Prove that $\{ \sum_{n=0}^k z^n \}_{k=0}^\infty$ does not converge uniformly?

Question

So as the title says, can anyone prove that $\{ \sum_{n=0}^k z^n \}_{k=0}^\infty$ does not converge uniformly 0n the disk $D(0,1)$? I think it would converge uniformly to $1/(1-z)$ since it is a geometric series, but professor posed the problem so I'm thinking that must not be correct. Thoughts?

Answer 1

Consider $f_n(z) = 1+z+\cdots + z^n = { 1-z^{n+1} \over 1-z}$. Then, for $|z|<1$, we have $f_n(z) \to f(z)={1 \over 1-z}$.

We have $d_n(z)=f(z)-f_n(z) = {z^n \over 1-z}$ (this is pointwise convergence).

Let $z_n = {n \over 1+n}$, and note that $|z_n| <1$ and $d_n(z_n) = {1+n \over (1-{1 \over 1+n})} (1-{1 \over n+1})^{n+1}$. Since $(1-{1 \over n+1})^{n+1} \to {1 \over e}$, we see that $d_n(z_n) \to \infty$.

In particular, $\lim_n \sup_{|z|<1} |f(z)-f_n(z)| = \infty$

Answer 2

Hint:

In $D(0,1),$

$$\left|\sum_{n=0}^k z^n - (1-z)^{-1}\right| = \left|\frac{-z^{k+1}}{1-z}\right| = \frac{|z|^{k+1}}{|1-z|} \geqslant \frac{|z|^{k+1}}{1 + |z|} \geqslant \frac{|z|^{k+1}}{2}.$$

Now consider the supremum over $D(0,1)$ in the limit as $k \to \infty.$

Answer 3

Realize that $\sum_{k=0}^{\infty} x = 1/(1-x)$ is unbounded at $x=1$ on $(0,1)$. Also realize that $\sum_{k=0}^n x$ is bounded by $n$ on $(0,1)$. From this, you can conclude that convergence cannot be uniform (do you see how?).