TITLE: Prove $\lim\limits_{n\to∞}{\sum\limits_{x=0}^n\binom nx(1+{\rm e}^{-(x+1)})^{n+1}\over\sum\limits_{x=0}^n\binom nx(1+{\rm e}^{-x})^{n + 1}}=\frac 13$
QUESTION [16 upvotes]: 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.

REPLY [11 votes]: 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,
$$
\lim_{n\to\infty}\frac{\sum\limits_{j=0}^n\binom{n}{j}\left[1+e^{-(j+1)}\right]^{n+1}}{\sum\limits_{j=0}^n\binom{n}{j}\left[1+e^{-j}\right]^{n+1}}=\frac13\tag4
$$