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!

So for given $a_n$, the next term $a_{n+1}$ is one of the roots of
$$X^2-(a_n+1)X-a_n, $$
i.e.,
$$\tag1\begin{align}a_{n+1}&=\frac{a_n+1\pm\sqrt{(a_n+1)^2+4a_n}}{2}\\&=\frac{a_n+1\pm\sqrt{(a_n+3)^2-8}}{2},\end{align}$$
which unfortunately gives us two choices. But there are some restrictions: As we (seem to) want a sequence of real numbers, we must have $(a_n+3)^2\ge 8$ for all $n$, i.e., $$\tag2a_n\ge 2\sqrt 2-3\approx -0.17\quad \text{or}\quad a_n\le -2\sqrt 2-3\approx -5.8.$$
If $a_n>0$ and we take the plus-branch in $(1)$, then we have $a_{n+1}\ge a_n+1$ from the first line in $(1)$.
As long as we take the plus-branch throughout, we thus find that $a_n\ge n$.
Assume we do not always take the plus-branch in $(1)$ and let $n$ be the first time we take the minus-branch. Then $$ \begin{align}a_{n+1}&=\frac{a_n+1-\sqrt{(a_n+3)^2-8}}2\\&=\frac{(a_n+1)^2-(a_n+3)^3+8}{2(a_n+1+\sqrt{(a_n+3)^2-8})}\\
&=\frac{-4a_n}{2(a_n+1+\sqrt{(a_n+3)^2-8})}\\
&<\frac{-4a_n}{2(a_n+1+\sqrt{(a_n+3)^2})}\\
&=-1+\frac{2}{a_n+2}\\&\le -1+\frac2{n+2}\\&\le -\frac13\\&<2\sqrt 2-3,\end{align}$$
contradicting $(2)$.
We conclude that we always take the plus-branch, i.e., 
$$a_{n+1}=\frac{a_n+1+\sqrt{(a_n+3)^2-8}}{2}$$
and (as seen above)
$$ a_n\ge n.$$
Now
$$\begin{align}a_{n+1}-a_n&=\frac{1-a_n+\sqrt{(a_n+3)^2-8}}2\\
&= 2+\frac{\sqrt{(a_n+3)^2-8}-(a_n+3)}2\\
&=2-\frac{4}{\sqrt{(a_n+3)^2-8}+(a_n+3)}\end{align}$$
and as $a_n\to\infty$, the desired result follows.
