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. 

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
$$\lim_{n \to \infty} n(1-a_n) =\lim_{x \to \infty} \frac{g^{-1}(h(x))}{1/x} = \frac{e}{2(e+1)}.$$
