If $a_0=1$ and $a_{n+1}=a_n +e^{-a_n}$ then does the limit of $a_n-\log{n}$ exist and if so then what is it? I don't  want the full answer but hints for solving this question.My idea to attempt the question so far is the following by back tracing if we calculate the limit of $\dfrac{e^{a_n}}{n}$  we will get our answer just by applying logarithm.As $\dfrac{e^{a_n}}{n}$  might be useful in the sense that we can use the given conditions of the problem and with Taylor series expression.But how should I proceed to get to $\dfrac{e^{a_n}}{n}$. I think we need to construct a new sequence.Is my approach correct or I need to think differently? Hints required
 A: 
Proposition 1. $$x+\frac{1}{e^x}\geq \log{\left(e^x+1\right)}, \forall x\geq0$$

Indeed, from $\forall x\geq 0$
$$\log{(1+x)}\leq x \Rightarrow
\log{\left(1+\frac{1}{e^x}\right)}\leq \frac{1}{e^x} \Rightarrow \\
x+\log{\left(1+\frac{1}{e^x}\right)}\leq x+\frac{1}{e^x} \Rightarrow \\
\log{e^x}+\log{\left(1+\frac{1}{e^x}\right)}\leq x+\frac{1}{e^x} \Rightarrow \\
\log{\left(e^x+1\right)}\leq x+\frac{1}{e^x}$$


Proposition 2. $$a_n\geq\log{(n+1)},\forall n\geq0$$

By induction $$a_0=1>\log{(0+1)}=0$$
$$a_1=1+\frac{1}{e}>\log{(1+1)}=0.693...$$
Now, let's assume $a_n\geq\log{(n+1)}$, then
$$a_{n+1}=a_n+\frac{1}{e^{a_n}} \overset{Prop. 1}{\geq} 
\log{(e^{a_n}+1)} \geq
\log{(n+1+1)}=\log{(n+2)}$$


Proposition 3. $a_n -\log{n}$ is decreasing and bounded/positive, thus converging.

Indeed
$$a_{n+1}-a_n=\frac{1}{e^{a_n}} \Rightarrow \\
\left(a_{n+1}-\log{(n+1)}\right)-\left(a_n-\log{(n+1)}\right)=\frac{1}{e^{a_n}} \Rightarrow\\
\left(a_{n+1}-\log{(n+1)}\right)-\left(a_n-\log{n}\right)=
\frac{1}{e^{a_n}} -\log{\left(1+\frac{1}{n}\right)} \leq ...$$
we know that
$$\frac{1}{n+1}\leq \log{\left(1+\frac{1}{n}\right)}$$
thus
$$...\leq \frac{1}{e^{a_n}} -\frac{1}{n+1}\overset{Prop.2}{\leq}0$$
As a result
$$\color{red}{0<} \log{(n+2)}-\log{(n+1)} \overset{Prop.2}{\leq} \color{red}{a_{n+1}-\log{(n+1)}\leq a_n-\log{n}}$$

The remaining part is to find the limit ...
