By defining these 3 matrices:
$\displaystyle M_1 = \begin{bmatrix} 1&0&0&0&0 \\ 1&1&0&0&0 \\ 1&1+x&1&0&0 \\ 1&1+x(1+x)&1+2x&1&0 \\ 1&1+x(1+x(1+x))&1+x(1+x)+x(1+2x)&1+3x&1 \end{bmatrix}$
$\displaystyle M_2 = \begin{bmatrix} 1&0&0&0&0 \\ x&1&0&0&0 \\ 1&x&1&0&0 \\ 0&1&x&1&0 \\ 0&0&1&x&1 \end{bmatrix}$
$\displaystyle M_3 = \begin{bmatrix} 1&1&1&1&1 \\ \frac{1}{2}(-1+x)x&1&1&1&1 \\ \frac{1}{2}(-1+x)x&\frac{1}{2}(-1+x)x&1&1&1 \\ 1&\frac{1}{2}(-1+x)x&\frac{1}{2}(-1+x)x&1&1 \\ 1&1&\frac{1}{2}(-1+x)x&\frac{1}{2}(-1+x)x&1 \end{bmatrix}$
and multiplying and dividing elementwise $M_1\cdot M_2 / M_3$ , we get the following matrix $A$:
$\displaystyle A = \begin{bmatrix} 1&0&0&0&0 \\ \frac{2}{-1+x}&1&0&0&0 \\ \frac{2}{(-1+x)x}&\frac{2(1+x)}{-1+x}&1&0&0 \\ 0&\frac{2(1+x(1+x))}{(-1+x)x}&\frac{2(1+2x)}{-1+x}&1&0 \\ 0&0&\frac{(2(1+x(1+x)+x(1+2x)))}{((-1+x)x)}&\frac{(2(1+3x))}{(-1 + x)}&1 \end{bmatrix}$
Then the matrix inverse of $A$ is a new matrix $B$ starting:
$\displaystyle B = \begin{bmatrix} 1&0&0 \\ \frac{-2}{-1 + x}&1&0 \\ (\frac{4}{(-1+x)^2} - \frac{2}{(-1+x)x} + \frac{4x}{(-1 + x)^2})&(\frac{-2}{-1+x} - \frac{2x}{-1+x})&1 \end{bmatrix}$
If we then calculate the ratios of the first column in matrix $B$ we get a sequence $c\;$:
$\displaystyle c = \frac {B(n,1)}{B(n+1,1)}, \; n=1,2,3\;...$
and if we define a sequence $b$:
$\displaystyle b = 1,\; 1+x,\; 1+2x,\; 1+3x \;...$
and multiply element-wise $c$ with $b$ and add $x$, we get a sequence $d\;$:
$\displaystyle d = x + b*c$
$\displaystyle d = \frac{1+x}{2}, \frac{x(2+x+x^2)}{1+x+2x^2}, \frac{1+x(6+x(7+2x(4+x)))}{2(1+x)(2+x+3x^2)}\;...$
My question is: Do the terms in this sequence $d$ tend to $\displaystyle \sqrt x\;$?
Example: $x=2\;$
$\displaystyle \frac{3}{2}, \frac{16}{11}, \frac{137}{96}, \frac{1642}{1157}, \frac{8429}{5950}, \frac{67952}{48001}, \frac{1509939}{1066976}, \frac{38701726}{27353183}, \frac{1124000429}{794502270}, \frac{36478904464}{25787223797} \;...$
$\displaystyle \frac{36478904464}{25787223797} = 1.41461$ which is close to $\displaystyle \sqrt 2 = 1.41421$
As a Mathematica program this is:
(*x is the value to calculate the square root of*) Clear[x, t, tt, ttt, M1, M2, M3, A, B, b, c, n, k, i, j, squarerootofx]; (*x=2; uncomment this to calculate sqrt of 2 or any other number*) (*nn is size of matrix*) nn = 5; (*Variant of the Pascal triangle*) t[n_, 1] = 1; t[n_, k_] := t[n, k] = If[n > 1, t[n - 1, k - 1] + x*t[n - 1, k], 0]; M1 = Table[Table[t[i, j], {j, 1, nn}], {i, 1, nn}]; MatrixForm[M1] tt[n_, k_] := tt[n, k] = If[Or[n == k, n == k + 2], 1, If[n == k + 1, x, 0]]; M2 = Table[Table[tt[i, j], {j, 1, nn}], {i, 1, nn}]; MatrixForm[M2] ttt[n_, k_] := ttt[n, k] = If[n == k, 1, If[Or[n == k + 1, n == k + 2], (x - 1)*x/2, 1]]; M3 = Table[Table[ttt[i, j], {j, 1, nn}], {i, 1, nn}]; MatrixForm[M3] (*Elementwise multiplication and division of the 3 matrices*) A = M1*M2/M3; MatrixForm[A] B = Inverse[A]; MatrixForm[B] b = Range[1, x*(nn - 2) + 1, x] c = Ratios[B[[All, 1]]]^-1; d = squarerootofx = x + b*c FullSimplify[squarerootofx] N[squarerootofx];