I read a document of wikipedia which describes Binomial series.
In this document,
I interested in the part of Conditions for convergence, number 2 that.
If $|x| = 1$, the series converges absolutely if and only if either $Re(α) > 0$ or $α = 0$, where $Re(α)$ denotes the real part of $α$.
Now, Consider a weakly simplified version $\cdots(*)$ of above statement.
$$\mathrm{If \ 0 < \alpha}\in \mathbb{R}, \ (1+x)^{\alpha} = \sum_{n=0}^\infty {\alpha \choose n}x^n \mathrm{\ \ \ for}\ -1 \leq x \leq 1.$$
My idea to prove this statement is,
i) Because of Taylor series convergence, it's obvious on $-1 < x < 1.$
ii) By using Abel's Theorem and $(1+x)^\alpha$ is continuous on $\mathbb{R}$, Show the series converges at $x=1$.
iii) Hence by a result of (ii), Put $x=1$ into $(1+x)^\alpha$ then
Since the series converges absolutely at $x=1$, the series also converges at $x=-1$.
But in step (ii), it's hard to show Abel's Theorem could be applied in this situation.
i.e. Is the series $\sum_{n=0}^\infty {\alpha \choose n}$ converging? $\cdots(**)$
Could you give me some idea of solution for $(**)$?
Or Is there another ideas of proof for $(*)$?
Thank you for your answers. Have a good day!