# Series converging uniformly in a subset of the unit disc

Suppose that $0 < a_{n+1} < a_n$ for all $n$ and $a_n \rightarrow 0$. Let $r > 0$. Show that $\sum_{n=0}^{\infty} a_nz^n$ converges uniformly on the set where $|z| \leq 1$ and $|z-1| \geq r$.

My thought process:

1) I understand $|z| \leq 1$ is all the points in the unit circle. I do not know what $|z-1| \geq r$ is, nor do I know what the intersection of those looks like. Basically, what does the set that this converges on look like?

2) This looks sort of like a geometric series, but the $a_n$ term is throwing me off. Typically this series would converge to $\frac{a}{1-z}$ if $a_n$ was just a constant, right? How does that change when $a_n$ is not constant, but converging to zero?

3) I think I understand the definition of uniform convergence for a function, but how does that connect to a series? In other words, for $\varepsilon >0$, I need to show that $|f_n(z) - f(z)| < \varepsilon$ for all $n>N$, but what is $f_n$ and $f$? I would assume $f_n$ is the partial sum and $f(z)=\frac{a}{1-z}$. In other words, I need to show $|\frac{a(1-z^n)}{1-z}-\frac{a}{1-z} |< \varepsilon$? If that is correct I'm again stuck because I don't know how the $a_n$ changes this series.

• I don't think is true since for $a_n=\frac{1}{n+1}$, the series $\sum_{n=0}^\infty a_nz^n$ doesn't converge uniformly on $|z|\leq 1$.
– Surb
Oct 2, 2015 at 18:56
• @Surb The series does however converge uniformly in the given subset of the unit disc.
– zhw.
Oct 2, 2015 at 21:13
• Since $\sum_{n=1}^\infty\frac{1}{n+1}$ diverge, how it can converge uniformly on $|z|\leq 1$ ?
– Surb
Oct 2, 2015 at 21:22

For $0<r<2$ we consider $\displaystyle D_r=\{z\in\mathbb{C}:\vert z\vert\le 1,\vert z-1\vert\ge r\}$
Also, let $S_n(z)=\sum_{k=0}^na_kz^k$. Now, it is clear that for $m>n$ we have \eqalign{(1-z)(S_m(z)-S_n(z))&=(1-z)\sum_{k=n+1}^ma_kz^k\cr &=\sum_{k=n+1}^ma_kz^k-\sum_{k=n+2}^{m+1}a_{k-1}z^k\cr &=\sum_{k=n+1}^ma_kz^k-\sum_{k=n+1}^{m}a_{k-1}z^k+a_nz^n-a_{m}z^{m+1}\cr &=\sum_{k=n+1}^m(a_k-a_{k-1})z^k+a_nz^n-a_{m}z^{m+1}} Hence if $\vert z\vert\le 1$ we have \eqalign{\vert (1-z)(S_m(z)-S_n(z))\vert&\le \sum_{k=n+1}^m\vert a_k-a_{k-1}\vert+ a_n +a_{m}\cr &=\sum_{k=n+1}^m( a_{k-1}-a_k)+ a_n +a_{m}=2a_n} Thus if $z\in D_r$ then $$\vert S_m(z)-S_n(z)\vert\le\frac{2a_n}{\vert z-1\vert}\le \frac{2a_n}{r}$$ So, we have proved that $$\forall\, (m,n),\quad m>n\Longrightarrow \sup_{z\in D_r} \vert S_m(z)-S_n(z)\vert\le \frac{2a_n}{r}$$ Thus the sequence of partial sums $(S_n)_n$ is a uniformly Cauchy sequence on $D_r$, and consequently. it is uniformly convergent on $D_r$.
This follows from the proof of Dirichlet's test, once we observe that for $|z| \le 1, |z-1|\le r,$ the following uniform bound holds:
$$\tag 1 |1 + z + z^2 + \cdots + z^n| = |(z^{n+1}-1)/(z-1)| \le 2/|z-1| \le 2/r.$$