On a certain sequence Let $\{y_n\}$ be a sequence defined by $y_n=x_n-\ln n$ , where $\{x_n\}$ is defined by $x_1=1 ; x_{n+1}=x_n + e^{-x_n} , \forall n\in \mathbb N$ . Then is $\{y_n\} $ bounded ? Is $\{y_n\}$ convergent ? I know that $\{x_n\}$ is strictly increasing and unbounded ; but I am unable to say anything about $\{y_n\}$ . Please help  .Thanks in advance  
 A: It is clear that the increasing sequence $x_n \to +\infty$ Put $z_n=\exp(x_n)$, $z_n\to +\infty$ also. We have that $\frac{z_{n+1}-z_n}{z_n}=\exp(1/z_n)-1\sim \frac{1}{z_n}$, this show that $u_n=z_{n+1}-z_n\to 1$. Using Cesaro's theorem, we get that $\frac{u_1+\cdots+u_n}{n} \to 1$, hence $z_n/n\to 1$, and $\log z_n-\log n=x_n-\log n\to 0$. 
A: By simple calculation, one can observe that $y_i$ is decreasing for small $i$s and it is likely to converge to $0$.
Indeed, $$\begin{align}y_{n+1}&=x_{n+1}-\ln{n+1}\\&=x_n+e^{-x_n}-\ln{n}-\ln\left({1+\frac{1}{n}}\right)\\&=y_n+e^{-y_n-\ln n}-\ln\left({1+\frac{1}{n}}\right)\\&=y_n+\frac{e^{-y_n}}{n}-\ln\left({1+\frac{1}{n}}\right)\end{align}$$
and we can derive few properties from this.


*

*$y_n>\frac{1}{n}$.


We prove this by mathematical induction. $y_1=1$, and suppose that $y_k>\frac{1}{k}$. Then since $x+\frac{e^{-x}}{n}$ is increasing for positive $x$, $$y_{k+1}=y_k+\frac{e^{-y_k}}{k}-\ln\left(1+\frac{1}{k}\right)\ge\frac{1}{k}+\frac{e^{-1/k}}{k}-\ln\left(1+\frac{1}{k}\right)>\frac{2}{k}-\ln\left(1+\frac{1}{k}\right)>\frac{1}{k+1}$$where the last inequality is from Taylor expansion of $\ln(1+x)$.


*$y_n<\frac{\ln n}{n}$ if $n\ge10$.


We also prove this by mathematical induction. $$y_{k+1}=y_k+\frac{e^{-y_k}}{k}-\ln\left(1+\frac{1}{k}\right)\le\frac{\ln k}{k}+\frac{e^{-\ln k/k}}{k}-\ln\left(1+\frac{1}{k}\right)<\frac{\ln (k+1)}{k+1}$$
By 1 and 2, we can conclude that $y$ converges to $0$.
Furthermore, one can manually check that $y_2<y_1$ and for $k>1$,$$y_{k+1}-y_k=\frac{e^{-y_k}}{k}-\ln\left(1+\frac{1}{k}\right)\le\frac{e^{-\ln k/k}}{k}-\ln\left(1+\frac{1}{k}\right)<0$$ so $y$ is strictly decreasing.
