# Can I use L'Hopital's to show $\lim_{x\to1^-}(1-x)[\frac{d}{dx}(1-x)\sum_{n=1}^\infty a_nx^n]=0$ for $a_n$ a bounded sequence of reals?

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.

• Without L'Hospital rules it clearly go to zero, do you really required l'Hospital rules ? – EDX Aug 1 '20 at 18:04
• @EDX Why does it clearly go to $0$? – Milo Moses Aug 1 '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 1 '20 at 18:22
• 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 1 '20 at 18:26
• 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 2 '20 at 12:49

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 $$-1[{}\neq 0]$$.
• @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 3 '20 at 9:11
• 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. – metamorphy Aug 3 '20 at 11:32
• How do you know that $H(-\log_2(-\log(x)))$ oscillates as $x$ tends to $1$? – Milo Moses Aug 3 '20 at 15:59