If only that numerator was a difference of squares!

We are going to show that the limit exists if and only if $a_1>-2$ . Then according to Peter Van der Linden's reasoning we can conclude that if this condition is met, $\displaystyle \lim_{n \to \infty} a_n = 2$ .

Case 1.: $a_1>2$

We prove first by induction that $a_n>2$ for all $n$ . Clearly for $n=1$ it is true, so assuming $a_n>2$ we obtain

$a_{n+1} = \frac{(a_n)^2 + 4}{a_n + 2} > \frac{2a_n + 4}{a_n + 2} = 2$

hence we are done.

$a_n > 2$ implies:

$2a_n > 4$

$(a_n)^2 + 2a_n > (a_n)^2 + 4$

$a_n > \frac{(a_n)^2 + 4}{a_n+2}$ (since $a_n + 2 > 0$ ),

which means $a_n > a_{n+1}$ so $a_n$ is strictly decreasing. Since our infinite sequence is bounded from below and strictly decreases, $\displaystyle \lim_{n \to \infty} a_n$ must exist.

Case 2.: $a_1 = 2$

This case yields $a_2 = \frac{(a_1)^2 + 4}{a_2 + 2} = 2$ so by induction $a_n = 2$ can be proven for all n. Thus $\displaystyle \lim_{n \to \infty} a_n = 2$ .

Case 3.: $-2 < a_1 < 2$

This time the same reasoning applies as in Case 1. , but all of the relation signs inverted. After $a_n < 2$ and the strict increasing property of the sequence have been proven, it follows that $\displaystyle \lim_{n \to \infty} a_n$ exists.

(Note that $a_n + 2 > 0$ still holds, so the division can be performed.)

Case 4.: $a_1 = -2$

This case is clearly not possible, otherwise the recursion would not be defined for $a_2$ .

Case 5.: $a_1 < -2$

We prove by induction that $a_n < -2$ . For $n=1$ it is clear, so assuming $a_n < -2$ we obtain

$a_n < 2$

$2a_n < 4$

$(a_n)^2 + 2a_n < (a_n)^2 + 4$

$a_n > \frac{(a_n)^2 + 4}{a_n+2}$ ( $a_n + 2 < 0$ $\Rightarrow$ The relation sign inverts.)

Hence $-2 > a_n > a_{n+1}$ , thus we are done.

We have seen that the limit (if exists) can only be 2, which means in this case $a_n < -2$ implies that the limit does not exist.

If $\lim_{n \rightarrow \infty} a_n$ exist, then it the row must converge to a certain value. Which means $a_{n+1} =a_n$ . So let's solve the equation by multiplying the denominator. This leads to $a_{n} ^2 +2 a_n = a_{n} ^2 +4$ . Now it's easy to solve and find $a_n=2$ .