${\displaystyle \begin{array}{c}\multicolumn{1}{c}{}\\ \multicolumn{1}{c}{X_1}& =& 2,X_2=1.75,X_2=1.732142857,X_4=1.73205081\\ \multicolumn{1}{c}{X_5}& =& 1.732050808,X_6=1.732050808,X_{7}1.732050808\\ \multicolumn{1}{c}{X_8}& =& 1.732050808\mathrm{and}\mathrm{so}\mathrm{on}\end{array}}$

We notice that when $X_n=1.732050808$, so is $X_{n+1}$. Squaring these terms we get $X_1^2=4,X_2^2=3.0625,...,X_5^2=3$ and the rest of the other terms are the same!!

This implies that when $X_n\approx \sqrt{3}$ so is $X_{n+1}$ and the values of $X_n$ tend to the limit $\sqrt{3}$. This special property can easily be proven. Assume that the limit exists, so $X_{n+1}=X_n=X$, then solve the equation

${\displaystyle X=\frac{1}{2}(X+\frac{3}{N})}$

If we test it for $N=3$, we see that $X_{29}=1.44224957$, which is what the calculator gives for the cube root of 3. Testing it for $N=8$, we get $X_1=2$, which is the right answer.