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