Show that $\sum_{k=0}^\infty \binom{\alpha}{k}x^k$, $|x| \leq 1$ converges

In this problem we are going to establish the binomial series

$$(1 + x)^{\alpha} = \sum_{k=0}^\infty \binom{\alpha}{k}x^k$$, $|x| \leq 1$, for any $\alpha$, by showing that $$\lim_{n\to \infty}{R_{n,0}(x)} = 0.$$ The proof involves several steps, and it uses the Cauchy form for the remainder, and the Lagrange forms for the reminder.

a) Use the quotient rule to show that the series converges for $|r| \leq 1$.

Ok, so what I did was, $$\frac{a_{k+1}}{a_k} = \frac{\binom{\alpha}{k+1}r^{k+1}}{\binom{\alpha}{k}r^k} = \frac{r^{k+1}\frac{\alpha!}{(k+1)!(\alpha - k -1)!}}{r^k \frac{\alpha !}{k! (\alpha - k)!}} = r \frac {k! (\alpha - k)! \alpha!}{(k+1)! (\alpha -k -1)! \alpha!}= \frac{r(\alpha - k)}{k+1}.$$ By the quotient rule, $\frac{r(\alpha - k)}{k+1}$ should be less than $1$, so that
$$\sum_{k=0}^\infty \binom{\alpha}{k}r^k$$ converges. An that is why I stated the problem: If $|r| \leq 1$, show that $$\frac{r(\alpha - k)}{k+1} \leq 1$$ where $\alpha \in \Bbb{R}, k \in \Bbb{N}$ , which is false...

• What does $k$ stands for? – CIJ Nov 20 '16 at 23:05
• What about k? . – hamam_Abdallah Nov 20 '16 at 23:05
• $k$ is a natural number – Trux Nov 20 '16 at 23:12
• It doesn't hold for $\alpha=\frac{k+1}{r}+k.$ – mfl Nov 20 '16 at 23:20
• where does $r$ come from ? – G Cab Nov 21 '16 at 0:00

This isn't true in all cases. Take $\alpha = 10$ and $k = 1$. You'll get $r\cdot \frac{9}{2} \leq 1$ which is not true for all $r$ for which $|r| \leq 1$. Are we missing information?

• i will edit my question – Trux Nov 20 '16 at 23:26

Let $r = 0.5, \alpha = 21, k = 1$

Then your inequality does not hold

$$0.5\frac{21 - 1}{1 + 1} \not\leq 1$$

EDIT are you sure you simplified correctly this:

$$\frac{\binom{\alpha}{k+1}r^{k+1}}{\binom{\alpha}{k}r^k}$$

I am not implying that you didn't. I just don't have paper on me and the problem could be here.

2nd EDIT

I don't know how $\binom{\alpha}{k}$ is defined for non-integer and/or non-positive alpha, but when $\alpha$ is a positive integer and when $k > \alpha$, do you know what is the value of $\binom{\alpha}{k}$?

Maybe the proof takes that into consideration. Because if you manage to be able to ignore the $\binom{\alpha}{k}$ and $\binom{\alpha}{k+1}$ (for sufficiently big $k$) then you are left with

$$\frac{\binom{\alpha}{k+1}r^{k+1}}{\binom{\alpha}{k}r^k} < 1 \iff r < 1$$ which you already know.

• Im pretty sure, but i will and the steps – Trux Nov 20 '16 at 23:58
• @Trux how is $\binom{\alpha}{k}$ defined for non-integer $\alpha$? – RGS Nov 21 '16 at 0:02
• according to wikipedia, $\binom{\alpha}{k} = \frac{\alpha (\alpha -1) (\alpha - 2) ...(\alpha - k + 1)}{k!}$ – Trux Nov 21 '16 at 0:10
• ok, I will have to give it a thought... – Trux Nov 21 '16 at 0:14