I have one complex series and one sequence. It is used in complex analysis in a part of my book where they are integrated. However, as you know in order to change limit and integration order it has to converge uniformly. I can only see why they converge pointwise, not uniformly, can some of you please explain why they converge uniformly?
The first one:
Here $\zeta$ are the points on a circle, and z is one point inside that circle.
First we define $A = \frac{1}{\zeta -z-h}$, and B is $\frac{1}{\zeta-z}$, we let h goes to zero and consider the expression:
$\frac{1}{(\zeta-z-h)*(\zeta-z)}(A^{n-1}+A^{n-2}*B....+A*B^{n-2}+B^{n-1})$, now for me it is easy to see that we get the pointwise limit $\frac{n}{(\zeta-z)^{n+1}}$. But how can I see that it converges uniformly to this expression for all $\zeta$?
The second one:
We have the same situation as above, but now $z_0$ is the center of the circle . z is still a point inside the circle and the $\zeta$'s are the point on the boundary.
I see that pointwise we have $\Sigma_{n=0}^\infty(\dfrac{z-z_0}{\zeta-z_0})^n=\dfrac{1}{1-\frac{z-z_0}{\zeta-z_0}}$. But how come the convergence is uniform for all $\zeta$?, is there a way to show that?
The first limit you gave amounts to $$ \lim_{h\to 0} \frac{f_\zeta(z+h)-f_\zeta(z)}{h} = f_\zeta'(z) $$ where $f_\zeta(z) = \frac{1}{(\zeta-z)^n}$. The rate of convergence of divided difference to the derivative is controlled by the supremum of the second derivative $f_\zeta''$ in a neighborhood $N$ of $z$. This derivative is uniformly bounded in $N$ with respect to $\zeta$, as long as $N$ is small enough to be at positive distance from the boundary. By the Taylor estimate $$ \left|\frac{f_\zeta(z+h)-f_\zeta(z)}{h} - f_\zeta'(z)\right| \le \frac{h}2 \sup_N |f_\zeta''| $$ the convergence is uniform.
For the second one, the Weierstrass M-test applies. Let $k=|z-z_0|/r$, where $r=|\zeta-z_0|$ is the radius of the circle, independent of $\zeta$. Note that $k<1$. Since $\sum_{n=0}^\infty k^n$ converges, the Weierstrass test gives uniform convergence.