Show that if $z \in \Bbb C$ is such that $|z| < 1$, then $\sum_{n=0} ^{\infty} {a \choose n}z^n$ converges for any $a \in \Bbb C$.

Question

Show that if $z \in \Bbb C$ is such that $|z| < 1$, then $\sum_{n=0} ^{\infty} {a \choose n}z^n$ converges for any $a \in \Bbb C$.

The fact that these are complex numbers is throwing me off. I know that as $n$ increases, the value of $z^n$ is getting smaller and smaller, therefore the binomial ${a \choose n}$ is being multiplied by smaller and smaller values.

Can I consider this a multiplication of two separate series? $\sum_{n=0}^{\infty} {a \choose n}$ and $\sum_{n=0}^{\infty} z^n$ ? Is that a sensible direction to head in?

Answer 1

$\sum_{n\geq 0}\binom{a}{n}z^n$ converges if $$ \lim_{n\to\infty}\Big|\frac{a_{n+1}}{a_n}\Big|<1$$ where $a_n=\binom{a}{n}z^n.$

Then compute $$\Big|\frac{a_{n+1}}{a_n}\Big|=\Big|\frac{a-n}{n+1}z\Big|$$ and note that the above limit converges to $|z|.$

Answer 2

Although the existing answer using the ratio test is more straightforward, I thought I'd post another approach which might be of interest.

The series $\sum_{n=0}^\infty \binom{a}{n} z^n$ is the Taylor series of the function $(1+z)^a$ centered at $z=0$. On the other hand, for any $a$, this function is holomorphic on $\mathbb{C}$ minus the ray along the real axis from $-1$ to $-\infty$ (where the branch cut is needed if $a$ is not an integer, and $-1$ must be excluded if $a$ is not an integer or if $a < 0$).

Therefore, by the theorem that any holomorphic function of one complex variable is analytic, it follows that this Taylor series has radius of convergence $\ge 1$, which is another way of saying the desired result that the series converges whenever $|z| < 1$. It also gives that the sum of this series is in fact equal to $(1+z)^a$ for $|z| < 1$.

(The obvious disadvantage of this approach is that you pretty much have to guess, or know beforehand, what the sum of the series actually is.)