add a proof for the sum:

Let $s(x)$ be the sum of the series. Then

$\begin{align} (1 + x)s'(x) &= (1 + x)\sum_{n = 0}^{\infty} n\binom{\alpha}{n}x^{n-1} \\ &= \sum_{n = 0}^{\infty} (n+1)\binom{\alpha}{n+1}x^n + \sum_{n = 0}^{\infty}n\binom{\alpha}{n}x^n \\ &= \sum_{n = 0}^{\infty} \left[(n+1)\binom{\alpha}{n+1} + n\binom{\alpha}{n}\right]x^n \\ &= \alpha \sum_{n = 0}^{\infty} \binom{\alpha}{n}x^n = \alpha \; s(x) \end{align}$

Furthermore, let $f(x) := (1+x)^{-\alpha} s(x)$. Then

$\displaystyle f'(x) = -\alpha(1+x)^{-\alpha - 1} s(x) + (1+x)^{-\alpha} s'(x) = -(1+x)^{-\alpha} s'(x) + (1+x)^{-\alpha} s'(x) = 0$

which implies that $f(x)$ is costant. Since $f(0) = 1$, we have $f(x) = 1$ and therefore $(1+x)^{-\alpha} s(x) = 1$. Multiply both sides by $(1+x)^{\alpha}$ and we are done.