# Prove $\lim\limits_{n\to∞}{\sum\limits_{x=0}^n\binom nx(1+{\rm e}^{-(x+1)})^{n+1}\over\sum\limits_{x=0}^n\binom nx(1+{\rm e}^{-x})^{n + 1}}=\frac 13$

I am trying to find limit of the following function: $$\lim_{n\rightarrow \infty}\frac{\sum\limits_{x = 0}^{n}\binom{n}{x}\left[1 + \mathrm{e}^{-(x+1)}\right]^{n + 1}}{\sum\limits_{x=0}^{n} \binom{n}{x}\left[1 + \mathrm{e}^{-x}\right]^{n + 1}}.$$

When I wrote a Python code for this, I saw that it converges to $1/3$.

I am not sure how to approach this. Could somebody give a pointer about how to go about it?

EDIT: I want just some positive lower bound on this. So even if limit cannot be evaluated exactly, it is okay if I get some lower bound on this.

$\def\e{\mathrm{e}}$For $n \in \mathbb{N}_+$, denote$$S_n = \sum_{k = 0}^n \binom{n}{k} (1 + \e^{-(k + 1)})^{n + 1},\ T_n = \sum_{k = 0}^n \binom{n}{k} (1 + \e^{-k})^{n + 1}.$$

First,\begin{align*} S_n &= \sum_{k = 0}^n \binom{n}{k} (1 + \e^{-(k + 1)})^{n + 1} = \sum_{k = 0}^n \binom{n}{k} \sum_{j = 0}^{n + 1} \binom{n + 1}{j} \e^{-(k + 1)j}\\ &= \sum_{k = 0}^n \sum_{j = 0}^{n + 1} \binom{n}{k} \binom{n + 1}{j} \e^{-(k + 1)j} = \sum_{k = 0}^n \binom{n}{k} + \sum_{k = 0}^n \sum_{j = 1}^{n + 1} \binom{n}{k} \binom{n + 1}{j} \e^{-(k + 1)j}\\ &= 2^n + \sum_{k = 0}^n \sum_{j = 0}^n \binom{n}{k} \binom{n + 1}{j + 1} \e^{-(k + 1)(j + 1)}, \tag{1} \end{align*} thus $S_n \geqslant 2^n$. Note that$$(k + 1)(j + 1) \geqslant k + j + 1 \Longleftrightarrow kj \geqslant 0,$$ thus from (1) there is\begin{align*} S_n &\leqslant 2^n + \sum_{k = 0}^n \sum_{j = 0}^n \binom{n}{k} \binom{n + 1}{j + 1} \e^{-(k + j + 1)}\\ &= 2^n + \e^{-1} \sum_{k = 0}^n \sum_{j = 0}^n \binom{n}{k} \e^{-k} \binom{n + 1}{j + 1} \e^{-j}\\ &= 2^n + \e^{-1} \left( \sum_{k = 0}^n \binom{n}{k} \e^{-k} \right)\left( \sum_{j = 0}^n \binom{n + 1}{j + 1} \e^{-j} \right)\\ &\leqslant 2^n + \e^{-1} (1 + \e^{-1})^n (1 + \e^{-1})^{n + 1}\\ &= 2^n + \e^{-1} (1 + \e^{-1})^{2n + 1}. \end{align*} Therefore,$$1 \leqslant \frac{S_n}{2^n} \leqslant 1 + \e^{-1} (1 + \e^{-1}) \left( \frac{1}{2} (1 + \e^{-1})^2 \right)^n.$$ Note that $\dfrac{1}{2} (1 + \e^{-1})^2 < 1$, thus $S_n \sim 2^n$ $(n \to \infty)$.

Next,\begin{align*} T_n &= \sum_{k = 0}^n \binom{n}{k} (1 + \e^{-k})^{n + 1} = \sum_{k = 0}^n \binom{n}{k} \sum_{j = 0}^{n + 1} \binom{n + 1}{j} \e^{-kj}\\ &= \sum_{k = 0}^n \sum_{j = 0}^{n + 1} \binom{n}{k} \binom{n + 1}{j} \e^{-kj}\\ &= \sum_{k = 0}^n \binom{n}{k} + \sum_{j = 0}^{n + 1} \binom{n + 1}{j} - 1 + \sum_{k = 1}^n \sum_{j = 1}^{n + 1} \binom{n}{k} \binom{n + 1}{j} \e^{-kj}\\ &= 2^n + 2^{n + 1} - 1 + \sum_{k = 1}^n \sum_{j = 1}^{n + 1} \binom{n}{k} \binom{n + 1}{j} \e^{-kj}\\ &= 3 × 2^n - 1 + \sum_{k = 0}^{n - 1} \sum_{j = 0}^n \binom{n}{k + 1} \binom{n + 1}{j + 1} \e^{-(k + 1)(j + 1)}, \tag{2} \end{align*} thus $T_n \geqslant 3 × 2^n - 1$. Also, analogously, from (2) there is\begin{align*} T_n &\leqslant 3 × 2^n + \sum_{k = 0}^{n - 1} \sum_{j = 0}^n \binom{n}{k + 1} \binom{n + 1}{j + 1} \e^{-(k + j + 1)}\\ &= 3 × 2^n + \e \left( \sum_{k = 0}^{n - 1} \binom{n}{k + 1} \e^{-(k + 1)} \right)\left( \sum_{j = 0}^n \binom{n + 1}{j + 1} \e^{-(j + 1)} \right)\\ &\leqslant 3 × 2^n + \e (1 + \e^{-1} )^n (1 + \e^{-1})^{n + 1}, \end{align*} which analogously implies that $T_n \sim 3 × 2^n$ $(n \to \infty)$.

