Consider the sequence of functions $\{F_n(\cdot)\}_{n \in \mathbb{N}}$ on $[0,1]$, where for each $n$, $F_n(\cdot)$ is defined as
\begin{equation}
F_n(x) = \sum_{i=2}^n \Big( x^{b_n} \frac{i}{(i+1)^{a_n}} - \frac{1}{i^{a_n}} \Big) \big( 1 - x^{b_n} \big)^{i-1},
\end{equation}
where $a_n$ and $b_n$ are defined as
\begin{equation}
a_n = \frac{1}{\log_2(n)} \text{ and } b_n = \log_2(n).
\end{equation}
The questions is: Does there exist an $\epsilon \in (0,1)$ (which does ont depend on $n$) such that for all $x \in [\epsilon, 1]$, it holds that
\begin{equation}
\liminf_{n \to \infty} F_n(x) > -\infty ?
\end{equation}
The figure below shows the plot of $F_{n=2^{21}}(\cdot)$ on $[0,1]$, and the "transition" is around $1/2$ (the inverse of the base of the logarithm defining $a_n$ and $b_n$). For $x$ close to $1$, we see that the curve is fairly flat and is close to $0$, and this is in fact a "typical" behavior for all $n$.
The next figure shows $F_{n=2^{21}}(\cdot)$ on $[0.9,1]$, which shows that when $x$ is close to $1$, $F_n(x)$ is in fact positive.
I think the important thing is that, in the definition of $F_n$ and when $i$ ranges from $2$ to $n$, how large the terms $x^{b_n} i$ are (and this would essentially determine how many terms in the partial sum in $F_n$ could be negative). Since $x^{b_n} = x^{\log_2(n)} = 1 / n^{\log_{0.5}(x)}$, when $x$ is much larger than $0.5$, $x^{b_n}$ decays slower than $1/n$. E.g., when $x = \sqrt{0.5}$, $x^{b_n} = 1 / \sqrt{n}$. This means that for $x$ close to $1$, the terms $x^{b_n} i / (i+1)^{a_n}$ should be larger than $1 / i^{a_n}$ for large $i$'s, and this prevents many terms being negative in the partial sum of $F_n$. I have not been able to have a proof of the existence of an $\epsilon$ such that $\liminf_{n \to \infty} F_n(x) > -\infty$ for $x \in [\epsilon,1]$. Anyone has an idea? Thanks very much.
