I'm trying to develop the real function $f(x) = (1 + x)^{\alpha}$, $\alpha \in \mathbb{R}$, but I'm stuck in the middle.
What I did until now:
First, we have that the $f$ is defined for $\forall x \in \mathbb{R}$.
Then, I examined if the $f$ is infinitely diferentiable:
$$f'(x) = \alpha(1 + x)^{\alpha - 1},$$
$$f'(x) = \alpha(\alpha - 1)(1 + x)^{\alpha - 2},$$
$$f'''(x) = \alpha(\alpha - 1)(\alpha - 2)(1 + x)^{\alpha - 3},$$
$$\vdots$$
$$f^{(n)}(x) = \alpha(\alpha - 1)(\alpha - 2) \cdots (\alpha - n + 1)(1 + x)^{\alpha - n}.$$
So, $f$ is infinitely diferentiable on the whole set $\mathbb{R}$.
Now, I chose the point $x_0 = 0$ in which we are going to develop $f$.
Because, $f$ is infinitely diferentiable, that implies it's also continuous with all its derivatives up to the $n$-th order and in the $n$-th order, in the arbitrarily neighborhood $U$ of the point $x_0 = 0$, and it has derivative of the $n + 1$-st order in $x_0$. So, now we have that all conditions of the following theorem,
Checking my understanding of the process of developing function into power series,
are fulfilled. Now we have that we can write MacLaurien formula for the $f$ (because $x_0 = 0$):
$$f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + R_n(x) = $$
$$= 1 + \frac{\alpha x}{1!} + \frac{\alpha(\alpha - 1) x^2}{2!} + \frac{\alpha(\alpha - 1)(\alpha - 2) x^3}{3!} + \cdots + \frac{\alpha \cdots (\alpha - n + 1) x^n}{n!} + R_n(x) =$$
$$= 1 + \sum\limits_{n = 1}^{\infty} \frac{\alpha(\alpha - 1) \cdots (\alpha - n + 1) x^n}{n!} + R_n(x).$$
My question:
I think that is a mistake that I have $1$ in the last row of the last equation. I don't think that it can be added to the remainder $R_n(x)$.
Please, could you tell me where I made a mistake and how to fix it?
Tuck it into your series by making it $\sum_{n=0}^\infty$
Since your sum is effectively $$\sum_{n=1}^\infty \binom \alpha n x^n$$ setting $n=0$ gives the $1$ you took out.