# Sequences and Series/Power series

Proposition (identity theorem for one-dimensional power series):

Let

$f(z):=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$ and $g(z):=\sum _{n=0}^{\infty }b_{n}(z-z_{0})^{n}$ be two (complex or real) power series that converge on $B_{\epsilon }(z_{0})$ for some $\epsilon >0$ . Suppose that $z_{0}$ is an accumulation point of the set $\{z\in B_{\epsilon }(z_{0})|f(z)=g(z)\}$ . Then we have $a_{n}=b_{n}$ for all $n\in \mathbb {N} _{0}$ .

Proof: Assume that not $a_{n}=b_{n}$ for all $n\in \mathbb {N} _{0}$ . Then there exists a least $n$ (call it $n_{0}$ ) such that $a_{n_{0}}\neq b_{n_{0}}$ . Consider the function

$h(z):=f(z)-g(z)=\sum _{n=0}^{\infty }(a_{n}-b_{n})(z-z_{0})^{n}$ ,

which is defined on at least $B_{\epsilon }(z_{0})$ . Since $a_{n}=b_{n}$ for $n , the power series $h$ starts at $(z-z_{0})^{n_{0}}$ . Therefore,

$j(z):={\frac {h(z)}{(z-z_{0})^{n_{0}}}}=\sum _{n=n_{0}}(a_{n}-b_{n})(z-z_{0})^{n-n_{0}}$ is a well-defined function on $B_{\epsilon }(z_{0})$ which is also continuous due to the continuity of power series. Moreover,

$j(0)=a_{n}-b_{n}\neq _{0}$ ,

and by continuity of $|j(z)|$ , there exists a $\delta >0$ such that $|j(z)|>0$ for all $z\in B_{\delta }(z_{0})$ . But by definition,

$h(z)=(z-z_{0})^{n}j(z)$ ,

so that we have for $z\in B_{\delta }(z_{0})\setminus \{z_{0}\}$ that $|h(z)|=|z-z_{0}|^{n}|j(z)|>0$ and consequently $h(z)\neq 0$ , and hence $f(z)\neq g(z)$ . But this contradicts the assumption that $z_{0}$ was an accumulation point of $\{z\in B_{\epsilon }(z_{0})|f(z)=g(z)\}$ . $\Box$ Example (falsity of the identity theorem for multi-dimensional power series):

For multi-dimensional power series, that is power series of the type

$h(z):=\sum _{\alpha \in \mathbb {N} _{0}^{d}}a_{\alpha }(z-z_{0})^{\alpha }$ for a $z_{0}=(z_{0,1},\ldots ,z_{0,d})\in \mathbb {C} ^{d}$ ,

the set $\{z|h(z)=0\}$ may have $z_{0}$ as an accumulation point even when $h$ does not vanish. An easy example (which works in any dimension $d\geq 2$ ) is $z_{0}=0$ and

$h(z)=z_{1}z_{2}$ . To do:The LHS needs to converge to $\alpha$ as $x=x(z)$ is chosen in the right way.

Theorem (Abel's theorem):

Let

$\sum _{n=1}^{\infty }a_{n}z^{n}$ be a real or complex power series of convergence radius $1$ , and suppose that

$\lim _{z\to 1}\sum _{n=1}^{\infty }a_{n}z^{n}=\alpha \in \mathbb {C}$ .

Then

$\sum _{n=1}^{\infty }a_{n}=\alpha$ .

{{proof|By Abelian partial summation, we have

$\sum _{1\leq n\leq x}a_{n}z^{n}=z^{x}A(x)-\ln(z)\int _{1}^{x}A(t)z^{t}dt$ for $|z|<1$ and $x\geq 1$ , where we denote as usual

$A(x):=\sum _{1\leq n\leq x}a_{n}$ .

Substituting $z=\exp(w)$ , we get

$\sum _{1\leq n\leq x}a_{n}\exp(wn)=\exp(wx)A(x)-w\int _{1}^{x}A(t)\exp(wt)dt$ .

We then put