TITLE: How to show that this series does not converge uniformly on the open unit disc?
QUESTION [7 upvotes]: Given the series $\sum_{k=0}^\infty z^k $, it is easy to see that it converges locally, but how do I go about showing that it does not also converge uniformly on the open unit disc? I know that for it to converge uniformly on the open disc that $sup{|g(z) - g_k(z)|}$, z element of open unit disc, must equal zero. However, I am finding it difficult to show that this series does not go to zero as k goes to infinity.
Edit:Fixed confusing terminology as mentioned in answer.

REPLY [7 votes]: The $n$th partial sum of your series is $S_n(z) = {\displaystyle\sum_{k=0}^n z^k = {1 - z^{n+1} \over 1 - z}}$, using the summation formula for geometric series. The sum of the overall series is ${\displaystyle \sum_{k=0}^{\infty} z^k = {1 \over 1 - z}}$.
To say the series converges uniformly to the limit on the disc is to say that $S_n(z)$ converges uniformly to ${\displaystyle {1 \over 1 - z}}$ on the disc. In other words, you have uniform convergence if for every $\epsilon > 0$ there is an $N$ such that if $n > N$ then for all $z$ with $|z| < 1$ you have
$$ |S_n(z) - {1 \over 1 - z}| < \epsilon$$
But 
$$|S_n(z) - {1 \over 1 - z}| = |{1 - z^{n+1} \over 1 - z} - {1 \over 1 - z}|$$
$$ = |{-z^{n+1} \over 1 - z}|$$
$$= {|z|^{n+1} \over |1 - z|}$$
So the question becomes, is it true that for every $\epsilon > 0$ there is an $N$ such that if $n > N$ then for all $z$ with $|z| < 1$ you have
$${|z|^{n+1} \over |1 - z|} < \epsilon $$
But this is clearly false. For any $n$, the limit ${\displaystyle \lim_{x \rightarrow 1^-}{x^{n+1} \over 1 - x} = \infty}$, so there is no $n$ for which ${|z|^{n+1} \over |1 - z|} < \epsilon $ holds for all $z$ with $0 < z < 1$, much less for all $z$ on the unit disc. So uniform convergence does not hold.

REPLY [2 votes]: Confine attention to real $x$ in the interval $0<x<1$. 
Let 
$$s_n(x)=1+x+x^2+\cdots +x^{n-1}.$$
If we use $s_n(x)$ to approximate the sum, the truncation error is $>x^n$.
Choose a positive $\epsilon$, where for convenience $\epsilon<1$.  We want to make the truncation error $<\epsilon$, so we need 
$$x^n <\epsilon,\qquad \text{or equivalently}\qquad n >\frac{|\ln(\epsilon)|}{|\ln(x)|}.$$
Since $\ln x \to 0$ as $x \to 1^{-}$, the required $n$ grows without bound as $x\to 1^{-}$.