How do you prove the uniform convergence of the following series?

Question

Consider the series $$ \phi(t)=\sum^{k\geq0}\frac{w^k}{1+kt} $$ where $w\in\mathbb{C}$ and $0<|w|<1$. This series is allegedly easily shown to be uniformly convergent in a compact subset of $\mathbb{C}-\{-1,-1/2,-1/3,\ldots\}$, but I don't see how. Can anybody fill in the gaps?

Answer 1

Consider the compact subset of the complement $$ K=\left\{-\frac1{k+k^{-k}}:k\in\mathbb{Z},k\ge2\right\}\cup\{0\}\tag{1} $$ No matter how large $n$ is, for $t=-\frac1{n+n^{-n}}$, we have $$ 1+nt=\frac1{n^{n+1}+1}\tag{2} $$ whereas for $k\ge n+1$, we have $$ 1+kt\le-\frac{1-n^{-n}}{n+n^{-n}}\le-\frac{3/4}{n+1}\tag{3} $$ Therefore, $$ \sum_{k\ge n}\frac{w^k}{1+kt}\ge w^n\left[\left(n^{n+1}+1\right)-\frac{4(n+1)}3\frac{w}{1-w}\right]\tag{4} $$ Since $(4)$ does not vanish, the sum does not converge uniformly on $K$.

Answer 2

Let $K = \{0\} \cup \{(-1 + i/(2w)^k)/k \mid k \ge 1\}$.

$K$ is compact and doesn'contain any $1/k$.

However, if $t_k = (-1 + i/(2w)^k)/k$ then $w^k/(1+kt_k) = -i2^k$,

and so
$|\phi(t_k)| \ge 2^k - \sum_{k' \neq k} |w^{k'}/(1+k't_k )| = 2^k - \sum_{k' \neq k} |kw^{k'}|/|k-k'+k'i/(2w)^k| $

Since $|k-k'+ki/(2w)^k| \ge |k-k'| \ge 1$,
$|\phi(t_k)| \ge 2^k - \sum_{k' \neq k} |kw^{k'}| \ge 2^k -k \sum_{k' \ge 0} |w|^{k'} = 2^k-k/(|w|-1)$.

As $k \to \infty$, $|\phi(t_k)| \to \infty$, so $\phi$ is not bounded on $K$.

If the convergence was uniform then $\phi$ would be continuous and bounded on $K$, so the convergence is not uniform.