# Thread: Proof that an infinite series is analytic

1. ## Proof that an infinite series is analytic

Prove that the function $f(z)=1+ \sum _{n=1}^{\infty} \frac{a(a-1)...(a-n+1)}{n!}z^n$ is analytic when $|z|<1$ and that its derivative is $\frac{a f(z)}{1+z}$. Then demonstrate that the derivative of $(1+z)^{-a} f(z)$ is worth $0$ and deduce that $f(z)=(1+z)^a$. Note that $(1+z^a)=exp(a Log(1+z))$.

My attempt: $f(z)= \sum _{n=1}^{\infty} \frac{(a-0)(a-1)(a-2)...[a-(n-1)]}{n!} z^n=az+\frac{a(a-1 z^2)}{2}+\frac{a(a-1)(a-2) z^3}{3}+...=\alpha z + \beta z^2 + \gamma z ^3 +...$ where $\alpha$, $\beta$, $\gamma$,... $\in \mathbb{R}$. So it seems to me that f is analytic for any value of z though I don't know how to formally prove it. So I don't understand why they bother with |z|<1.
I'd like to tackle this exercise and eventually finish it with your help. Thanks for any comment/tip.

2. Originally Posted by arbolis
Prove that the function $f(z)=1+ \sum _{n=1}^{\infty} \frac{a(a-1)...(a-n+1)}{n!}z^n$ is analytic when $|z|<1$ and that its derivative is $\frac{a f(z)}{1+z}$. Then demonstrate that the derivative of $(1+z)^{-a} f(z)$ is worth $0$ and deduce that $f(z)=(1+z)^a$. Note that $(1+z^a)=exp(a Log(1+z))$.

My attempt: $f(z)= \sum _{n=1}^{\infty} \frac{(a-0)(a-1)(a-2)...[a-(n-1)]}{n!} z^n=az+\frac{a(a-1 z^2)}{2}+\frac{a(a-1)(a-2) z^3}{3}+...=\alpha z + \beta z^2 + \gamma z ^3 +...$ where $\alpha$, $\beta$, $\gamma$,... $\in \mathbb{R}$. So it seems to me that f is analytic for any value of z though I don't know how to formally prove it. So I don't understand why they bother with |z|<1.
I'd like to tackle this exercise and eventually finish it with your help. Thanks for any comment/tip.

Put $a_n:=\frac{\alpha(\alpha-1)\cdot\ldots\cdot(\alpha-n+1)\,z^n}{n!}$ , then after making some algebra we get $\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{\alph a -n}{n+1}\right||z|\xrightarrow [n\to\infty]{}|z|$ , and

then the series converges absolutely for $|z|<1$ and it's thus an analytic function there.

Tonio

3. Oh thanks a lot. I will try to do the algebra on my own. So you applied the root test and saw that the radius of convergence depends on |z|. And if |z|<1 then the series converges absolutely and the function is therefore analytic.
Is there a theorem that states that any absolutely convergent power series is an analytic function? I personally don't find it "that obvious".
Correct me if I'm wrong: Any analytic function can be written as an infinite series. Must these infinite series converge absolutely? Or maybe a simple convergence is enough for some series to be analytic functions?