convergence of sequence $t_{n+1}=t_n+e^{-t_n}$

Question

How to prove that sequence given by its general term

$t_{n+1}=t_n+e^{-t_n}$

converges/diverges? I wrote out a few terms but am stuck and don't know how to proceed.

Answer 1

The limit, if it exists, satisfies $L=L+e^{-L}$, which for finite $L$ implies $e^{-L}=0$, which has no finite solution anyway. Since $t_{n+1}>t_n$, the sequence diverges to $+\infty$.

Answer 2

The general method is to compute a lot of terms and graph the results. In this case, the graph looks similar to the graph of $\,\log(n).\,$ The recurrence $\,t_{n+1}=t_n+e^{-t_n}\,$ is a discrete approximation to the differential equation $\,dy/dx = e^{-y}\,$ and $\,dx/dy = e^y\,$ with solution $\, x = e^y+c\,$ for some constant $\,c.\,$ This suggests a solution $\,t_n \approx \log(n) -c.\,$ Define $\, x := 1/n, \, y := \log(x).\,$ The asymptotic solution is $\,t_n \approx -y -(c+ \frac12 y)x - \frac1{24}(4(1+3c+3c^2) + (6+12c)y + 3y^2)x^2 + O(x^3).\, $ Each coefficient of $\,x^n\,$ is a degree $\,n\,$ polynomial in $\,y\,$ whose coefficients are polynomials in $\,c.\,$ The constant $\,c\,$ depends on initial values. For example, if $\,t_1 = 0,\,$ then $\,c = -0.29024716805205.$

A nice expansion for the asymptotic solution using continued logarithms is

$$ t_n \approx f(1,n),\;\;\text{ where }\;\; f(k,n) = \log(n - c - a_k + \frac12 f(k+1,n)), $$

and where $\,a_1,a_2,a_3,a_4,\dots = 0,1/3,17/36,3239/6480,\dots\,$ are given by a complicated recursion.

Answer 3

There is also a sort of practical approach. Suppose you think there is an upper bound $B.$ Suppose you start out at $t_0 = Z.$ Well, $e^{-t}$ is strictly decreasing. Each step, as long as $t_n < B,$ increases by at least $e^{-B}.$ In order to guarantee getting past $B,$ it is enough to go $B - Z,$ and this will happen in $\frac{B-Z}{e^{-B}} = (B - Z)e^B$ steps. That is, using the ceiling function, define $$ N = 1 + \left\lceil (B-Z)e^B \right\rceil \; \; . $$ We have $$ t_N > B $$