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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.

Blue
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Milo Moses
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  • Without L'Hospital rules it clearly go to zero, do you really required l'Hospital rules ? – EDX Aug 01 '20 at 18:04
  • @EDX Why does it clearly go to $0$? – Milo Moses Aug 01 '20 at 18:05
  • 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- $ – A learner Aug 01 '20 at 18:22
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    Where, you first use L'hospital rule in your approach, it can be made totally wrong, by taking $a_n=\frac{1}{n^2} $ – A learner Aug 01 '20 at 18:26
  • You should use some more information on about "$a_n$". – A learner Aug 01 '20 at 18:27
  • I wonder '$a_n$' is such $ \sum a_n x^n$ converges – EDX Aug 01 '20 at 18:31
  • $\sum a_n x^n$ can be assumed to diverge, since it it converges then the H-L tauberian theorem can be used to recover tons of information about $a_n$ which is sufficient for me – Milo Moses Aug 01 '20 at 18:34
  • Moreover, we can assume that $a_n$ does not tend to $0$ uniformly – Milo Moses Aug 01 '20 at 18:35
  • @EDX if $\sum a_n x^n $ is uniformly convergent, then we can take lim $x\to 1- $ inside the summation, and we can get a convergent series. – A learner Aug 02 '20 at 04:56
  • @Milo Moses you can take also this as, $\frac{\sum a_n x^n }{\frac{1}{1-x}} $, you can see , as you apply L'hospital rule, you can never get finite derived term, as the series is of infinite terms and denominator term gets it's power higher to higher, so, by this, is there any conclusion you get ever? – A learner Aug 02 '20 at 05:05
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    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) – EDX Aug 02 '20 at 12:49
  • @EDX What do you mean $a_n=1$ is a good counterexample? – zhw. Aug 03 '20 at 19:47

1 Answers1

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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 $-1[{}\neq 0]$.

metamorphy
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  • Just a question what happens as the coefficients are in a decreasing order or an increasing order ? – Barackouda Aug 03 '20 at 08:33
  • @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. – metamorphy Aug 03 '20 at 09:11
  • Thanks to confirm my intuition !I was not aware about the Abel's theorem. (+1) – Barackouda Aug 03 '20 at 09:53
  • How do you know that $H(-\log_2(-\log(x)))$ oscillates as $x$ tends to $1$? – Milo Moses Aug 03 '20 at 15:59