<s>[INST] Convergence to $\sqrt{2}$ It is a very good way to approximate $\sqrt{2}$ using the following;
Let $D_{k}$ and $N_{k}$ be the denominator and the numerator of the $k$th term, respectively.
Let $D_1=2$ and $N_1=3$, and for $k\geq2$, $D_k=D_{k-1}+N_{k-1}$, and $N_{k}=D_{k-1}+D_{k}$.
To clarify, here is the sequence;
$\frac{3}{2},\frac{2+2+3}{2+3},\frac{2+3+2+3+2+2+3}{2+3+2+2+3},\cdots$ that is $\frac{3}{2},\frac{7}{5},\frac{17}{12},\cdots$


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*Why the sequence converges to $\sqrt{2}$ even the initial numerator and denominator were other positive integers?

*How to find the $j$th term (i.e. $\frac{N_{j}}{D_{j}}$) without finding the preceding terms, say the $45$th term?

*Is there a similar way to approximate the square root of any other positive integer, say $\sqrt{3}$?
I do know many ways to approximate square roots, such as Newton's method. But I am asking about a similar way.
 [/INST] You have a pair of linear recurrence relations.  You can write it as $$\begin {pmatrix} N_k\\D_k \end {pmatrix}=\begin {pmatrix} 1&2\\1&1 \end {pmatrix}\begin {pmatrix} N_{k-1}\\D_{k-1} \end {pmatrix}$$  You find the eigenvalues and eigenvectors of the matrix.  The eigenvector $\begin {pmatrix} \sqrt 2\\1 \end {pmatrix}$corresponding to the eigenvalue $\sqrt 2 +1$ will dominate, so you will converge to $\sqrt 2$ for any starting condition that does not have this eigenvector with a $0$ coefficient.
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