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!

We first need to show $\{a_n\}$ is an increasing sequence, and the argument of this part could adopt the proof above by Prof. Blatter. Once this is overcome, then $a_n > 0$ for all $n \ge 1$. Next ,  $a_{n} = (a_n - a_{n-1})+ (a_{n-1} - a_{n-2}) +...=(a_2 -a_1) + a_1 = (1+\dfrac{a_{n-1}}{a_n})+ (1 + \dfrac{a_{n-2}}{a_{n-1}}) + (1 + \dfrac{a_1}{a_2}) + a_1 > (n-1)+1 = n$. Thus we have: $ 1< \dfrac{a_{n+1}}{a_n}  = 1 + \dfrac{1}{a_n} + \dfrac{1}{a_{n+1}} < 1+\dfrac{1}{n}+\dfrac{1}{n+1}$ and this shows $\dfrac{a_{n+1}}{a_n} \to 1$ as $n \to \infty$. This implies $a_{n+1} - a_{n} = 1 + \dfrac{a_n}{a_{n+1}}\to 1+1 = 2$ as claimed. 
