Convergence of a crazy power series "Let $\alpha$ be a given real number, $\alpha>0$ and $\alpha \notin \Bbb{N}$. Proof that the series
$$\sum_{n=1}^{+\infty}\frac{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-(n-1))}{n!}x^n$$
converges for $|x| \leq 1$ and diverges for $|x|>1$."
It's easy to show that this power series converges for $|x|<1$. We have just to use the formula for the convergence radius
$$R=\lim _{n \to +\infty} \left| \frac{a_n}{a_{n+1}}\right|$$
for $a_n=\frac{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-(n-1))}{n!}$, in this case. Using that, we find $R=1$. The trouble is to show that $\sum_{n=1}^{+\infty}a_nx^n$ converges for $|x|=1\dots$
 A: In order to show that $\sum_{n=1}^{+\infty}a_nx^n$ converges for $|x|=1$, i.e., for $x=1$ or $x=-1$ we can use Raabe's Test to show that both of these series,
$$\sum_{n=1}^{+\infty}a_n \text{ and} \sum_{n=1}^{+\infty}a_n(-1)^n$$
converges.
First, note that $\left| \frac{a_{n+1}}{a_n}\right|=|\frac{a_{n+1}(-1)^{n+1}}{a_n(-1)^n}|$. So, the limit involved in Raabe's Test gives the same computation for both of these series.
$$\lim_{n\to+\infty}n\left(1-\left|\frac{a_{n+1}}{a_n}\right|\right)=\lim_{n\to+\infty}n\left(1-\left|\frac{\alpha\dots(\alpha-n)}{(n+1)!}\cdot\frac{n!}{\alpha\dots(\alpha-(n-1)}\right|\right)$$
$$\lim_{n\to+\infty}n\left(1-\left|\frac{\alpha-n}{n+1}\right|\right)=\lim_{n\to+\infty}n\left(1-\frac{n-\alpha}{n+1}\right)=\lim_{n\to+\infty}n\left(\frac{n+1+\alpha-n}{n+1}\right)$$
$$=1+\alpha>1,\text{ since $\alpha>0.$}$$
Thus, by Raabe's Test, the series converges for $|x|=1$. The series diverges for $|x|>1$ obviously because these $x$ are outside the convergence radius.
Note that $a_n$ is never zero, since $\alpha \notin \Bbb{N}$.
A: Here is a direct approach I wanted to share with you. 
First, let $\alpha>1$. Then with $k=\lfloor\alpha\rfloor$, we have
$$
\begin{split}
\frac{|a_n|}{\alpha(\alpha-1)} &= \frac{\alpha-2}{1}\cdot\frac{\alpha-3}{2}\cdots\frac{\alpha-k}{k-1}
\cdot\frac{k-\alpha+1}{k}\cdot\frac{k-\alpha+2}{k+1}\cdots\frac{k-\alpha+(n-1-k)}{n-2}\cdot\frac1{(n-1)n}\\
&\leq\frac{\alpha-2}{1}\cdot\frac{\alpha-3}{2}\cdots\frac{\alpha-k}{k-1}\cdot\frac1{(n-1)n} 
\end{split}
$$
making the convergence obvious.
Now assume $0<\alpha<1$.
Then we have
$$
\begin{split}
\frac{|a_n|}{\alpha(1-\alpha)} &= \frac{2-\alpha}{1}\cdot\frac{3-\alpha}{2}\cdots\frac{n-1-\alpha}{n-2}\cdot\frac1{(n-1)n} \\
&=\big(1+\beta\big)\big(1+\frac\beta2\big)\cdots\big(1+\frac\beta{n-1}\big)\cdot\frac1{(n-1)n}
\end{split}
$$
where $\beta=1-\alpha\in(0,1)$. From the Maclaurin series of $\log(1+x)$, i.e., from Mercator's series, we have the inequality
$$
\log(1+x)< x \qquad \textrm{for} \quad 0<x\leq1 , 
$$
which implies
$$
\log\big(1+\beta\big)+\log\big(1+\frac\beta2\big)+\ldots\log\big(1+\frac\beta{n-1}\big)\leq\beta\big(1+2+\ldots+\frac1{n-1}\big) .
$$
Then taking into account
$$
2+\ldots+\frac1{n-1}\leq\int_1^{n-1}\frac{\mathrm{d}x}x = \log(n-1) ,
$$
we have
$$
\log\big(1+\beta\big)+\log\big(1+\frac\beta2\big)+\ldots\log\big(1+\frac\beta{n-1}\big)\leq\beta\big(1+\log(n-1)\big) 
\leq\beta\log(ne),
$$
or
$$
\big(1+\beta\big)\big(1+\frac\beta2\big)\cdots\big(1+\frac\beta{n-1}\big)
\leq e^{\,\beta} n^{\,\beta} .
$$
Finally, this gives
$$
|a_n| \leq \frac{C\,n^{\,\beta}}{n^{\,2}} = \frac{C}{n^{1+\alpha}} ,
$$
for some constant $C>0$.
