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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.

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  • $\begingroup$ Without L'Hospital rules it clearly go to zero, do you really required l'Hospital rules ? $\endgroup$ – EDX Aug 1 '20 at 18:04
  • $\begingroup$ @EDX Why does it clearly go to $0$? $\endgroup$ – Milo Moses Aug 1 '20 at 18:05
  • $\begingroup$ How could you use L'hospital rule,,,is it in $\frac{0}{0} $ form or $\frac{\infty}{\infty} $ form,. You should know that whether $\sum a_n x^n $ converge or diverge, as $x \to 1- $ $\endgroup$ – A learner Aug 1 '20 at 18:22
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    $\begingroup$ Where, you first use L'hospital rule in your approach, it can be made totally wrong, by taking $a_n=\frac{1}{n^2} $ $\endgroup$ – A learner Aug 1 '20 at 18:26
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    $\begingroup$ It is very unclear about what the condition on $a_n$ because being bounded is not sufficient $a_n=1$ is a good example. Uniformarly convergent of $na_n$ (the derivated serie) on $[\epsilon,1]$ with $\epsilon>0$ is clearly sufficient. The interest in your exercise is to set a condition for $a_n$ such the result you want to show can be true without being too obvious (as uniform convergence) $\endgroup$ – EDX Aug 2 '20 at 12:49
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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 $-1[{}\neq 0]$.

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  • $\begingroup$ Just a question what happens as the coefficients are in a decreasing order or an increasing order ? $\endgroup$ – Erik Satie Aug 3 '20 at 8:33
  • $\begingroup$ @user698573: You mean, $a_n$ is increasing (say) and bounded? Then the claim holds (let $a_0=0$): $$(1-x)\big((1-x)f(x)\big)'=(1-x)\sum_{n=1}^\infty n(a_n-a_{n-1})x^{n-1}\underset{x\to 1^-}{\longrightarrow}0$$ by Abel's theorem, since $n(a_n-a_{n-1})\to 0$ as $n\to\infty$ because $\sum_n(a_n-a_{n-1})$ converges. $\endgroup$ – metamorphy Aug 3 '20 at 9:11
  • $\begingroup$ Thanks to confirm my intuition !I was not aware about the Abel's theorem. (+1) $\endgroup$ – Erik Satie Aug 3 '20 at 9:53
  • $\begingroup$ Instead of $k\mapsto 2^k$, one could take a faster-growing sequence, say $k\mapsto 2^{2^k}$. This would make $g(x)$ oscillate between $−1$ and $1$ very steeply. Looks like the stuff under the limit can be made unbounded at any desired rate. $\endgroup$ – metamorphy Aug 3 '20 at 11:32
  • $\begingroup$ How do you know that $H(-\log_2(-\log(x)))$ oscillates as $x$ tends to $1$? $\endgroup$ – Milo Moses Aug 3 '20 at 15:59

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