Therefore,$$\lim_{n \to \infty} \frac{\displaystyle\sum_{k = 0}^n \binom{n}{k} (1 + \e^{-(k + 1)})^{n + 1}}{\displaystyle\sum_{k = 0}^n \binom{n}{k} (1 + \e^{-k})^{n + 1}} = \lim_{n \to \infty} \frac{S_n}{T_n} = \lim_{n \to \infty} \frac{2^n}{3 × 2^n} = \frac{1}{3}.$$

Preliminary Inequalities

Using the formula $\frac{x^{n+1}-1}{x-1}=x^n+\cdots+1$, we get \begin{align} \left(1+e^{-k}\right)^{n+1}-1 &=e^{-k}\!\left(\left(1+e^{-k}\right)^n+\cdots+1\right)\\[6pt] &\le(n+1)e^{-k}\left(1+e^{-k}\right)^n\tag1 \end{align} Therefore, \begin{align} \sum_{k=1}^{n+1}\binom{n+1}{k}\left(\left(1+e^{-k}\right)^{n+1}-1\right) &\le\color{#C00}{\sum_{k=1}^{n+1}\binom{n+1}{k}}(n+1)\,\color{#C00}{e^{-k}}\color{#090}{\left(1+e^{-k}\right)^n}\\ &\le(n+1)\color{#C00}{\left(1+\frac1e\right)^{n+1}}\color{#090}{\left(1+\frac1e\right)^n}\\[3pt] &=(n+1)\left(1+\frac1e\right)^{2n+1}\\[9pt] &=o\!\left(2^n\right)\tag2 \end{align} since $\left(1+\frac1e\right)^2\lt2$.

Application of $\boldsymbol{(2)}$ \begin{align} \frac{\sum\limits_{j=0}^n\binom{n}{j}\left[1+e^{-(j+1)}\right]^{n+1}}{\sum\limits_{j=0}^n\binom{n}{j}\left[1+e^{-j}\right]^{n+1}} &=\frac{\sum\limits_{j=0}^n\binom{n}{j}\sum\limits_{k=0}^{n+1}\binom{n+1}{k}e^{-k(j+1)}}{\sum\limits_{j=0}^n\binom{n}{j}\sum\limits_{k=0}^{n+1}\binom{n+1}{k}e^{-kj}}\tag{3a}\\ &=\frac{\sum\limits_{k=0}^{n+1}\binom{n+1}{k}e^{-k}\left(1+e^{-k}\right)^n}{\sum\limits_{k=0}^{n+1}\binom{n+1}{k}\left(1+e^{-k}\right)^n}\tag{3b}\\ &=\frac{\sum\limits_{k=0}^{n+1}\binom{n+1}{k}\left(1+e^{-k}\right)^{n+1}}{\sum\limits_{k=0}^{n+1}\binom{n+1}{k}\left(1+e^{-k}\right)^n}-1\tag{3c}\\ &=\frac{\overbrace{\ \ \ 2^{n+1}\ \ \ }^{k=0}+\overbrace{2^{n+1}+o\!\left(2^n\right)}^{k\ge1}}{\underbrace{\,\ \ \ \ 2^{n\vphantom{+1}}\ \ \ \ \,}_{k=0}+\underbrace{2^{n+1}+o\!\left(2^n\right)}_{k\ge1}}-1\tag{3d}\\ &=\frac{4+o(1)}{3+o(1)}-1\tag{3e}\\[18pt] &=\frac13+o(1)\tag{3f} \end{align} Explanation:
$\text{(3a)}$: Binomial Theorem
$\text{(3b)}$: Binomial Theorem
$\text{(3c)}$: algebra
$\text{(3d)}$: apply $(2)$
$\text{(3e)}$: divide numerator and denominator by $2^n$
$\text{(3f)}$: simplification

Thus, $$\lim_{n\to\infty}\frac{\sum\limits_{j=0}^n\binom{n}{j}\left[1+e^{-(j+1)}\right]^{n+1}}{\sum\limits_{j=0}^n\binom{n}{j}\left[1+e^{-j}\right]^{n+1}}=\frac13\tag4$$