Generalized binomial theorem

You have that if

$$y=(1+x)^\alpha$$

then

$$y'(0)=\alpha$$ $$y''(0)=\alpha(\alpha-1)$$ $$\cdots=\cdots$$ $$y^{(n)}(0)=\alpha(\alpha-1)\cdots(\alpha-n+1)$$

$$y^{(n)}(0)=\frac{\alpha!}{(\alpha-n)!}$$

You can prove this last generalization by induction on $n$. The last equation is written for the sake of "order", but if you're not comfortable with it, you can use the $\Gamma$ function to denote the factorials.

Then

$$\mathcal{T}_k(y,0)=\sum\limits_{n=0}^k \frac{\alpha!}{(\alpha-n)!}\frac{x^n}{n!}$$

Then using the binomial coefficient notation

$$\mathcal{T}_k(y,0)=\sum\limits_{n=0}^k {\alpha \choose n}x^n$$

What you need now is that the series converges for $|x|<1$.

In my opinion you can argue as follows:

Fix $x$. Then the series

$$\lim\limits_{k \to \infty} \mathcal{T}_k(y,0)=\sum\limits_{n=0}^\infty {\alpha \choose n}x^n$$

will converge when

$$ \lim\limits_{n \to \infty} \left|\frac{{\alpha \choose n+1}}{{\alpha \choose n}}\frac{x^{n+1}}{x^n}\right|<1$$

$$ \lim\limits_{n \to \infty} \left|\frac{{n+1}}{n-\alpha}x\right|<1$$

$$ \left|x\right|<1$$

ADD: Here's a part of an answer to a similar question:

Assume

$${\left( {1 + x} \right)^{\alpha}} = \sum\limits_{k = 0}^\infty {{a_k}{x^k}} $$

Since if two series $$\eqalign{ & \sum {{a_k}{{\left( {x - a} \right)}^k}} \cr & \sum {{b_k}{{\left( {x - a} \right)}^k}} \cr} $$

sum up to the same function then

$${a_k} = {b_k} = \frac{{{f^{\left( k \right)}}\left( a \right)}}{{k!}}$$

for every $k \leq 0$, we can assume:

$$a_k = \dfrac{f^{(k)}(0)}{k!}$$

Putting $y = {\left( {1 + x} \right)^{\alpha}}$ we get

$$y'(0) = \alpha$$ $$y''(0) = \alpha(\alpha-1)$$ $$y'''(0) = \alpha(\alpha-1)(\alpha-2)$$ $$y^{{IV}}(0) = \alpha(\alpha-1)(\alpha-2)(\alpha-3)$$

We can prove in general that

$$y^{(k)}= \alpha(\alpha-1)\cdots(\alpha-k+1)$$

or put in terms of factorials

$$y^{(k)}(0)= \frac{\alpha!}{(\alpha-k)!}$$

This makes

$$a_k = \frac{\alpha!}{k!(\alpha-k)!}$$

which is what we wanted.

$${\left( {1 + x} \right)^\alpha } = \sum\limits_{k = 0}^\infty {\alpha \choose k} {{x^k}} $$

You can prove this in a more rigorous manner by differential equations:

1. Set $f(x) = \displaystyle \sum\limits_{k = 0}^\infty {\alpha \choose k} {{x^k}}$ and prove the radius of convergence is 1.
2. Show that $f(x)$ is the solution to the ODE $$y' - \frac{\alpha }{{x + 1}}y = 0$$ with initial condition $f(0)=1$.
3. By the theorem that the solution to the linear equation

$$y'+P(x)y=R(x)$$

with initial conditions $f(a) = b$ is unique, you can prove the assertion. (prove that $y = {\left( {1 + x} \right)^{\alpha}}$ also satifies the equation and you're done.)