the infinite series $\sum z^n$ for $z \in \mathbb{C}$ does not converge uniformly on $|z| <1$

Question

The series $\sum_{n=0}^{\infty} z^n$ does not converge uniformly on $D(0,1) = \{ z \in \mathbb{C} : |z|<1 \}$. Here is my attempt:

Suppose to the contrary that the series $\sum_{n=0}^{\infty} z^n$ converges uniformly on $D(0,1)$. Since uniform convergence implies pointwise convergence, the series converges to \begin{equation*} \sum_{n=0}^{\infty} z^n = \frac{1}{1-z} \end{equation*} for all $z \in D(0, 1)$. Choose $z = 1 - \frac{1}{k}$ for $k \in \mathbb{N}$ so that $z \in D(0, 1)$. Then, we have \begin{equation*} \sum_{n=0}^{\infty} \left(1 - \frac{1}{k}\right)^n = \frac{1}{1 - \left(1 - \frac{1}{k}\right)} = k. \end{equation*} However, the series $\sum_{n=0}^{\infty} \left(1 - \frac{1}{k}\right)^n$ diverges as $k \to \infty$. This is a contradiction.

Now I know that I have to use the supremum norm, but why the attempt does not work?

Although the sequence converges to $1$, I thought that it still lives in $D(0,1)$.

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Edit:

I changed the summation from $\sum_{n=0}^{\infty}\left(1-\frac{1}{n}\right)^n$ to $\sum_{n=0}^{\infty}\left(1-\frac{1}{k}\right)^n$.

Answer 1

If the functions $f_n: \overline{D}(0,1) \to \mathbb{C}$ are continuous on the closed disk $\overline{D}(0,1)$ and the series $\sum f_n$ is uniformly convergent on the open disk $D(0,1)$, then the series must be uniformly convergent on the closed disk as well.

From that it follows that the sum $S(z) = \sum_{n=0}^\infty f_n(z)$ is continuous on $\overline{D}(0,1)$. Hence, for any point $z_B$ on the boundary, the limit of $S(z)$ as $z \to z_B$ exists. Consequently, showing that the limit does not exist or equivalently that the series diverges at $z_B$ is a valid way to prove the series is not uniformly convergent on the open disk.

In this case, the functions $z \mapsto z^n$ are continuous on the closed disk. Taking the sequence $z_k = (1 - 1/k)$ we have $z_k \to 1 \in \overline{D}(0,1)$ and, as you have shown, $\sum_{n=0}^\infty z_k^n $ diverges as $k \to \infty$. Hence, the series $\sum z^n$ cannot be uniformly convergent on $D(0,1)$.

To prove that the uniform convergence on the open disk implies uniform convergence on the closed disk we use the Cauchy criterion. Denote the partial sum as $S_n(z) = \sum_{k=0}^n f_k(z)$. Uniform convergence on $D(0,1)$ means that for any $\epsilon > 0$ there exists $N(\epsilon)$ such that $|S_m(z) - S_n(z)| < \epsilon$ for all $m \geqslant n \geqslant N(\epsilon)$ and for all $z \in D(0,1)$. Since the partial sums are continuous, for any point $z_B$ on the boundary we have

$$|S_m(z_B) - S_n(z_B)| = \lim _{z \to z_B}|S_m(z) - S_n(z)| \leqslant \epsilon$$

for all $m\geqslant n \geqslant N(\epsilon)$, which shows that the uniform convergence extends to the closed disk.