Show that a power series is analytic inside its radius of convergence

Question

Let $f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k$ with radius of convergence $R$ then $f$ is analytic on the open disk around $z_0$ with radius $R$.

What I was thinking about is an approach based on this sketch:

I want to find a new power series based on a new disk with center $\widetilde{z_0}$ and variable $z$. Therefore I am using $z-z_0 = (\widetilde{z_0}-z_0)+(z-\widetilde{z_0})$ which yields

$$ f(z) = \sum_{k=0}^\infty a_k\left( (\widetilde{z_0}-z_0)+(z-\widetilde{z_0}) \right)^k = \sum_{k=0}^\infty a_k\sum_{n=0}^k\binom{k}{n}(\widetilde{z_0}-z_0)^n(z-\widetilde{z_0})^{k-n}. $$

I am looking for power series of the form $\sum_{k=0}^\infty b_k(z-\widetilde{z_0})^k$ but haven't been able to come up with the next step. What is your suggestion based on the method of using the binomial formula (explicitly no derivatives of power series etc. if possible)?

Answer 1

Here are two variants. We start with proceeding with your approach.

Since \begin{align*} \sum_{k=0}^{\infty}\sum_{n=0}^k\alpha_{k,n}=\sum_{0\leq n\leq k<\infty}\alpha_{k,n} =\sum_{n=0}^\infty\sum_{k=n}^\infty\alpha_{k,n}\tag{1} \end{align*}

we can write \begin{align*} f(z)&=\sum_{k=0}^\infty a_k\sum_{n=0}^k\binom{k}{n}(\widetilde{z_0}-z_0)^n(z-\widetilde{z_0})^{k-n}\\ &=\sum_{k=0}^\infty a_k\sum_{n=0}^k\binom{k}{n}(\widetilde{z_0}-z_0)^{k-n}(z-\widetilde{z_0})^{n}\tag{2}\\ &=\sum_{n=0}^\infty\left(\sum_{k=n}^\infty a_k\binom{k}{n}(\widetilde{z_0}-z_0)^{k-n}\right)(z-\widetilde{z_0})^{n}\tag{3}\\ \end{align*}

Comment:

• In (2) we change the index summation $n \rightarrow k-n$

• In (3) we change the order of summation according to (1)

Another variant could be based upon Taylor series expansion.

Since the Tayler series of $f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k$ expanded at $z=\widetilde{z_0}$ is \begin{align*} f(z)=\sum_{n=0}^{\infty}\frac{f^{(n)}\left(\widetilde{z_0}\right)}{n!}(z-\widetilde{z_0})^{n} \end{align*}

we obtain \begin{align*} f(z)&=\sum_{n=0}^{\infty}\frac{f^{(n)}\left(\widetilde{z_0}\right)}{n!}(z-\widetilde{z_0})^{n}\\ &=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\left.\left(\frac{d^n}{dz^n}\right)\left(\sum_{k=0}^\infty a_k(z-z_0)^k\right)\right|_{z=\widetilde{z_0}}\right)(z-\widetilde{z_0})^{n}\\ &=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\left.\left(\sum_{k=0}^\infty a_kk(k-1)\cdots(k-n+1)(z-z_0)^{k-n}\right)\right|_{z=\widetilde{z_0}}\right)(z-\widetilde{z_0})^{n}\\ &=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\sum_{k=n}^\infty a_kk(k-1)\cdots(k-n+1)(\widetilde{z_0}-z_0)^{k-n}\right)(z-\widetilde{z_0})^{n}\\ &=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\sum_{k=n}^\infty a_k\frac{k!}{(k-n)!}(\widetilde{z_0}-z_0)^{k-n}\right)(z-\widetilde{z_0})^{n}\\ &=\sum_{n=0}^{\infty}\left(\sum_{k=n}^\infty a_k\binom{k}{n}(\widetilde{z_0}-z_0)^{k-n}\right)(z-\widetilde{z_0})^{n}\\ \end{align*}

Answer 2

Modify a little using the symmetry of the binomial expansion: \begin{align} f(z) & = \sum_{k=0}^\infty a_k\left( (\widetilde{z_0}-z_0)+(z-\widetilde{z_0}) \right)^k \\ & = \sum_{k=0}^\infty a_k\sum_{n=0}^k\binom{k}{n}(\widetilde{z_0}-z_0)^{k-n}(z-\widetilde{z_0})^{n}. \end{align} The coefficient of $(z-\widetilde{z_0})^0$ is $$ a_0(\widetilde{z_0}-z_0)^0+a_1\binom{1}{0}(\widetilde{z_0}-z_0)^1+a_2\binom{2}{0}(\widetilde{z_0}-z_0)^2+\cdots $$ The coefficient of $(z-\widetilde{z_0})^1$ is $$ a_1\binom{1}{1}(z-\widetilde{z_0})^0+a_2\binom{2}{1}(\widetilde{z_0}-z_0)^1+a_3\binom{3}{1}(\widetilde{z_0}-z_0)^2+\cdots $$ Looks like a pattern. The coefficient of $(z-\widetilde{z_0})^{j}$ appears to be $$ a_j\binom{j}{j}(z-\widetilde{z_0})^0+a_{j+1}\binom{j+1}{j}(\widetilde{z_0}-z_0)^1+a_{j+2}\binom{j+2}{j}(\widetilde{z_0}-z_0)^{2}+\cdots. $$