I am given the question:
True or False? $$ \begin{align} \sqrt{7} &= \sqrt{1 + 6} \\ &= 1 + \frac{1}{2}(6) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}6^2 + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}6^3 - \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{4!}6^4 + \cdots \end{align} $$
I know that the answer is false, but I am struggling to understand why. I was recently taught that $\left(1 + x\right)^r, \forall r \in \mathbb{R}$,$f(x) = \left(1+x\right)^r \Rightarrow \sum_{n=0}^\infty \binom{r}{n}x^n$.
Is it because $(1 + 6)^{\frac{1}{2}}$ means something else than $f(6) = (1 + x)^{\frac{1}{2}}$?
The problem is that the series $$\sum_{n=0}^\infty \binom{1/2}{n}x^n$$ diverges when $n=6$, which can be seen using the ratio test.