Proof: Assume that not
for all
. Then there exists a least
(call it
) such that
. Consider the function
,
which is defined on at least
. Since
for
, the power series
starts at
. 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/faa13e6d6c29fcb01195a9f0b1be66a012a1983c)
is a well-defined function on
which is also continuous due to the continuity of power series. Moreover,
,
and by continuity of
, there exists a
such that
for all
. But by definition,
,
so that we have for
that
and consequently
, and hence
. But this contradicts the assumption that
was an accumulation point of
.
Example (falsity of the identity theorem for multi-dimensional power series):
For multi-dimensional power series, that is power series of the type
for a
,
the set
may have
as an accumulation point even when
does not vanish. An easy example (which works in any dimension
) is
and
.
Theorem (Abel's theorem):
Let
![{\displaystyle \sum _{n=1}^{\infty }a_{n}z^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f313c0c77f5e0005084d639614977e9903a3c693)
be a real or complex power series of convergence radius
, and suppose that
.
Then
.
{{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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e3194d54c2becce72062e010154b6544de0792c)
for
and
, where we denote as usual
.
Substituting
, we get
.
We then put