# sequence $x_{n+1}=x_n+e^{-2018x_n}$ [closed]

Let the sequence $(x_n)_{n \geq 0}$ defined by \begin{cases} x_0=2017 \\ x_{n+1}=x_n+e^{-2018x_n} \end{cases} for every $n \ge 0$. Calculate $$\lim_{n\to \infty}x_n \text{ and } \lim_{n\to \infty}\frac{\ln(2018+n)}{x_n}$$

I proved that $\displaystyle \lim_{n\to \infty}x_n=\infty$, but I don't realise how to find the second limit. I tried Cesaro-Stolz lema, but it doesn't work. Please help me!

## closed as off-topic by Carl Mummert, Namaste, Xander Henderson, The Phenotype, Mohammad Riazi-KermaniMar 3 '18 at 2:09

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• are you sure that $$\lim_{n\to \infty}x_n=+\infty$$? – Dr. Sonnhard Graubner Mar 2 '18 at 19:52
• @DonAntonio The sequence $x_n$ is strictly increasing. – Patrick Stevens Mar 2 '18 at 20:09
• @Dr. Sonnhard Graubner A finite limit $a$ would have to satisfy $a=a+e^{-2018a}$. I found the strategy "read-understand-think-answer/comment" rather advantageous, so far. – Professor Vector Mar 2 '18 at 20:16
• What is the source of this problem? The coefficients make it seem like an exam or contest problem. – Carl Mummert Mar 2 '18 at 23:27

$$x_{n+1}=x_n+e^{-2018x_n}$$ means $$e^{2018x_n}\,(x_{n+1}-x_n)=1,$$ i.e. $$\sum^{n-1}_{k=0}e^{2018x_k}\,(x_{k+1}-x_k)=n.\tag{crucial}$$ Since $e^{2018x}$ is monotone increasing, this means $$\int^{x_n}_{x_0}e^{2018x}\,dx\ge n,$$ implying $$\frac{e^{2018x_n}-e^{2018\cdot2017}}{2018}\ge n.$$ This gives $$e^{2018x_n}\ge2018\,n+e^{2018\cdot2017},$$ and thus $$x_n\ge\frac1{2018}\ln(2018\,n+e^{2018\cdot2017})\tag{lower}$$ and $$x_{n+1}-x_n=e^{-2018x_n}\le\frac1{2018\,n+e^{2018\cdot2017}}.$$ This is one of the mean aspects of this problem: $x_n$ diverges, but very slowly.
Whatever, we know $$e^{2018\,x_{n+1}}\le e^{2018\,x_n}\cdot \exp\left(\frac{2018}{2018\,n+e^{2018\cdot2017}}\right),$$ that is $$e^{2018\,x_{n+1}}\le C\,e^{2018\,x_n},$$ where $C$ is a constant very near to (but greater than) $1$. But then, (crucial) implies $$\sum^{n-1}_{k=0}e^{2018x_{k+1}}\,(x_{k+1}-x_k)\le C\,n,$$ so $$\int^{x_n}_{x_0}e^{2018x}\,dx=\frac{e^{2018x_n}-e^{2018\cdot2017}}{2018}\le C\,n,$$ leading to $$x_n\le\frac1{2018}\ln(2018\,C\,n+e^{2018\cdot2017})\tag{upper}$$
From (lower) and (upper), it's clear that $$\lim_{n\to \infty}\frac{\ln(2018+n)}{x_n}=2018.$$