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 $n<n_{0}$, 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

$h(z)=z_{1}z_{2}$.

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 ${\displaystyle }$