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

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.

Answer 1

Here is the beginning of a potentially helpful idea which I unfortunately cannot flesh out at the moment. I will come back in a while and try to do so.

Consider the functions $$ f(x) = x + e^{-x}, \quad g(x) = \ln(1 + e^x). $$ Then $a_n = f \circ f \circ \cdots \circ f(0) = f^n(0)$ and $g^n(0) = \ln(n+1)$, so the condition $$ e^{-a_n} \leq \ln\left(\frac{n+1}{n} \right) = \ln(n+1) - \ln(n) $$ is equivalent to $$ f^{n+1}(0) - f^n(0) \leq g^n(0) - g^{n-1}(0). $$ This inequality could be established by establishing $$ g^n(x) - g^{n-1}(x) -( f^{n+1}(x) - f^n(x)) \geq 0 \quad \quad \quad (*) $$ for all $x \geq 0$ and simply taking $x=0$. Since this expression converges to $0$ as $x \to \infty$, it is sufficient to show that the derivative is negative. I believe this can be done by induction on $n$: by the chain rule $$ \frac{d}{dx} f^{n+1}(x) = (f^n)'(f(x))f'(x) = (1 - e^{-x})(f^n)'(f(x)) $$ and therefore $$ \frac{d}{dx} (f^{n+1}(x) - f^n(x)) = (1-e^{-x})((f^n)' - (f^{n-1})')(f(x)). $$ Similarly, $$ \frac{d}{dx}(g^n(x) - g^{n-1}(x)) = \frac{e^x}{1+e^x}((g^{n-1})'-(g^{n-2})')(g(x)). $$ I'm not sure where to go from here yet.

Answer 2

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.