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. 
 A: $\def\e{\mathrm{e}}$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,$$
\lim_{n \to \infty} \frac{\displaystyle\sum_{k = 0}^n \binom{n}{k} (1 + \e^{-(k + 1)})^{n + 1}}{\displaystyle\sum_{k = 0}^n \binom{n}{k} (1 + \e^{-k})^{n + 1}} = \lim_{n \to \infty} \frac{S_n}{T_n} = \lim_{n \to \infty} \frac{2^n}{3 × 2^n} = \frac{1}{3}.
$$
A: 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,
$$
\lim_{n\to\infty}\frac{\sum\limits_{j=0}^n\binom{n}{j}\left[1+e^{-(j+1)}\right]^{n+1}}{\sum\limits_{j=0}^n\binom{n}{j}\left[1+e^{-j}\right]^{n+1}}=\frac13\tag4
$$
