I am attempting to prove that if $a_n$ is a bounded sequence of real numbers then
$$\lim_{x\to1^-}(1-x)\left[\frac{d}{dx}(1-x)\sum_{n=1}^{\infty}a_nx^n\right]=0$$
My approach is to first make some algebraic manipulations, namely we see that
\begin{align*}
1&=\lim_{x\to1^-}\frac{(1-x)\sum_{n=1}^{\infty}a_nx^n}{(1-x)\sum_{n=1}^{\infty}a_nx^n}\\
&=\lim_{x\to1^-}\frac{1}{(1-x)\sum_{n=1}^{\infty}a_nx^n}\left(\frac{1-x}{\frac{1}{\sum_{n=1}^{\infty}a_nx^n}}\right)\\
\end{align*}
The reason I want to do this is that if I were able to apply L'Hopital's rule to
$$\frac{1-x}{\frac{1}{\sum_{n=1}^{\infty}a_nx^n}}$$
then I would get that
\begin{align*}
1&=\lim_{x\to1^-}\frac{1}{(1-x)\sum_{n=1}^{\infty}a_nx^n}\left(\frac{-1}{-\frac{\sum_{n=1}^{\infty}na_nx^{n-1}}{\left(\sum_{n=1}^{\infty}a_nx^n\right)^2}}\right)\\
&=\lim_{x\to1^-}\frac{\sum_{n=1}^{\infty}a_nx^n}{(1-x)\sum_{n=1}^{\infty}na_nx^{n-1}}\\
\end{align*}
From there we can subtract $1$ from both sides and multiply top and bottom by $(1-x)$ to get that
$$\lim_{x\to1^-}\frac{\left(1-x\right)\sum_{n=1}^{\infty}a_{n}x^{n}-\left(1-x\right)^2\sum_{n=1}^{\infty}na_{n}x^{n-1}}{\left(1-x\right)^2\sum_{n=1}^{\infty}na_{n}x^{n-1}}=0$$
Since
$$\left(1-x\right)^2\sum_{n=1}^{\infty}na_{n}x^{n-1}$$
is bounded, the only way for this quantity to go to zero would be for
$$\left(1-x\right)\sum_{n=1}^{\infty}a_{n}x^{n}-\left(1-x\right)^2\sum_{n=1}^{\infty}na_{n}x^{n-1}=(1-x)\left[\frac{d}{dx}(1-x)\sum_{n=1}^{\infty}a_nx^n\right]$$
to go to $0$, thus yielding what we want.
I am not sure if this use of L'Hopitals is (or can be) justified, since the limit of
$$\frac{-1}{-\frac{\sum_{n=1}^{\infty}na_nx^{n-1}}{\left(\sum_{n=1}^{\infty}a_nx^n\right)^2}}$$ as $x\to1^-$ is not required to exist. Is there any way I can make this argument rigorous?
EDIT: If I had the pair of inequalities
$$\limsup_{x\to 1^-}k(x)\frac{f(x)}{g(x)}\leq \limsup_{x\to 1^-}k(x)\frac{f'(x)}{g'(x)}$$
$$\liminf_{x\to 1^-}k(x)\frac{f'(x)}{g'(x)} \leq \liminf_{x\to 1^-}k(x)\frac{f(x)}{g(x)}$$
for differentiable functions $f$, $g$ and $k$ on $[0,1)$ then I could resolve my issue. On wikipedia it states that
$$\liminf_{x\to1^-}\frac{f'(x)}{g'(x)}\leq \liminf_{x\to1^-}\frac{f(x)}{g(x)} \leq \limsup_{x\to1^-}\frac{f(x)}{g(x)}\leq \limsup_{x\to1^-}\frac{f'(x)}{g'(x)}$$
but I can't complete the argument for when the factor of $k(x)$ is added.
Best Answer
My "favourite" counterexample works again. Consider $a_n=(-1)^k$ for $2^k\leqslant n<2^{k+1}$, $k\geqslant 0$.
Then, for $f(x):=\sum_{n=1}^\infty a_n x^n$, we get $g(x):=(1-x)f(x)=x+2\sum_{k=1}^\infty(-1)^k x^{2^k}$. Now let $$h(x)=g(x)+G(\log x),\qquad G(t)=\sum_{n=1}^\infty\frac{2^n-1}{2^n+1}\frac{t^n}{n!}.$$ Then it is easy to check that $h(x)=-h(x^2)$. That is, the function $H(t)=h(e^{-2^{-t}})$ (defined for all real values of $t$) is periodic: $H(t)=H(t+2)$. It is nonconstant, and in fact the linked answer shows that $$H(t)=\frac{2}{\log 2}\sum_{n\in\mathbb{Z}}\Gamma\left(\frac{2n+1}{\log 2}i\pi\right)e^{(2n+1)i\pi t}.$$
Gathering it all, we get $(1-x)f(x)=H\big(-\log_2(-\log x)\big)-G(\log x)$ and $$(1-x)\frac{d}{dx}\big((1-x)f(x)\big)=-\frac{1-x}{x\log x\log 2}H'\big(-\log_2(-\log x)\big)-\frac{1-x}{x}G'(\log x).$$ At $x\to1^-$, the second term vanishes, but the first one oscillates, since $\frac{1-x}{\log x}$ tends to $-1[{}\neq 0]$.