I'm having trouble with uniform convergence. I need to prove that $$\sum_{n=0}^\infty \left(\frac{z-1}{z+1}\right)^n $$ converges locally uniformly in the half-plane $Re z >0$ and find its sum.
I'm not too sure what the difference between locally uniform convergence and just regular uniform convergence is. I'm clear with the definition but do not see how they are any different.
So far I've been trying to prove just uniform convergence: $|f_n (x)-f(x)| =| \sum_{k=0}^\infty \left(\frac{z-1}{z+1}\right)^k- \sum_{k=n}^\infty \left(\frac{z-1}{z+1}\right)^k |$ = |$\sum_{k=n+1}^\infty \left(\frac{z-1}{z+1}\right)^n$| = $\frac{1}{2}|\frac{(z-1)^{n+1}}{(z+1)^n}| $
the supremum of that is infinite as $z\to\infty$ and $n\to\infty$ which is not what I want.
Appreciate any help!
Writing $z = x + iy$, we have that $$ \left| \frac{z-1}{z+1} \right|^2 = \frac{(x-1)^2+y^2}{(x+1)^2 + y^2} < 1.$$ Thus, the series converges pointwise. You can deduce local uniform convergence by using continuity of absolute value. Fixing $z_0 \in \text{Re}{z} > 0$, choose $s$ with $0 < s < 1$ and $r > 0$ so that for all $z \in B(z_0,r)$, $$ \left| \frac{z-1}{z+1} \right|^2 \leq s.$$ Then for any particular $N \in \mathbb{N}$, $$ \left|\sum_{n = N}^{\infty} \left(\frac{z-1}{z+1} \right)^n \right| \leq \sum_{n = N}^{\infty} \left| \frac{z-1}{z+1} \right|^n \leq \sum_{n = N}^{\infty}s^n.$$ Given $\varepsilon > 0$, to make the sum of the tail less than $\varepsilon$ (in absolute value) for any $z \in B(z_0,r)$, choose $N$ to make $$\sum_{n = N}^{\infty}s^n$$ less than $\varepsilon$.
Of course, using the formula for the sum of an infinite geometric series, the pointwise limit of the series is given by $$ \frac{1}{1 - \frac{z-1}{z+1}} = \frac{1}{\frac{z+1}{z+1} - \frac{z-1}{z+1}} = \frac{z+1}{2}.$$