You have the following recurrence formula:
$$a_0=1,\quad a_{n+1}=a_n+e^{-a_n}, \quad n\in\mathbb{N_0}$$
Now define sequence:
$$b_n=a_n-\ln n,\quad n\in\mathbb{N}$$
Prove that:
$$0 \lt b_{n+1} \lt b_n, \quad n\in\mathbb{N}\tag{1}$$
...and determine:
$$\lim_{n\to\infty}b_n$$
My try:
Firstly, let us make some observations about $a_n$. It obviously holds that:
$$a_{n + 1} > a_{n} > 1, \quad n \in \mathbb{N}$$
Now, suppose $(a_n)_{n \in \mathbb{N}}$ has a limit $a \in \mathbb{R}$ as $n$ tends to infinity. Then, we would have:
$$a = a + e^{-a}$$
which has no solution for $a \in \mathbb{R}$. So we conclude $\lim_{n\to\infty}a_n = +\infty$. Also, we can write $b_n$ in the following way:
$$b_{n + 1} = b_{n} + e^{-a_n} - \ ln{\frac{n + 1}{n}}$$.
From this, and if $(1)$ holds it is quite trivial to prove that $\lim_{n\to\infty}b_n = 0$. Now, I have not managed to prove $(1)$ since every inequality I tried to use was not nearly strong enough. Any help would be highly appreciated.
