Implications of $\lim\limits_{n\to∞}\left[nP_j-\smash[b]{\frac{\exp(x_j)}{1+\frac1n\sum\limits_{k=1}^n\exp(x_k)}}\right]=0\ (\forall j\in\mathbb N)$ Consider the following limit holds $\forall j\in \mathbb{N}$:
$$
\lim_{n\rightarrow \infty}\left[ nP_j-\frac{\exp(x_j)}{1+\frac{1}{n}\smash[b]{\sum\limits_{k=1}^n} \exp(x_k)}\right]=0,
$$
where


*

*$P_j\in [0,1]$ $\forall j\in \mathbb{N}$,

*$x_j\in [-\overline{M}, \overline{M}]$ $\forall j\in \mathbb{N}$, with $\overline{M}<\infty$.


I want to show that
$$
\lim_{n\rightarrow \infty}\left[\frac{1}{n}\sum_{j=1}^n n P_j-\frac{1}{n}\sum_{j=1}^ n\frac{\exp(x_j)}{1+\frac{1}{n}\smash[b]{\sum\limits_{k=1}^n} \exp(x_k)}\right]=0.
$$
Any hint?
 A: $\def\e{\mathrm{e}}$It will be proved that$$
\lim_{n \to ∞} \left( \frac{1}{n} \sum_{j = 1}^n nP_j - \frac{1}{n} \sum_{j = 1}^n \frac{\e^{x_j}}{\smash[b]{\displaystyle 1 + \frac{1}{n} \sum_{k = 1}^n \e^{x_k}}} \right) = \color{red}{-1}.
$$
Denote $y_j = \e^{x_j}$, $\displaystyle T_n = \frac{1}{n} \sum_{j = 1}^n y_j$. Since $|x_j| \leqslant \overline{M}$ for all $j$, then there exists $M > 0$ such that $0 \leqslant y_j \leqslant M$ for all $j$. Thus for any fixed $j$,$$
0 \leqslant \frac{\e^{x_j}}{\displaystyle 1 + \frac{1}{n} \sum_{k = 1}^n \e^{x_k}} = \frac{y_j}{1 + T_n} \leqslant y_j \leqslant M, \quad \forall n \in \mathbb{N}_+
$$
then$$
0 = \lim_{n \to ∞} \frac{1}{n} · \lim_{n \to ∞} \left( nP_j - \frac{y_j}{1 + T_n} \right) = \lim_{n \to ∞} \left( P_j - \frac{1}{n} · \frac{y_j}{1 + T_n} \right) = P_j,
$$
which implies$$
0 = \lim_{n \to ∞} \left( nP_j - \frac{y_j}{1 + T_n} \right) = -y_j \lim_{n \to ∞} \frac{1}{1 + T_n},
$$
then $T_n > 0 \Rightarrow \lim\limits_{n \to ∞} T_n = +\infty$. Therefore,\begin{align*}
&\mathrel{\phantom{=}}{} \lim_{n \to ∞} \left( \frac{1}{n} \sum_{j = 1}^n nP_j - \frac{1}{n} \sum_{j = 1}^k \frac{\e^{x_j}}{\smash[b]{\displaystyle 1 + \frac{1}{n} \sum_{k = 1}^n \e^{x_k}}} \right) = -\lim_{n \to ∞} \frac{1}{n} \sum_{j = 1}^n \frac{\e^{x_j}}{\displaystyle 1 + \frac{1}{n} \sum_{k = 1}^n \e^{x_k}}\\
&= -\lim_{n \to ∞} \frac{\displaystyle \frac{1}{n} \sum_{j = 1}^n \e^{x_j}}{\displaystyle 1 + \frac{1}{n} \sum_{k = 1}^n \e^{x_k}} = -\lim_{n \to ∞} \frac{\displaystyle \frac{1}{n} \sum_{j = 1}^n y_j}{\displaystyle 1 + \frac{1}{n} \sum_{k = 1}^n y_k} = -\lim_{n \to ∞} \frac{T_n}{1 + T_n} = -1.
\end{align*}
