Finding the limit of the sequence with $a_1=a$, $a_2=b$, and $a_{n}=\sqrt{a_{n-1} a_{n-2}}$ Let there be a recursive sequence that begins with two terms, $a_1 = a$ and $a_2 = b$. The third term, $a_3$, is created by taking the geometric mean ($\sqrt{a \times b}$) of the previous two terms. The fourth term, $a_4$, is once again created by taking the geometric mean of the previous two terms. This process is repeated indefinitely.
For example, if $a = 1$ and $b = 8$,
$a_3 = \sqrt{1 \times 8}=  \sqrt8 =2.82843... $
$a_4 = 4.75683...$
$a_5 = 3.66802...$
$a_6 = 4.17710...$
...and so on.
If you'll notice, as the terms go on, they are slowly converging towards one number. In this case, it's 4. This number will (tentatively) be called the "limit" ($L$).
My question is, what is the general rule for finding $L$ in terms of $a$ and $b$?
 A: Considering the first few terms: 
$$a, b, a^{\frac 12}b^{\frac 12}, a^{\frac 14}b^{\frac 34}, a^{\frac 38}b^{\frac 58}, $$
we see that the index of either $a$ or $b$ of each term is the arithmetic mean of the indices of the  $a$ or $b$ of the previous 2 terms.  
We therefore can start by finding the limit of the recursive sequence of arithmetic means:
$$m, n, \frac 12(m+n), \frac 12(n+\frac 12(m+n)),...$$
We can find the arithmetic mean of the previous 2 terms simply by subtracting from the second term half the signed difference between them, taking second minus first.
Let $d = n - m$, the sequence can be written as follows:
$$m, n, n-\frac d2, n-\frac d2+\frac d4, n-\frac d2+\frac d4, n-\frac d2+\frac d4-\frac d8,...$$
Note that each new difference is half the previous difference in magnitude and the signs of these differences alternate.
And we can rewrite and compute the infinite series:
$$n-\frac d2+\frac d4-\frac d8+... = n-\Bigl(\frac d2+\Bigl(\frac d2\Bigr)\Bigl(\frac {-1}{2}\Bigr)+\Bigl(\frac d2\Bigr)\Bigl(\frac {-1}{2}\Bigr)^2+\Bigl(\frac d2\Bigr)\Bigl(\frac {-1}{2}\Bigr)^3+...\Bigr)$$
$$=n-\frac {\frac d2}{1+\frac 12}$$
$$=n-\frac d3$$
$$=n-\frac {n-m}{3}$$
$$=\frac {m+2n}{3}$$
Now substitute $m=1$ and $n=0$ to get $1/3$ for the index of $a$, and substitute $m=0$ and $n=1$ to get $2/3$ for the index of $b$.
