# Can we find this limit?

Let $\{a_n\}_{n=1}^{\infty}$ be a sequence such that $a_1=1$ and $a_{n+1}=\sqrt{a_n^2+a_n}$ for $n\geq 0$.

Is it possible to find $\displaystyle\lim_{n\to\infty}\frac{a_n}{n}$ ?

I have no any idea.

The answer to your question is positive. The limit exists and equals $\frac{1}{2}$.

Note first that the sequence $(a_n)_{n \geq 1}$ is increasing $(a_{n+1}>a_n)$ and that it diverges to $+\infty$. Indeed, since $f(x) = \sqrt{x^2+x}$ is continuous on $\mathbb{R}_+$, if $(a_n)_{n \geq 1}$ were to converge to some real number $L$, $L$ would have to be a fixed point of $f$. But the unique fixed point of $f$ is $0$, and the increasing $(a_n)_{n \geq 1}$ with $a_1=1$ cannot converge to $0$. Therefore, $\lim\limits_{n \to \infty}a_n = +\infty$.

Set $a_0=0$ and write $\frac{a_n}{n} = \frac{1}{n}\sum\limits_{k=0}^{n-1}(a_{k+1}-a_k) = \frac{1}{n}\sum\limits_{k=0}^{n-1}b_k$ where we denote by $(b_n)_{n \geq 0}$ the sequence of differences having $\frac{a_n}{n}$ as its Cesàro mean.

Now, $b_n = a_{n+1} - a_n = \sqrt{a^2_n+a_n} - a_n = \frac{a_n}{\sqrt{a^2_n+a_n} + a_n} = \frac{1}{1+\sqrt{1+1/a_n}}$ converges to $1/2$ as $n \to \infty$ since $\lim\limits_{n \to \infty}a_n = +\infty$, so, by the Cesàro mean theorem, $\frac{a_n}{n}$ also converges to the same limit.