Given a recursion $a_{n+1}=\sqrt{a_n^2+a_n}$ with $a_1=1.$ Prove that $\lim{\left(a_n\right)}'=\frac12$ without using $a_n\sim\frac n2-\frac14\ln n$

Question

Given a recursion $a_{n+ 1}= \sqrt{a_{n}^{2}+ a_{n}}$ with $a_{1}= 1.$ Prove that $\lim{\left ( a_{n} \right )}'= \frac{1}{2}$ without using the result I've got $$a_{n}\sim\frac{n}{2}- \frac{1}{4}\ln n\,{\rm as}\,n\rightarrow\infty$$ Source: StackMath/@haidangel_ Estimations of some new recurrence sequences

of course I used L'Hopital rule to remake the problem, we may need other approaches. All your nice comments and solutions are welcome and appreciated. I need to the help, thanks a real lot !

Answer 1

There is no derivative for an integer-valued variable $n$, but there's an obvious replacement: the difference. Since $a_{n+1}\ge a_n\ge1$, we must have $a_n\to\infty$ as $n\to\infty$ (otherwise, the monotone sequence would have a finite limit $a\ge1$ satisfying $a=\sqrt{a^2+a}$, that's impossible). Then, $$a_{n+1}-a_n=\frac{a_n}{\sqrt{a^2_n+a_n}+a_n}=\frac1{\sqrt{1+\frac1{a_n}}+1}\to\frac12$$ as $n\to\infty$.