Let $b_n$ be a complex sequence that is absolutely bounded for large $n$ and $x>0$ such that $$\lim_{x\rightarrow 1} \left| \sum_{n=1}^\infty b_nx^n \right| = \infty \, .$$
Can anyone construct a sequence $b_n$ for which $$\lim_{x\rightarrow 1} \left| \frac{\sum_{n=1}^\infty |b_n|x^n}{\sum_{n=1}^\infty b_nx^n} \right| = \infty \, ?$$
I haven't found one :-(
Is it even possible and if not, how could you proof that the ratio is $<\infty$?
The function $$ B(x) = \frac1{1+x} + \frac1{\sqrt{1-x}} = \sum_{n=0}^\infty b_n x^n $$ provides such a sequence $b_n$. First, we note that $$ \sum_{n=0}^\infty b_n x^n = \frac1{1+x} + \frac1{\sqrt{1-x}} = \sum_{n=0}^\infty (-1)^n x^n + \sum_{n=0}^\infty 2^{-2n} \binom{2n}n x^n, $$ so $b_n = (-1)^n + 2^{-2n} \binom{2n}n$, which is bounded since $2^{-2n} \binom{2n}n$ tends to $0$ as $n\to\infty$. On the other hand, $0 \le 2^{-2n} \binom{2n}n \le 1$ for $n\ge0$, and so \begin{align*} |b_n| = \bigg| (-1)^n + 2^{-2n} \binom{2n}n \bigg| &= \bigg| (-1)^n(-1)^n + (-1)^n2^{-2n} \binom{2n}n \bigg| \\ &= 1 + (-1)^n2^{-2n} \binom{2n}n = (-1)^n b_n. \end{align*} Therefore $$ \sum_{n=0}^\infty |b_n| x^n = \sum_{n=0}^\infty (-1)^n b_n x^n = B(-x) = \frac1{1-x} + \frac1{\sqrt{1+x}}. $$ So finally, $$ \bigg| \frac{\sum_{n=1}^\infty |b_n|x^n}{\sum_{n=1}^\infty b_nx^n} \bigg| = \bigg| \frac{B(-x)}{B(x)} \bigg| = \frac{(1+x-\sqrt{1-x}) (x+1) (1-x+\sqrt{1+x})}{x (x+3) \sqrt{1-x^2}} $$ (after some easy algebra), which makes it clear that $$ \lim_{x\to1-} \bigg| \frac{\sum_{n=1}^\infty |b_n|x^n}{\sum_{n=1}^\infty b_nx^n} \bigg| = \frac{2\cdot2\cdot\sqrt2}{1\cdot4} \lim_{x\to1-} \frac1{\sqrt{1-x^2}} = \infty. $$ (Note how the addition of $1/\sqrt{1-x}$ to $B(x)$ forced $\sum_{n=1}^\infty b_nx^n$ to tend to $\infty$ as $x\to1-$, but not so fast that it kept up with the $1/(1-x)$ speed of divergence of $B(-x)$.)