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**Corollary**: If ##f## is holomorphic in an open set ##Ω##, then ##f## has infinitely many complex derivatives in ##Ω##. Moreover, if ##C⊂Ω## is a circle whose interior is also contained in ##Ω##, then

##f^{(n)}(z)=\dfrac{n!}{2πi}∫_C \dfrac{f(ζ)}{(ζ−z)^{n+1}}dζ## for all ##z## in the interior of ##C##

I'll write the complete proof copied from Stein's textbook and then I'll ask my question:

**Proof**The proof is by induction on ##n##, the case ##n=0## being simply the Cauchy integral formula. Suppose that ##f## has up to ##n−1## complex derivatives and that ##f^{(n−1)}(z)=\dfrac{(n−1)!}{2πi}∫_C \dfrac{f(ζ)}{(ζ−z)^n}dζ##.

Now for ##h## small, the difference quotient for ##f^{((n−1)}## takes the form

(1) ##f^{(n−1)}(z+h)−f^{(n−1)}(z)h=\dfrac{(n−1)!}{2πi}∫_C f(ζ)\dfrac{1}{h}[\dfrac{1}{(ζ−z−h)^n}-\dfrac{1}{(ζ−z)^n}]dζ##

We now recall that ##A^n−B^n=(A−B)[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}]##

With ##A=\dfrac{1}{(ζ−z−h)}## and ##B=\dfrac{1}{(ζ−z)}##, we see that the term in brackets in equation (1) is equal to

##\dfrac{h}{(ζ−z−h)(ζ−z)}[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}]##. But observe that if ##h## is small, then ##z+h## and ##z## stay at a finite distance from the boundary circle ##C##, so in the limit as ##h## tends to ##0##, we find that the quotient converges to

##\dfrac{(n−1)!}{2πi}∫_C f(ζ)[\dfrac{1}{(ζ−z)^2}][\dfrac{n}{(ζ−z)^{n−1}}]dζ=\dfrac{n!}{2πi}∫_C \dfrac{f(ζ)}{(ζ−z)^{n+1}}dζ##,

which completes the induction argument and proves the theorem.

**My question is**: how is it that ##\dfrac{h}{(ζ−z−h)(ζ−z)}[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}]## converges to ##[\dfrac{1}{(ζ−z)^2}][\dfrac{n}{(ζ−z)^{n−1}}]## given the condition of ##h## being sufficiently small?

I've tried to manipulate the first expression in order to get to the latter but I couldn't arrive to anything. I would be greatly appreciate if someone could clearly explain me this.