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!!

  • $\begingroup$ 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

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 $1$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.