# 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?

• Cesaro mean theorem ? – Gabriel Romon May 8 '18 at 12:47
• Thank you. Not exactly Cesaro mean, because $nP_j-\frac{\exp(x_j)}{1+\frac{1}{n}\sum_{k=1}^n \exp(x_k)}$ depends on $n$ and $j$ – TEX May 8 '18 at 12:50
• I am getting more confused: it it true that by the assumption above $\lim_{n\rightarrow \infty}P_j=\lim_{n\rightarrow \infty} \frac{\exp(x_j)}{n+\sum_{k=1}^n\exp(x_k)}$. Now, let's look at $\lim_{n\rightarrow \infty} \frac{\exp(x_j)}{n+\sum_{k=1}^n\exp(x_k)}$. We can see that $\lim_{n\rightarrow \infty} \frac{\exp(x_j)}{n+\sum_{k=1}^n\exp(x_k)}$ should be zero because $\frac{\exp(-\bar{M})}{n(1+\exp(\bar{M}))}\leq \frac{\exp(x_j)}{n+\sum_{k=1}^n\exp(x_k)}\leq \frac{\exp(\bar{M})}{n(1+\exp(-\bar{M}))}$. And hence, by assumption, $\lim_{n\rightarrow \infty}P_j=0$. Correct? – TEX May 8 '18 at 15:00

## 1 Answer

$\def\e{\mathrm{e}}$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*}

• Thanks. Why we can claim $\lim_{n\rightarrow \infty}(nP_j-\frac{y_j}{1+T_n})=-y_j\lim_{n\rightarrow \infty}\frac{1}{1+T_n}$? – TEX May 13 '18 at 17:37
• $\lim_{n\rightarrow \infty} (nP_j-\frac{y_j}{1+T_n})=y_j\lim_{n\rightarrow \infty} (n\frac{P_j}{y_j}-\frac{1}{1+T_n})=?$ – TEX May 13 '18 at 17:38
• Also, how can you conclude that $\lim_{n\rightarrow \infty} \frac{1}{n}\sum n P_j=0$? – TEX May 13 '18 at 18:49
• @CGT To all three questions, it's all because $P_j=0$ for all $j$. – Saad May 14 '18 at 0:11
• Thanks! I have posted a related question math.stackexchange.com/questions/2779710/…, if you have time, thank you – TEX May 14 '18 at 6:08