Given $a_1=1$ and $ a_{n+1}=a_n + 1 +\frac{a_n}{a_{n+1}}$ prove $\lim\limits_{n\to\infty }( a_{n+1} - a_n)=2$ Given a sequence $(a_n)_{n\geq 1}$  such that $a_1=1$ and $ a_{n+1}=a_n + 1 +\frac{a_n}{a_{n+1}}$ where $ n\geq1$
prove that $\lim\limits_{n\to\infty }( a_{n+1} - a_n)=2$ 
I tried to prove that $a_n$ is increasing but it I just can't get around to it.  Thank you in advance!

Let's address first the necessity that $a_n \geq \frac{1}{2}$ for any $n$ so that the sequence is well defined. Suppose that $a_n \geq \frac{1}{2}$, then there are two options for $a_{n+1}$:
$$a_{n+1}=\frac{1}{2}(1+a_n\pm \sqrt{a_n^2+6a_n+1})$$
The positive option is good and it satisfies $a_{n+1}\geq \frac{1}{2}$. Now, let's prove that if we choose $a_{n+1}=\frac{1}{2}(1+a_n- \sqrt{a_n^2+6a_n+1})$, then $a_{n+2}$ won't be defined anymore. We have:
$$a_{n+2}-a_{n+2}(a_{n+1}+1)-a_{n+1}=0$$
and the discriminant is:
$$\Delta = a_{n+1}^2+6a_{n+1}+1=a_{n+1}(a_n+1)+a_n+6a_{n+1}+1=a_{n+1}(7+a_n)+a_n+1$$
$$=\frac{1}{2}\left[(a_n+1)(a_n+9)-(a_n+7)\sqrt{a_n^2+6a_n+1}\right]$$
We are going to prove that this discriminant is negative:
$$(a_n+1)(a_n+9)<(a_n+7)\sqrt{a_n^2+6a_n+1}$$
After squaring and expanding, this simplifies to:
$$a_n^2+8a_n >2$$
which is true since $a_n \geq \frac{1}{2}$. Now, that we have proved we always have to choose the option greater than $\frac{1}{2}$, we can start proving the question.
Since we established we must have $a_n\geq \frac{1}{2}$ for any $n$, from the recurrence formula, $a_n$ is increasing and:
$$a_{n+1}-a_n=1+\frac{a_n}{a_{n+1}} < 2$$
therefore the sequence $a_{n+1}-a_n$ is bounded. Also
$$a_{n+1}^2 = a_{n+1}a_n +a_{n+1}+a_{n} \geq a_{n+1}a_n+2a_{n} = a_n(a_{n+1}+2) > a_na_{n+2}$$
This implies that
$$(a_{n+2}-a_{n+1})-(a_{n+1}-a_n) = \frac{a_{n+1}^2-a_na_{n+2}}{a_{n+1}a_{n+2}} > 0$$
the sequnce $a_{n+1}-a_n$ is bounded and increasing (thus convergent) and $\frac{a_n}{a_{n+1}}$ is also bounded above by $1$ and increasing (it converges to $1$). Therefore:
$$\lim_{n \to \infty} (a_{n+1}-a_{n}) = 1+\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 2$$
