Euler sequence, limit of an related sequence Study the convergence of the sequence $ \left( a_n\right)_{n\in\mathbb{N^*} }$ defined by 
$$
a_{n}+e^{a_{n}}=\left(  1+\frac{1}{n}\right)  ^{n}+1,~\forall n\in
\mathbb{N}
^{\ast}
$$
and find its limit. Moreover, deduce the values of limit
$$
\lim_{n\rightarrow\infty} n(1-a_n).
$$
I don't know how to obtain the condition of Bolzano-Weierstrass Theorem: a monotone and bounded sequence is convergent. The exponential function keeps me from proceeding; that's where I get stuck. 
 A: Clearly
$$ a_n+e^{a_n}=\left(1+\frac1n\right)^{n}+1\le e+1 $$
and $\{a_n\}$ is bounded above. Also note
$$ a_{n+1}-a_n+e^{a_{n+1}}-e^{a_n}=\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^{n}>0, \tag{1} $$
By the Lagrange Mean Value Theorem,
$$ e^{a_{n+1}}-e^{a_n}=e^{c_n}(a_{n+1}-a_n), \text{ where } c_n \text{ is between } a_n \text{ and }a_{n+1}.\tag{2}$$
From these two arguments, we have $a_{n}\le a_{n+1}$. Thus $\{a_n\}$ is bounded and monotonic and hence $\lim_{n\to\infty}a_n=a$ exists. Letting $n\to\infty$ gives
$$ a+e^a=e+1 $$
which implies $a=1$ since $f(x)=x+e^x$ is strictly increasing. For the second argument, from (1) and (2), we have
$$ a_{n+1}-a_n=(1+e^{c_n})^{-1}\left(\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^{n}\right).$$
By the Stolz–Cesàro theorem, we have
\begin{eqnarray}
\lim_{n\to\infty}n(1-a_n)&=&\lim_{n\to\infty}\frac{1-a_n}{\frac{1}{n}}\\
&=&\lim_{n\to\infty}\frac{(1-a_{n+1})-(1-a_n)}{\frac{1}{n+1}-\frac{1}{n}}\\
&=&\lim_{n\to\infty}n(n+1)(a_{n+1}-a_n)\\
&=&\lim_{n\to\infty}n(n+1)(1+e^{c_n})^{-1}\left(\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^{n}\right)\\
&=&\frac{1}{1+e}\lim_{n\to\infty}n(n+1)\left(\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^{n}\right)\\
&=&\frac{1}{1+e}\frac{e}{2}\\
&=&\frac{e}{2(1+e)}.
\end{eqnarray}
The limit can be calculated in the following way. Let $x=\frac{1}{n}$ and then
\begin{eqnarray}
&&\lim_{n\to\infty}n(n+1)\left(\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^{n}\right)\\
&=&\lim_{x\to0}\frac{x+1}{x^2}\left(\left(1+\frac x{x+1}\right)^{\frac{1}{x}+1}-\left(1+x\right)^{\frac 1x}\right).
\end{eqnarray}
Noting
$$ \left(1+\frac x{x+1}\right)^{\frac{1}{x}+1}=e-\frac{ex}{2}+\frac{23e}{24}x^2+O(|x|^3),\left(1+x\right)^{\frac{1}{x}}=e-\frac{ex}{2}+\frac{11e}{24}x^2+O(|x|^3)
$$
we have
\begin{eqnarray}
&&\lim_{n\to\infty}n(n+1)\left(\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^{n}\right)\\
&=&\lim_{x\to0}\frac{x+1}{x^2}\left(\left(1+\frac x{x+1}\right)^{\frac{1}{x}+1}-\left(1+x\right)^{\frac 1x}\right)\\
&=&\lim_{x\to0}\frac{x+1}{x^2}\frac{ex^2}{2}\\
&=&\frac{e}{2}.
\end{eqnarray}
A: Since both the function $f(x) = x + e^x$ and the sequence $\left(1 + \frac{1}{n}\right)^n + 1$ are strictly increasing, it follows that the sequence $a_n$ must be strictly increasing; it is bounded because $\left(1 + \frac{1}{n}\right)^n + 1 \leq e + 1 = f(1)$ is bounded.
If $a = \lim a_n$, then $a$ must satisfy $a + e^a = 1 +e$, which implies that $a = 1$ (because $x + e^x$ is strictly increasing, hence injective).
For the second limit, let $g(x) = 1-x + e^{1-x}$ and $h(x) = \left(1 + \frac{1}{x}\right)^x$, so that $1 - a_n = g^{-1}(h(n))$. If we calculate
$$\lim_{x \to \infty} \frac{g^{-1}(h(x))}{1/x} = \lim_{x \to \infty} \frac{h'(x)}{g'(g^{-1}(h(x))) \cdot (-1/x^2)} = \lim_{x \to \infty} \frac{x^2 h'(x)}{-g'(g^{-1}(h(x))},$$
by L'Hôpital. Now $g'(y) = - (1 + e^{1-y})$ and $h'(x) = \left(1 + \frac{1}{x}\right)^x\left(\ln\left(1 + \frac{1}{x}\right) - \frac{1}{x+1}\right)$. Recalling that $g^{-1}(h(x)) \to 0$ as $x \to \infty$, we get that  $-g'(g^{-1}(h(x)) \to e + 1$. Moreover, 
$$x^2 h'(x) = \left(1 + \frac{1}{x}\right)^x \cdot x^2\left(\ln\left(1 + \frac{1}{x}\right) - \frac{1}{x+1}\right).$$ 
Looking good, since $\left(1 + \frac{1}{x}\right)^x \to e$. For the other term, another round of L'Hôpital yields
$$\lim_{x \to \infty} \frac{\ln\left(1 + \frac{1}{x}\right) - \frac{1}{x+1}}{1/x^2} = \lim_{x \to \infty} \frac{-1/(x(x+1)^2)}{-2/x^3} =\lim_{x \to \infty} \frac{x^3}{2x(x+1)^2} = \frac{1}{2}.$$
Collecting all the loose terms, we see that
$$\lim_{n \to \infty} n(1-a_n) =\lim_{x \to \infty} \frac{g^{-1}(h(x))}{1/x} = \frac{e}{2(e+1)}.$$
