# Sequences and Series/Power series

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

Let

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

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

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

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

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

${\displaystyle j(z):={\frac {h(z)}{(z-z_{0})^{n_{0}}}}=\sum _{n=n_{0}}(a_{n}-b_{n})(z-z_{0})^{n-n_{0}}}$

is a well-defined function on ${\displaystyle B_{\epsilon }(z_{0})}$ which is also continuous due to the continuity of power series. Moreover,

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

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

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

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

Example (falsity of the identity theorem for multi-dimensional power series):

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

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

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

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

Theorem (Abel's theorem):

Let

${\displaystyle \sum _{n=1}^{\infty }a_{n}z^{n}}$

be a real or complex power series of convergence radius ${\displaystyle 1}$, and suppose that

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

Then

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

{{proof|By Abelian partial summation, we have

${\displaystyle \sum _{1\leq n\leq x}a_{n}z^{n}=z^{x}A(x)-\ln(z)\int _{1}^{x}A(t)z^{t}dt}$

for ${\displaystyle |z|<1}$ and ${\displaystyle x\geq 1}$, where we denote as usual

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

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

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

We then put