I have the following question:


Does there exist a non-negative function $g$ on $(0,1)$ such that
$$1\leq F(x):=\dfrac{\displaystyle\sum\_{k=0}^{\infty}a\_{k}\,(k+1)^{2}\,x^{k}}{\displaystyle\sum\_{k=0}^{\infty}(k+1)\,x^{k}}\leq 2,\;\forall\;x\in (0,1),$$ and $$\displaystyle\lim\_{x\rightarrow 1^{-}}F(x)\;\text{ does not exist?}$$
Here $$a\_{k}:=\int\_{0}^{1}g(x)\,x^{2k+1}dx.$$


Note that if $g\equiv2$ then the first condition satisfies, but the second condition doesn't.



Construct a function with certain growth property
Such a function $g$ exists.


Indeed, we have
\begin{equation\*}
 \sum\_{k=0}^\infty(k+1)\,x^k=\frac1{(1-x)^2}
\end{equation\*}
and
\begin{align\*}
 \sum\_{k=0}^\infty a\_k(k+1)^2\,x^k&=
 \int\_0^1 du\,g(u)\,\sum\_{k=0}^\infty x^ku^{2k+1} \\ 
 &= \int\_0^1 du\,g(u)u\,\frac{1+xu^2}{(1-xu^2)^3} \\ 
 &=\frac12\int\_0^1 dv\,h(v)\,\frac{1+xv}{(1-xv)^3},
\end{align\*}
where $h(v):=g(\sqrt v)\ge0$ for $v\in(0,1)$. So,
\begin{equation\*}
 F\_h(x):=F(x)=(1-x)^2S(x)/2, \tag{-1}
\end{equation\*}
where
\begin{equation\*}
S(x):=S\_h(x):=\int\_0^1 dv\,h(v)\,\frac{1+xv}{(1-xv)^3}. \tag{0}
\end{equation\*}
So, for $h(x)=h\_2(x):=2$, we have $F\_h=1$ and hence
the condition
\begin{equation\*}
 1\le F(x)\le2 \tag{1}
\end{equation\*}
for $x\in(0,1)$ holds.


Consider now $\tilde h:=h\_2+ph=2+ph$, where $p\in(0,\infty)$ and $h$ is a measurable nonnegative function such that $F\_h$ is bounded (on $(0,1)$) and the limit $F\_h(1-)$ does not exist. If the $p$ is small enough, then condition (1) will hold for $F\_{\tilde h}=1+pF\_h$ in place of $F$ and the limit $F\_{\tilde h}(1-)$ will not exist.


Thus, it suffices to construct a measurable nonnegative function $h$ on $(0,1)$ such that $F\_h$ is bounded and the limit $F\_h(1-)$ does not exist. Let us show that the formula
\begin{equation\*}
 h(v):=\sum\_{n\ge1}h\_n1(a\_n<v<b\_n)
\end{equation\*}
defines such a function, where
\begin{equation\*}
 a\_n:=1-q\_n,\quad 0<b\_n-a\_n=o(q\_n),\quad h\_n:=\frac{q\_n}{b\_n-a\_n},\quad q\_n:=\frac1{2^{2^n}}, \tag{2}
\end{equation\*}
so that $q\_n^2=q\_{n+1}$.


Take now any $v\_0\in(0,1)$. Since $h$ is bounded on $(0,v\_0)$, we have $(1-x)^2\int\_0^{v\_0} dv\,h(v)\,\frac{1+xv}{(1-xv)^3}\to0$ as $x\uparrow1$. So, by (-1) and (0),
\begin{equation\*}
 F\_h(x)\sim G(x):=G\_h(x):=(1-x)^2\int\_0^1 dv\,h(v)\,\frac1{(1-xv)^3}
\end{equation\*}
as $x\uparrow1$.
Thus, it suffices to show that the function $G$ is bounded (on $(0,1)$) and the limit $G(1-)$ does not exist.
Note that
\begin{equation\*}
 G(x)=(1-x)^2\sum\_{n\ge1} h\_n \int\_{a\_n}^{b\_n} \frac{dv}{(1-xv)^3}. 
\end{equation\*}
For any natural $n\_0$, the sum $\sum\_{n\_0\ge n\ge1} h\_n \int\_{a\_n}^{b\_n} \frac{dv}{(1-xv)^3}\le\sum\_{n\_0\ge n\ge1} h\_n \int\_{a\_n}^{b\_n} \frac{dv}{(1-v)^3}$ is bounded and hence $(1-x)^2\sum\_{n\_0\ge n\ge1} h\_n \int\_{a\_n}^{b\_n} \frac{dv}{(1-xv)^3}\to0$ as $x\uparrow1$.
Also, in view of the first two conditions in (2), it is easy to see that $1-xv\sim1-xa\_n$ as $n\to\infty$ uniformly in $x\in(0,1)$ and $v\in(a\_n,b\_n)$.
So, letting
\begin{equation\*}
 q:=1-x,
\end{equation\*}
we see that
\begin{equation\*}
 G(x)\sim H(q):=\sum\_{n\ge1}\frac{q^2 q\_n}{(q+q\_n)^3}
\end{equation\*}
as $x\uparrow1$ or, equivalently, $q\downarrow0$.
Thus, it suffices to show that the function $H$ is bounded (on $(0,1)$) and the limit $H(0+)$ does not exist.


For natural $m\to\infty$ and $q\in[q\_{m+1},q\_m]$, write
\begin{equation\*}
 H(q)=H\_{<m}+H\_m+H\_{m+1}+H\_{>m+1}, 
\end{equation\*}
where
\begin{equation\*}
 H\_{<m}:=\sum\_{n<m}\frac{q^2 q\_n}{(q+q\_n)^3}\sim\sum\_{n<m}\frac{q^2}{q\_n^2}
 \sim\frac{q^2}{q\_{m-1}^2}=\frac{q^2}{q\_m}\le q\_m\to0,
\end{equation\*}
\begin{equation\*}
 H\_{>m+1}:=\sum\_{n>m+1}\frac{q^2 q\_n}{(q+q\_n)^3}\sim\sum\_{n>m+1}\frac{q\_n}q
 \sim\frac{q\_{m+2}}q\le\frac{q\_{m+2}}{q\_{m+1}}\to0,
\end{equation\*}
\begin{equation\*}
 H\_m:=\frac{q^2 q\_m}{(q+q\_m)^3}\le\frac{q^2}{q\_m^2}\le1,
\end{equation\*}
\begin{equation\*}
 H\_{m+1}:=\frac{q^2 q\_{m+1}}{(q+q\_{m+1})^3}\le\frac{q\_{m+1}}q\le1. 
\end{equation\*}  

So, $H$ is indeed bounded.


Finally, for $q=q\_{m+1}$
\begin{equation\*}
H\_m+H\_{m+1}=\frac{q\_{m+1}^2q\_m}{(q\_{m+1}+q\_m)^3}+\frac18\to\frac18,
\end{equation\*}
whereas
for $q=2q\_{m+1}\big[\in[q\_{m+1},q\_m]\big]$
\begin{equation\*}
H\_m+H\_{m+1}=\frac{4q\_{m+1}^2q\_m}{(2q\_{m+1}+q\_m)^3}+\frac4{27}\to\frac4{27},
\end{equation\*}
which yields
$H(q\_m)\to\frac18$ and $H(2q\_m)\to\frac4{27}$. Thus, the limit $H(0+)$ does not exist.


