# Given $a_1 =1, a_{n+1}=a_n+e^{-a_n}$. Prove that $b_n:=a_n-\ln(n)$ converges

Given the sequence $a_1 =1, a_{n+1}=a_n+e^{-a_n}$. Prove that $b_n:=a_n-\ln(n)$ converges.

I have just known $\lim a_n =\infty$. I have tried to prove $b_{n+1} \leq b_n$ but I didn't make it. Help me, thank you so much

• There's an answer which you can follow at math.stackexchange.com/questions/2674093/… – Patrick Stevens Mar 7 '18 at 7:21
• What is definition of $b_n$? – Arjang Mar 7 '18 at 8:42
• @Arjang $b_n:=a_n-\ln n$ – Robert Z Mar 7 '18 at 8:46
• @RobertZ : "Prove that bn:=an−ln(n) is convergent " or "Prove that bn:=an−ln(n) converges"? havent seen the former in literature. – Arjang Mar 7 '18 at 8:48
• @RobertZ Hmmm, the difference is tiny and this uses basically the tools used to solve the other question but ok, I reopened it (and then this might be reclosed for lack of context...). – Did Mar 7 '18 at 9:05

By following Professor Vector's answer given in sequence $x_{n+1}=x_n+e^{-2018x_n}$ (see Patrick Stevens' hint) we have that $e^{a_n}(a_{n+1}-a_n)=1$ and $$e^{a_n}-e=\int_{a_1}^{a_n}e^t dt\geq \sum_{k=1}^{n-1}e^{a_k}(a_{k+1}-a_k)=n-1$$ and therefore $a_n\geq \ln(n-1+e)$. Then $$b_n=a_n-\ln(n)\geq \ln\left(1+\frac{e-1}{n}\right)>0.$$ Moreover (this part is NOT contained in the linked answer) $b_{n+1}\leq b_n$ iff $$e^{-a_n}=a_{n+1}-a_n\leq \ln(n+1)-\ln(n)=\ln(1+1/n)$$ which is implied by (recall that $\ln(1+x)\geq x(1-x)$ for $x>0$), $$\frac{1}{n-1+e}\leq \frac{1}{n}\left(1-\frac{1}{n}\right)$$ that is $$n^2\leq (n-1)(n-1+e)-1=n^2+(e-2)n+1-e$$ which holds for $n>2$.
Hence we may conclude that $b_n$ is eventually decreasing and positive and it follows that it tends to a non negative limit.