Suppose we have two infinite series which depend on some number $x$:
$$A(x)=\sum_{n=1}^{\infty} a_nx^n,$$
$$B(x)=\sum_{n=1}^{\infty} b_nx^n.$$
Here $a_n,b_n$ are positive real numbers such that for all $x \in (0,1)$ we have that $A(x)$ and $B(x)$ are convergent sequences and $A(1), B(1)$ are both divergent sequences.
Define $A_k(x) = \sum_{n=2^k}^{2^{k+1}} a_n x^n$ and similarly $B_k(x)=\sum_{n=2^k}^{2^{k+1}} b_n x^n$. Suppose we have the property that the following limit holds:
$$
\lim_{k\rightarrow \infty} \frac{A_k(1)}{B_k(1)}=c,
$$
I would then like to show that $\lim_{x\rightarrow 1} \frac{A(x)}{B(x)}=c$. This seems true and I have checked it numerically for some examples but I can't seem to be able to prove it.. As a very simple example we can just take $a_n$ and $b_n$ to be constant and equal to $a$ and $b$. For this example it is trivial to check that the statement indeed holds.
Best Answer
The question begs to consider $b_n\equiv 1$, $a_n=n$ if $n=2^k$ and $a_n=0$ otherwise: $$B(x)=\frac{1}{1-x},\qquad A(x)=\sum_{k=0}^{\infty}2^k x^{2^k}.$$ In the above notation, $A_k(1)/B_k(1)\underset{k\to\infty}{\longrightarrow}2$. The asymptotics of $A(x)$ when $x\uparrow 1$ is what I like to analyse (perhaps not the simplest way) using $$e^{-y}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)y^{-s}\,ds\qquad(y,c>0)$$ which gives, this time for $c>1$, $$A(e^{-y})=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\Gamma(s)y^{-s}}{1-2^{1-s}}\,ds.$$ This is equal to the (infinite) sum of residues of the integrand at its poles (which are all simple, making the residues easy to evaluate): $$A(e^{-y})=\frac{1}{y\ln2}\sum_{n\in\mathbb{Z}}\exp\Big(-\frac{2n\pi i}{\ln2}\ln y\Big)\Gamma\Big(1+\frac{2n\pi i}{\ln2}\Big)-\sum_{n=0}^{\infty}\frac{(-y)^n}{n!(2^{n+1}-1)}.$$ So, when $x\uparrow 1$, $(1-x)A(x)$ oscillates around $1/\ln2$ (with very small amplitude).
I should note here that the oscillating nature of $A(x)+R(x)$, where $$R(x)=\sum_{n=0}^{\infty}\frac{(\ln x)^n}{n!(2^{n+1}-1)},$$ can be seen an elementary way, following the lines of Hardy cited in this excellent answer; namely, $F(x)=A(x)+R(x)$ satisfies $F(x)=2F(x^2)$, i.e. $(-\ln x)F(x)$ is a periodic function of $\ln(-\ln x)$ (with period $\ln 2$). When $x\uparrow 1$ we have $R(x)\to 1$ and $\ln(-\ln x)\to-\infty$.
Thus, it appears we've been considering a counterexample.