Why does non-decreasing entropy imply actual convergence to that max entropy distribution? Let $X_n$ be i.i.d with finite variance. Let $\bar X_n=\frac 1n \sum_{i=1}^nX_i$. It is a famous result that the continuous/differential entropy of the normalized average is non-decreasing. $$\mathrm H(\sqrt n\bar X_n)\leq \mathrm H(\sqrt{n+1}\bar X_{n+1})$$
This inequality suggests the sequence converges in distribution to the max entropy distribution subject to expectation and variance constraints, known to be the corresponding normal distribution. However, there is no strictness (indeed equality can occur if all the process is already Gaussian).
Question. What prevents this monotonic sequence of entropies from converging to a smaller limit than the maximal entropy?
Added. I would like to understand the intuition behind this fact, hopefully alongside a proof sketch.
 A: $\newcommand{\de}{\delta}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\R}{\mathbb R}\newcommand{\vpi}{\varphi} $Without loss of generality, the variance of $X_1$ is $1$.
Let $H:=\mathrm H$, which let us assume to denote the differential entropy, so that $H(X)=\int_{\mathbb R} f(x)\ln\frac1{f(x)}\,dx$ for a random variable $X$ with pdf $f$.
If the pdf of $X_1$ is bounded or, more generally, the pdf of $\bar X_k$ is bounded for some natural $k$, then, by a local limit theorem (see e.g. Theorem 7 in Section 2 of Chapter VII), the pdf (say $f_n$) of $\sqrt n\,\bar X_n$ converges uniformly on $\R$ to the standard normal pdf (say $\vpi$) as $n\to\infty$. Note that $\vpi\le1/\sqrt{2\pi}<1$. So, for all large enough $n$ we have $f_n\le1$ and hence $f_n\ln\frac1{f_n}\ge0$.
So, by Fatou's lemma,
\begin{equation*}
    \liminf_{n\to\infty}H(\sqrt n\,\bar X_n)=\liminf_{n\to\infty}\int_\R f_n\ln\frac1{f_n}
    \ge\int_\R \vpi\ln\frac1\vpi=H(Z)   
\end{equation*}
if $Z\sim N(0,1)$.
On the other hand, $H(\sqrt n\,\bar X_n)\le H(Z)$ for all $n$, since the variance of $\sqrt n\,\bar X_n$ is $1$ and $Z$ maximizes the differential entropy among all absolutely continuous random variables with variance $1$.
So,
\begin{equation*}
    \lim_{n\to\infty}H(\sqrt n\,\bar X_n)=H(Z).  \tag{1}\label{1} 
\end{equation*}

Without the condition that the pdf of $\bar X_k$ be bounded for some natural $k$, the conclusion \eqref{1} can fail to hold, in a rather dramatic manner.
An example when $H(\sqrt n\,\bar X_n)=-\infty$ for all $n$, so that \eqref{1} fails to hold, is given by the formula
\begin{equation*}
    f=\frac1s\,\sum_{k\ge1}c_k\,1_{[x_k,3x_k/2]}
\end{equation*}
for the pdf of $X_1$, where $s:=\sum_{k\ge1}c_k x_k/2=\pi^2/12$,
$c_k:=e^{2^k}/k^2$ and $x_k:=e^{-2^k}$; of course, in this example the pdf of $\bar X_k$ is not bounded for any natural $k$.
The main idea here is that the pdf (say $f_n$) of $Y_n:=\sqrt n\,\bar X_n$ may remain very large in (say) a neighborhood of $0$, so much so that $H(Y_n)=\int f_n\ln\frac1{f_n}=-\infty$.
Details on this example: Let $\de_k:=x_k/2$. Consider the class $G$ of all pdf's $g$  such that
\begin{equation*}
    g\ge\sum_{j\ge1}a_j\,1_{[u_j,u_j+\de_j]}, \tag{2}\label{2} 
\end{equation*}
where
\begin{equation*}
    a_j:=\frac{C_1}{j^p\,\de_j}\quad\text{and}\quad u_j:=\frac{\al x_j}2 \tag{3}\label{3} 
\end{equation*}
for some real $C_1>0$, some real $p>0$, some real $\al\ge1$, and all $j\ge1$.
Clearly, \eqref{1} implies that $f\in G$. Moreover, the pdf's in the class $G$ are very lacunary, which allows us to control the convolutions of any two pdf's in $G$.
More specifically, let us show that -- crucially -- the class $G$ is closed w.r. to the convolution:
Take any $g\in G$, so that \eqref{2} and \eqref{3} hold, and then take any $h\in G$, so that
\begin{equation*}
    h\ge\sum_{j\ge1}b_j\,1_{[v_j,v_j+\de_j]}, %\tag{2a}\label{2a} 
\end{equation*}
where
\begin{equation*}
    b_j:=\frac{C_2}{j^q\,\de_j}\quad\text{and}\quad v_j:=\frac{\be x_j}2 %\tag{3a}\label{3a} 
\end{equation*}
for some real $C_2>0$, some real $q>0$, some real $\be\ge1$, and all $j\ge1$.
Note that $1_{[u_j,u_j+\de_j]}*1_{[v_j,v_j+\de_j]}\ge\frac12\,1_{[w_j,w_j+\de_j]}$, where $w_j:=\frac{(\al+\be+1) x_j}2$. So,
\begin{equation*}
    g*h\ge\sum_{j\ge1}d_j\,1_{[w_j,w_j+\de_j]}, 
\end{equation*}
where $d_j:=a_jb_j\de_j=\frac{C_1C_2/2}{j^{p+q}\de_j}$. Thus, indeed the class $G$ is closed w.r. to the convolution.
So, to complete the consideration of the example, it remains to show that
\begin{equation*}
    H(g):=-\int_\R g\ln g=-\infty \tag{4}\label{4} 
\end{equation*}
for any $g$ such that \eqref{2} and \eqref{3} hold. Take indeed any such $g$. Clearly, for some natural $j_p$ and all natural $j\ge j_p$ we have $a_j\ge1$ and hence $g\ge1$ on the interval $[u_j,u_j+\de_j]$. Also, the intervals $[u_j,u_j+\de_j]$ are pairwise disjoint.
Also, $t\ln t$ is increasing in $t\ge1$.
It follows that
\begin{equation*}
    \int_\R g\ln g\,1(g\ge1)\ge \sum_{j\ge j_p}a_j\,\de_j\,\ln a_j 
    =\sum_{j\ge j_p}\frac{C_1}{j^p}\,\ln\frac{2C_1}{j^p\,e^{-2^j}}=\infty. \tag{5}\label{5}
\end{equation*}
On the other hand, $t\ln t\ge-1/e$ for all real $t>0$. So,
\begin{equation*}
    \int_\R g\ln g\,1(g<1)\ge-\frac1e\, \int_{[0,u_1+\de_1]}1>-\infty. \tag{6}\label{6}
\end{equation*}
Now \eqref{4} follows from \eqref{5} and \eqref{6}. $\quad\Box$
