Corollary from Cauchy's integral formula

[mahler1]

I have doubts about the proof given in Stein's complex analysis textbook from the following Cauchy's integral formula corollary:

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.

 

[STEMucator]

I have doubts about the proof given in Stein's complex analysis textbook from the following Cauchy's integral formula corollary:

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.

$\frac{h}{(ζ−z−h)(ζ−z)}[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}] \space (1)$

From what I can see, the $h$ in the numerator of this expression will first cancel the $\frac{1}{h}$ term in the integrand.

As $h$ gets small, $(1) → [\dfrac{1}{(ζ−z)(ζ−z)}][\dfrac{n}{(ζ−z)^{n−1}}]$.

If you plug in $A$ and $B$ into the formula for $A^n - B^n$, what is the result if $h$ is small?

EDIT: As another hint, take a look at the first two terms $A^{n−1} + A^{n−2}B$ and the last two terms $AB^{n−2} + B^{n−1}$.

If you add the first two terms when $h$ is small, the result is $\frac{2}{(ζ-z)^{n-1}}$. If you add the last two terms, what do you get? Notice you're really just adding $\frac{1}{(ζ-z)^{n-1}}$ a total of $n$ times.

 

[mahler1]

$\frac{h}{(ζ−z−h)(ζ−z)}[A^{n−1}+A^{n−2}B+⋅⋅⋅+AB^{n−2}+B^{n−1}] \space (1)$

From what I can see, the $h$ in the numerator of this expression will first cancel the $\frac{1}{h}$ term in the integrand.

As $h$ gets small, $(1) → [\dfrac{1}{(ζ−z)(ζ−z)}][\dfrac{n}{(ζ−z)^{n−1}}]$.

If you plug in $A$ and $B$ into the formula for $A^n - B^n$, what is the result if $h$ is small?

EDIT: As another hint, take a look at the first two terms $A^{n−1} + A^{n−2}B$ and the last two terms $AB^{n−2} + B^{n−1}$.

If you add the first two terms when $h$ is small, the result is $\frac{2}{(ζ-z)^{n-1}}$. If you add the last two terms, what do you get? Notice you're really just adding $\frac{1}{(ζ-z)^{n-1}}$ a total of $n$ times.

Oh, what a fool of me, I should have put $h[\dfrac{1}{(ζ−z)^2}][\dfrac{n}{(ζ−z)^{n−1}}]$ but, as you've said, then $h$ will cancel with $\dfrac{1}{h}$. Thanks for your answer, now I see why it converges to that expression.