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!
 A: 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:
$$\lim_{n \to \infty} (a_{n+1}-a_{n}) = 1+\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 2$$
A: Solving the given recursion for $a_{n+1}$ gives
$$a_1=1,\qquad a_{n+1}=f(a_n)\quad(n\geq1)$$
with
$$f(x)={1\over2}\bigl(1+x+\sqrt{x^2+6x+1}\,\bigr)$$
(Atticus has shown that we have to choose the $+$-solution in order to avoid complex $a_n$). It follows that
$$f(x)-x={1\over2}+{1\over2}\bigl(\sqrt{x^2+6x+1}-x\bigr)={1\over2}+{1\over2}{6+{1\over x}\over\sqrt{1+{6\over x}+{1\over x^2}}+1}>{1\over2}\ .$$
This shows that $a_{n+1}-a_n>{1\over2}$ for all $n\geq1$; hence $\lim_{n\to\infty} a_n=\infty$. As a consequence we also have
$$\lim_{n\to\infty}(a_{n+1}-a_n)=\lim_{n\to\infty}\bigl(f(a_n)-a_n\bigr))=\lim_{x\to\infty}\bigl(f(x)-x\bigr)=2\ .$$
A: 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. 
A: 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.
