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Hi here's my question:

Show that the following power series has radius of convergence r=1:

$\sum_{k=0}^\infty \binom{\alpha}{k}x^k$

for $\alpha>0$.

I'm so thrown by the $\binom{\alpha}{k}$. So far I've tried 2 approaches and think both are wrong:

1) Using that $\binom{\alpha}{k} = \alpha(\alpha-1)...(\alpha-(k-1))/k!$ = $\alpha_k/k!$

$\sum_{k=0}^\infty \binom{\alpha}{k}x^k$ = $\sum_{k=0}^\infty\alpha_kx^k/k!$

I know that $\sum_{k=0}^\infty x^k/k!$ has radius of convergence r=$\infty$

but I'm not sure where I can go from there or if I can at all.

2) For $\sum_{k=0}^\infty\alpha_k x^k/k!$ ,
$L=\lim_{k\to \infty}sup(|\alpha_k/k!|)^\frac1k = 1$

So since $r=0$ if $L=+\infty$ ; $r=+\infty$ if $L=0$ and $r=\frac1L$ if $0<L,+\infty$.

Then $r=\frac11 = 1$

hence radius of convergence is 1. However this is more of a working backwards so I feel I need more detail on the second line ($L=$) but am unsure of how to do it.

Thank you!!

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  • $\begingroup$ Try the ratio test: using your formula for $\binom{\alpha}{k}$, it should be easy to simplify $\binom{\alpha}{k+1} / \binom{\alpha}{k}$. $\endgroup$ – Daniel Schepler Jun 29 '17 at 21:57
  • $\begingroup$ Am I missing something.. For all $a,$ eventually we will hit a $k$ where $k>a$ and ${a\choose k}$ is undefined. What happens then? $\endgroup$ – Doug M Jun 29 '17 at 22:11
  • $\begingroup$ We can get an upper bound for $|\binom {a}{k}|$ in terms of factorials and apply Stirling's Formula to the factorials, to get an upper bound for $|\binom {a}{k}|^{1/k} .$ But it is much easier to use the A by Jose Carlos Santos. $\endgroup$ – DanielWainfleet Jun 30 '17 at 6:17
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I assume that $\alpha\notin\mathbb Z$.

Then\begin{align*}\frac{\binom\alpha{n+1}}{\binom\alpha n}&=\frac{\frac1{(n+1)!}\alpha(\alpha-1)\ldots(\alpha-n)}{\frac1{n!}\alpha(\alpha-1)(\alpha-(n-1))}\\&=\frac{\alpha-n}{n+1}\end{align*}and therefore $\displaystyle\lim_{n\in\mathbb N}\left|\frac{\binom\alpha{n+1}}{\binom\alpha n}\right|=1$. So, the radius of convergence is $1$.

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  • $\begingroup$ To ew22 : For $\alpha \not \in \{0\}\cup \mathbb N $ (so that there are infinitely many non-zero co-efficients) : If $0<|x|=1-r<1$ then for all but finitely many $k$ we have $|\binom {\alpha}{k+1}x^{k+1}/\binom {\alpha}{k}x^k|< (1+r/2)|x|=1-r/2-r^2/2.$ If $|x|=1+r>1$ then for all but finitely many $k$ the absolute ratio of successive terms of the series is more than $(1-r/2)|x| = (1-r/2)(1+r),$ which is more than $1.$ $\endgroup$ – DanielWainfleet Jun 30 '17 at 6:30

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