I am working on exercise 5.15 in Rudin's RCA. The problem is to find an unbounded sequence $\{s_{n}\}$ of complex numbers such that the sequence $\{\sigma_{i}\}$ defined by $$\sigma_{i}=(1-r_{i})\sum_{j=0}^{\infty}s_{j}r_{i}^{j}$$ converges, where $\{r_{i}\}$ is a sequence in the interval $(0,1)$ that converges to $1$. I was able to find an unbounded sequence (in brutal fashion) such that a subsequence of $\{\sigma_{i}\}$ converges to zero, but I haven't been able to show that $\{\sigma_{i}\}$ has the same limit.
What follows is my attempt.
Let $\{i_{n}\}$ be a strictly increasing sequence of positive integers such that $$ 1-r_{i_{n}}<\frac{2}{n^{2}(n+1)}. $$ Let $\{j_{n}\}$ be a strictly increasing sequence of positive integers such that every integer $i$ at most $i_{n}$ satisfies $$ r^{j_{n}}_{i}<\frac{1}{n^{3}}. $$ For each $j_{n}$, let $s_{j_{n}}$ be $n$, and let $\{s_{j}\}$ be the resulting sequence, where zeros fill in the gaps in $\{j_{n}\}$. By design $$ \sum_{k=1}^{\infty}s_{j_{k}}r_{i_{n}}^{j_{k}}=\sum_{k=1}^{i_{n}}s_{j_{k}}r_{i_{n}}^{j_{k}}+\sum_{i_{n}+1}^{\infty}s_{j_{k}}r_{i_{n}}^{j_{k}}<\frac{n(n+1)}{2}+\sum_{k=1}^{\infty}\frac{1}{k^{2}}, $$ so $$ \sigma_{i_{n}}=(1-r_{i_{n}})\sum_{j=0}^{\infty}s_{j}r_{i_{n}}^{j}<\frac{1}{n}+\frac{2}{n^{2}(n+1)}\sum_{k=1}^{\infty}\frac{1}{k^{2}}<\frac{1}{i}+\frac{2}{i^{2}(i+1)}\sum_{k=1}^{\infty}\frac{1}{k^{2}}. $$ The subsequence $\{\sigma_{i_{n}}\}$ converges to zero, but I can't say the same of its parent.
Let $s_n=(-1)^{n+1}n$. It's easy to see that
$$r_n\in(0,1)\implies\sum_{j=0}^\infty s_nr_n^j=\frac{r_n}{(1+r_n)^2}$$
And as $n\to\infty$,
$$\sigma_n=\frac{(1-r_n)r_n}{(1+r_n)^2}\stackrel{n\to\infty}\longrightarrow0$$