indirect proof of identity theorem of power series

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{a}_n{(z-z_0)}^n={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{b}_n{(z-z_0)}^n$ (1)

is valid in the set of points $z$ presumed in the theorem (http://planetmath.org/IdentityTheoremOfPowerSeries) to be proved.

Antithesis:  There are integers $n$ such that  $a_n\ne b_n$;  let $\nu$ ($\geqq 0$) be least of them.

We can choose from the point set an infinite sequence  $z_1,z_2,z_3,\mathrm{\dots }$  which converges to $z_0$ with  $z_n\ne z_0$  for every $n$.  Let $z$ in the equation (1) belong to  $\{z_1,z_2,z_3,\mathrm{\dots }\}$  and let’s divide both of (1) by ${(z-z_0)}^{\nu }$ which is distinct from zero; we then have

$\underset{f(z)}{\underbrace{a_{\nu }+a_{\nu +1}(z-z_0)+a_{\nu +2}{(z-z_0)}^2+\mathrm{\dots }}}=\underset{g(z)}{\underbrace{b_{\nu }+b_{\nu +1}(z-z_0)+b_{\nu +2}{(z-z_0)}^2+\mathrm{\dots }}}$ (2)

Let here $z$ to tend $z_0$ along the points $z_1,z_2,z_3,\mathrm{\dots }$, i.e. we take the limits ${lim}_{n\to \mathrm{\infty }}f(z_n)$ and ${lim}_{n\to \mathrm{\infty }}g(z_n)$.  Because the sum of power series is always a continuous function, we see that in (2),

$$\mathrm{left}\mathrm{side}⟶f(z_0)=a_{\nu }\mathit{\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}}\mathrm{right}\mathrm{side}⟶g(z_0)=b_{\nu }$$

But all the time, the left and of (2) are equal, and thus also the limits.  So we must have  $a_{\nu }=b_{\nu }$,  contrary to the antithesis.  We conclude that the antithesis is wrong.  This settles the proof.

Note.  I learned this proof from my venerable teacher, the number-theorist Kustaa Inkeri (1908–1997).

Title	indirect proof of identity theorem of power series
Canonical name	IndirectProofOfIdentityTheoremOfPowerSeries
Date of creation	2013-03-22 16:47:48
Last modified on	2013-03-22 16:47:48
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	9
Author	pahio (2872)
Entry type	Proof
Classification	msc 40A30
Classification	msc 30B10