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.
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.
