<s>[INST] How prove this $b_{n}>b_{n+1}$, if $a_{1}=1,a_{n+1}=a_{n}+e^{-a_{n}},b_{n}=a_{n}-\ln{n}$ Question:

let sequence 
  $$a_{1}=1,a_{n+1}=a_{n}+e^{-a_{n}}$$
  let $$b_{n}=a_{n}-\ln{n},n\in Z$$
show that:
  $$ b_{n}>b_{n+1}$$
  where $$\ln{n}=\log{n}$$

my idea: since $$b_{n}=a_{n}-\ln{n}$$
then we have
$$b_{n+1}=b_{n}+\dfrac{1}{n}e^{-b_{n}}-\ln{\dfrac{n+1}{n}}$$
then we only prove
$$e^{-b_{n}}\le\ln{\left(1+\dfrac{1}{n}\right)^n}$$
and we only prove
$$b_{n}\ge -\ln{\ln{\left(1+\dfrac{1}{n}\right)^n}}?$$
then I can't.
 [/INST] Proof:
Fact 1: It holds that $a_n > - \ln \ln \frac{n+1}{n}$. (The proof is given at the end.)
From Fact 1, we have
\begin{align}
b_n - b_{n+1} &= a_n - \ln n - (a_{n+1} - \ln (n+1))\\
&= a_n - \ln n - a_n - \mathrm{e}^{-a_n} + \ln (n+1)\\
&= \ln \frac{n+1}{n} - \mathrm{e}^{-a_n}\\
&> 0.
\end{align}
(Q. E. D.)
$\phantom{2}$
Proof of Fact 1: We use the mathematical induction.
For $n = 1$, we need to prove that $a_1 > - \ln \ln 2$. It is true.
Assume that the inequality is true for $n = k$ ($k\ge 1$), i.e., $a_k > -\ln \ln \frac{k+1}{k}$.
Let us prove that the inequality is also true for $n = k + 1$, i.e., $a_{k+1} > -\ln \ln \frac{k+2}{k+1}$.
Since $x\mapsto x + \mathrm{e}^{-x}$ is strictly increasing on $(0, \infty)$, we have
\begin{align}
a_{k+1} &= a_k + \mathrm{e}^{-a_k}\\
& > -\ln \ln \frac{k+1}{k} + \mathrm{e}^{-(-\ln \ln \frac{k+1}{k})}\\
&= -\ln \ln \frac{k+1}{k} + \ln \frac{k+1}{k}.
\end{align}
Thus, it suffices to prove that
$$-\ln \ln \frac{k+1}{k} + \ln \frac{k+1}{k} > -\ln \ln \frac{k+2}{k+1}$$
or
$$\frac{k+1}{k}\ln \frac{k+2}{k+1} > \ln \frac{k+1}{k}$$
or
$$\frac{1}{k}\ln \frac{k+2}{k+1} > \ln \frac{k+1}{k} - \ln \frac{k+2}{k+1} = \ln\left(1 + \frac{1}{k^2+2k}\right).$$
Since $\ln (1+x) < x$ for $x > 0$, it suffices to prove that
$$\frac{1}{k}\ln \frac{k+2}{k+1} > \frac{1}{k^2+2k}$$
or
$$\ln\left(1 + \frac{1}{k+1}\right) > \frac{1}{k+2} = \frac{\frac{1}{1+k}}{1 + \frac{1}{k+1}}$$
which follows from $\frac{x}{1+x} < \ln (1+x)$ for $x > 0$. We are done.
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